Consider the process in which a closed system is heated under constant volume. There is no PV work because the volume of the system is fixed. We assume there is no shaft work either. Then, the first law gives,
Therefore, the amount of heat that is exchanged under constant volume is equal to the change of internal energy, provided that no shaft work is present. Recall that shaft work, if present, will be made obvious through the presence of mechanical devices such as impellers, pumps, compressors, turbines, and the like. If a problem makes no mention of any such devices, it will be assumed that no shaft work is involved, without explicitly stating this assumption each time.
Example 3.5: Constant-Volume Cooling
A sealed metal cylinder contains steam at 1 bar, 500 °C. How much heat must be removed at constant volume in order to produce saturated vapor?
SolutionOutline: This is a constant-volume process; therefore, the heat is equal to the change of internal energy between the initial (A) and final (B) state:
Q = UB − UA.
The initial state is known. In the final state we know the specific volume (must be equal to the specific volume in the initial state) and that the state is saturated vapor. These two pieces of information are sufficient to fix the final state. Once the final state is known, the heat is calculated by the above equation.
Numerical substitutions: From the steam tables at the initial state we read:
VA = 3.5656 m3/kg, UA = 3132.2 kJ/kg.
At the final state water is saturated vapor with VB = VA = 3.5656 m3/kg. The desired temperature is between 78 °C and 80 °C. By interpolation we find:
TB = 78.8 °C, UB = 2480.1 kJ/kg.
The amount of heat that must be exchanged with the surroundings is
A common way to exchange heat is under constant pressure. The expansion/ contraction of volume that accompanies heating/cooling involves the exchange of PV work with the surroundings, and this must be accounted for in the energy balance. We assume that the PV work is exchanged reversibly. Experimentally, this means that heat must be added or removed at a gentle rate to allow temperature to equilibrate and the expansion or contraction to take place in an controlled manner. Applying eq. (3.4) and noting that pressure is constant, the work is
with subscripts A and B referring to the initial and final state, respectively. Substituting this result in eq. (3.9) and solving for the heat, we find,
In the last result, the right-hand side is the difference of the term U + PV evaluated at the final state B, and the same term evaluated at the initial state A (recall that PB = PA = P for this process). This grouping of state functions is itself a state function. This motivates the definition of a new thermodynamic property, enthalpy:
Using this definition, the heat in constant-pressure process is
Therefore, in constant-pressure heating of a closed system, the amount of heat that is exchanged is equal to the change in enthalpy. Enthalpy is a state function and has the same units as internal energy (kJ/kg or J/mol). It is the relevant thermodynamic in the energy balance of open systems, as we will see in Chapter 6. As the large majority of chemical processes involves open systems, such as units with multiple inlet and outlet flow streams, enthalpy is used far more than internal energy and is tabulated more extensively. In fact, tabulations of internal energy are uncommon because it can be calculated from tabulations of enthalpy and volume.
Example 3.6: Enthalpy of Steam
Use the steam tables to obtain the enthalpy of steam at 1 bar, 300 °C, and compare the value calculated by application of the definition, eq. (3.12).
Solution From the steam tables at P = 1 bar, T = 300 °C, we find
H = 3074.5 kJ/kg.
At the same pressure and temperature we also find
V = 2.6389 m3/kg, U = 2810.7 kJ/kg.
To apply eq. (3.12) we first calculate the product PV:
H = U + PV = (2810.7 kJ/kg) + (263.89 kJ/kg) = 3074.54 kJ/kg.
The calculated value agrees with the tabulated value within the precision of the table. Small round-off errors should be expected due to truncation in the tables.
Example 3.7: Constant-Pressure Cooling of Steam
8.5 kg of steam at 600 °C, 15 bar, are cooled at constant pressure by removing 6200 kJ of heat. Determine the final temperature and the amount of PV work that is exchanged with the surroundings.
Solution To specify the final state we need two intensive variables. The final pressure is known since it is constant during the process and equal to the initial pressure. The missing piece of information is the enthalpy of the final state, which will be obtained by energy balance.
For constant-pressure heating, Q = H2 − H1, which solved for H2 gives
In the final state we know pressure (P2 = P1 = 15 bar) and enthalpy (H2 = 2965.2 kJ/kg). By inspection in the steam tables at P = 15 bar, the final temperature must be between 250 °C and 300 °C. The exact temperature is obtained by interpolation:
Since pressure and temperature are known at the final state, the internal energy is obtained directly from the table. This requires one additional interpolation:
Finally, the work is
W = 2727.6 − 3294.5 −(−729.4) = 162.5 kJ/kg.
The work calculated here is per kg of steam. The total amount of work for the entire amount of steam is W = (8.5 kg)(162.5 kJ/kg) = 1381.4 kJ. The work is positive, which means, it is done by the surroundings and is absorbed by the system.
Example 3.8: Work and Heat are Path Functions
Steam undergoes the following reversible process in a closed system: from initial conditions 10 bar, 400 °C, to 550 °C under constant pressure, then to 8 bar under constant volume. Determine the energy balances. How would the energy balances change if steam from the same initial state were first cooled at constant volume to 8 bar, then heated at constant pressure to the same final state?
Solution The process is shown by the path 1 → 2 → 3 in Figure 3-6. The properties for states 1 and 2 are obtained directly form the steam tables; the state at 3 requires interpolation. The information is summarized in the table below:
Figure 3-6: Energy balances along different paths (Example 3.8).
The energy balances are performed along each branch separately.
1 → 2: In this constant-pressure segment, heat is equal to the enthalpy change:
Alternate Path When the process is conducted by first applying a constant-V path followed by a constant-pressure one, the combined path is the one marked as 1 → 4 → 3 in Figure 3-6. The calculation can be done by following the same procedure as with the previous path. However, the calculation can be simplified by noting that several required quantities have computed already.
The change in internal is the same as in the previous case since the initial and final states are the same:
The results for the two cases are summarized below:
Comments
• Internal energy is a state function: Its difference between two states is determined by the states alone and not by the path that is used to connect these states.
• Work and heat are path functions: Their value depends on the entire path that connects two states.
• That work is a path function is demonstrated by the fact that work corresponds to the area under the path. Clearly, the path 1 → 2 → 3 involves more work because it covers a larger area. Since the sum of work plus heat is independent of the path (it is equal to the change of the internal energy), we conclude that heat must be a path function as well.
A classical mechanical system is characterized by a set of mechanical state variables: velocity, elevation in a gravitational field, electrical charges, and so forth. If these variables are given and the external fields are known, the system is fully specified and its behavior at any instant of time, past or future, can be calculated. Thermodynamic systems are characterized by an additional state variable: temperature. Unlike mechanical variables, which describe external interactions, temperature characterizes the internal state of macroscopic matter. Temperature is a measure of the energy stored inside matter in various forms, which we collectively call internal energy. Temperature gives rise to another type of energy exchange that is not encountered among purely mechanical systems, heat: when two systems, each at its own temperature, are put into thermal contact, energy in the form of heat flows from the higher temperature to the lower temperature. For the complete energy balance, heat and internal energy must both be accounted for.
The incorporation of heat effects into the energy balance constitutes one of the fundamental principles of thermodynamics known as the first law. This chapter is devoted to the mathematical formulation of the first law and in the definition of two thermodynamic properties, internal energy and enthalpy, both of which are important in the calculation of energy balances.
Instructional Objectives In this chapter we will formulate the mathematical statement of the first law for a closed system and will learn how to:
1. Do energy balances in closed systems.
2. Distinguish between path and state functions.
3. Use the steam tables to calculate internal energy and enthalpy.
4. Use heat capacities to calculate changes in internal energy and enthalpy.
5. Apply the energy balance to systems undergoing vaporization or condensation.
6. Perform calculations of internal energy and enthalpy in the ideal-gas state.
3.1 Energy and Mechanical Work
Heat, work and energy are measured in the same units but they represent different physical entities. We begin with work and energy, which are familiar concepts from mechanics. Energy is the ability of a system to produce work, namely, ability to cause the displacement of a force. Consider an object with mass m resting on the floor (Figure 3-1). To lift it to the top of the table at height Δz above ground, we must supply an amount of work equal to W = mgΔz. When the mass is resting on the desk, its potential energy has increased by ΔEp = mgΔz, and the work done to lift the object is now part of the energy of the body and is stored as potential energy. The energy added to the system will stay with it for as long it remains at rest on the table. This could be hours, days, or millennia: potential energy is preserved for as long as the state of the system, in this case elevation, is preserved. The more general conclusion to be drawn from this analysis is that energy is a property of the state and as such, a storable quantity. We can recover it by letting the mass drop to the floor and use its kinetic energy to, say, catapult a small projectile into the air, or to load a spring, or to push a nail into the floor.
Figure 3-1: Potential energy and mechanical work. From state A to state B the potential energy increases by the amount of work needed to lift the weight. This work must be supplied from the surroundings, that is, from a force external to the system.
Let us focus on work now. Work takes place when a force is displaced. If we lift the weight ourselves, this work is supplied through the action of our muscle. Once the object is placed on the table, there is no more displacement, thus no more work: work is exchanged during a process. It characterizes, not the state of the system, but the transition of the system between states (from the floor to the top of the desk). In thermodynamic terms, the picture illustrated in Figure 3-1 would be described as follows: A system initially in equilibrium state A undergoes a process that brings it to final state B; during this process the system exchanges work with the surroundings so that the energy change of the system is equal to the amount of work exchanged:
This equation reads, “as a result of the process, the energy of the system in state B has increased relative to A by the amount of work added.” Also notice that work represents energy that passes from one system into another. In this example, our muscle (chemical energy) transfers work to the lifted object, where it is converted into potential energy. As a transfer quantity, work is characterized by a direction, “from” (the muscle) “to” (the system). Finally, when work passes into a new system, it is converted into some form of energy (in our example, potential), which now is property of the state. Let us summarize:
• Energy is storable; it remains in the system for as long as the state of the system is preserved. It is a state function.
• Work is energy in transit: it appears when energy is passed from one system into another.
• Work is not a storable quantity as such.1 Work that enters a system must be stored in some form of energy.
1. By this we mean that it is incorrect to say that a system “stores work”; the proper statement is, the system “stores energy.”
• Work is associated with a direction “from” one system in which it originates, “to” another, where it is transferred to.
To indicate the direction of the transfer we adopt a sign convention:
Sign convention for work: Work is positive if it enters the system, negative if it exits.
When the surroundings do the work, the work is positive because it is absorbed by the system. When the work is negative, it is done by the system and absorbed by the surroundings. Clearly, the sign depends on the definition of the system since work that enters one system (the “system”) exits the other (the “surroundings”) and vice versa. To avoid ambiguity, the definition of the system must be made clear when we report positive or negative values of work.
Note
Forms of Energy
As a matter of classification, energy comes in two basic forms, kinetic, and potential. Kinetic is energy stored in the motion of a moving object. Potential energy is a more general category and encompasses several forms that arise from various conservative forces, such as gravitational fields, electric charges, spring forces, and others. All forms of potential energy have the common characteristic that they are described by a potential function Φ(x, y, z) that depends on the space coordinates, x, y, z, and has the property that the force in a given direction is equal to the negative derivative of the potential with respect to the corresponding coordinate:
As an example, in the gravitational potential Φ(x, y, z) = mgz, the force (weight) in the z direction is
with the negative sign indicating that the weight points downwards. The forces in the x and y direction are both zero. The intermolecular potential introduced in Section 1.1 is another example of potential energy.
3.2 Shaft Work and PV Work
A direct method by which to exchange work with a fluid is through the use of a mechanical device. When wind blows through a turbine, it produces work that is manifested in the rotational motion of the turbine. A centrifugal pump imparts work to a fluid through the rotation of an impeller. This type of work is called shaft work. It will generally be recognizable through the presence of a rotating or reciprocating shaft that absorbs work from a fluid or imparts work to it.
Another more subtle form of work is associated with movement of the system boundaries. Thermodynamic systems make mechanical contact through the pressure that is exerted on the boundaries that separate them. At equilibrium, this pressure is equal on both sides of the boundary. If a pressure imbalance arises between the system and its surroundings, the boundaries of the system must move in response to the mechanical force. Such imbalance may arise from the application of a mechanical force that acts to compress or expand the system, or through the application of heat, which causes the volume to expand or contract. The movement of boundaries involves the exchange of work, which we call PV work.
To obtain an equation for the PV work, consider the compression of a gas in a cylinder fitted with a piston whose area is a (Fig. 3-2b). Compression is done through the application of an external pressure Pex, which produces a force on the piston, F = Pexa, and causes the piston to move by dx. The amount of work associated with this process is equal to the force on the piston, Pexa, multiplied by the displacement dx. Noting that the product adx is equal to the change of volume, −dV, the amount of work is
Figure 3-2: Examples of work: (a) shaft work (wind turbine); (b) PV work for compression/ expansion of a gas in a cylinder; (c) PV work associated with volume changes.
The negative sign is consistent with the convention adopted for work: in compression, dV is negative and the work is positive (it is added to the system); in expansion, dV is positive and the work is negative (work is transferred from the system to the surroundings).
Equation (3.2) gives the work in terms of the external pressure that causes the change. It is more convenient, however, to relate the work to the pressure of the system. For compression, the external pressure must be higher than the pressure of the system, P, which means, Pex = P + δP, where δP is a positive increment; for expansion, δP is negative. If the process is conducted in a quasi-static manner, then δP → 0 and Pex → P. Such process is called mechanically reversible and in this case the PV work is
This is the differential amount of work along a small step of a process that causes a volume change. The total amount of work involved in taking the system from an initial state A to final state B along a reversible path is obtained by integration:
Figure 3-3: Graphical interpretation of reversible PV work.
This result is general and not limited to gases, nor to objects of cylindrical shape. It applies to any system, gas, liquid, or solid, whose volume changes by compression or expansion, under or against the opposing pressure of the surroundings. The only requirement is that the process must be mechanically reversible.
Reversible PV work has a simple graphical interpretation on the PV graph: it is equal in absolute value to the area under the path of the process when this is plotted on the PV graph (Figure 3-3). The sign indicates whether the work is added to or removed from the system. For a path that moves in the direction of increasing V, the area is by convention positive and the work is negative: the system expands and does work against the surroundings. For a process that moves in the direction of decreasing V (compression), the area is negative and the work is positive.
An important conclusion is that PV work depends on the entire path that connects the two states. Different paths between the same two states generally correspond to different areas, and thus different amounts of work. Therefore, the PV work is a path function whose value depends not only on the initial and final states, but on the entire path. This is in contrast to thermodynamic properties (state functions) whose value depends on the state alone.
Example 3.1: PV Work in Expansion
A cylinder fitted with a piston contains 1 liter of gas. The piston has a 1-in diameter and weighs 5 kg. How much work is needed to expand the gas reversibly to twice its volume against the pressure of the atmosphere (1.013 bar)?
Solution The gas expands against the combined pressure of the piston and the atmosphere. The weight of the piston is Mg = (5 kg)(9.81 m/s2)= 49.05 N and the area of the piston is πD2/4 = π(2.54 × 10−2 m)2 = 5.067 × 10−4 m2. The pressure exerted by the piston is
The total pressure in the cylinder is
P = P′ + P0 = 0.968 bar + 1.013 bar = 1.981 bar.
The expansion is to take place against an external pressure of 1.981 bar, which remains constant during the process. If the volume doubles, the change in volume is 1 L = 1000 cm3 = 10−3 m3. The corresponding amount of work is
The negative sign indicates that this work is done by the system on to the surroundings.
Example 3.2: PV Work Using the Soave-Redlich-Kwong Equation of State
Ethylene is compressed reversibly in a closed system. The compression is conducted isothermally at 350 K, from initial pressure 20 bar to final pressure 55 bar. Calculate the work using the SRK equation of state.
Solution First, we write the SRK equation in the form,
To obtain the work, this pressure must be integrated at constant temperature as a function of V:
Numerical substitutions The critical parameters of ethylene and the acentric factor are:
Tc = 282.35 K, Pc = 50.418 bar, ω = 0.0866.
Using these values, the constants a and b are:
a = 0.404344 J m3/mol2, b = 4.03395 × 10−5 m3/mol.
The volume V1 must be obtained by solving the SRK equation. As in Example 2.11, we solve for the compressibility factor first. At the initial state (20 bar, 350 K), the equation for Z is
Z3 − Z2 + 0.0670103Z − 0.00264794 = 0.
This equation has one real root, Z = 0.931084, from which we obtain the molar volume of the initial state:
In the final state, the equation for the compressibility factor is
Z3 − Z2 + 0.180579Z − 0.020025 = 0.
This equation has one real root, Z = 0.806979, and the corresponding molar volume is
Finally, by numerical substitution into eq. [A], the isothermal work is
W = 385.6 J/mol.
The work is positive, as it should be, since volume decreases.
Example 3.3: Isothermal Compression of Steam
Steam undergoes reversible isothermal compression at 350 °C, from initial pressure 20 bar to 40 bar. Calculate the amount of work.
Solution This problem is very similar to the previous one (see Example 3.2). Here, however, the pressure-volume relationship is given in tabular form and the integration of the PV work must be done numerically. First, we collect values of the specific volume V at 350 °C at various pressures between the final and initial state:
One way to perform the integration is using the trapezoidal rule within each pressure interval. An alternative method, which is generally more accurate, is to fit an equation to the data and perform the integration analytically using the fitted equation. We will illustrate the use of both methods.
Trapezoidal rule: In this method, the integral of n tabulated values {xi, fi} of a function f(x) is given by the following approximation:
Using xi = Vi, yi = Pi, we construct the following table:
From the sum of the last column we find,
The work is given by the negative of the above integral:
Fitting the data: The data are fitted to a quadratic polynomial in V,
P = a0 + a1V + a2V2
with coefficients
a0 = 81.4217, a1 = −801.366, a2 = 2598.63.
We use the quadratic equation to calculate the integral between V1 and V2:
Using the fitted parameters and V1 = 0.1386 m3/kg, V2 = 0.06647 m3/kg, we find
W = 1.998 bar m3/kg = 199.8 kJ/kg.
Figure 3-4 shows the data and the fitted line. In this case both methods give nearly identical results because the tabulation includes closely spaced points.
Figure 3-4: Calculation of isothermal PV work at 350 °C from steam tables. The solid line is a quadratic fit to the points from the tables (see Example 3.3).
3.3 Internal Energy and Heat
Molecules possess energy of various forms. All molecules possess kinetic energy due to the motion of the center of mass in space (translational kinetic energy). In addition to translational kinetic energy, polyatomic molecules possess rotational kinetic energy, manifested as motion relative to the center of mass, and vibrational energy, as motion of the constituent atoms relative to the equilibrium length of their chemical bonds. Molecules also possess potential energy as a result of interactions between different molecules as well as between atoms of the same molecule. These interactions can be loosely visualized as springs whose compression or extension requires energy.2 For our purposes it is not necessary to consider these molecular modes of energy storage in any detail. It suffices to say that matter, regardless of chemical composition or phase, is capable of storing energy internally. We refer to these combined storage modes as internal energy and we will use the symbol U. Like all forms of energy, internal energy is a state function and for a pure substance, it is a function of pressure and temperature.
2. To apply the spring analogy to intermolecular interactions we imagine the spring constant to be variable with distance: the intermolecular spring stiffens when highly compressed but becomes very soft when it is overextended.
The molecular nature of matter gives rise to a different type of energy transfer, heat. The mean kinetic energy of molecules in a hot substance is higher than that in a cold substance. The nature of molecular collisions is such, that when molecules with low kinetic energy are mixed with molecules of higher energy, energy is exchanged in a way that gives all molecules the same energy on average. The effect of collisions then is to spread the energy among molecules, much like the cue ball distributes its energy among all balls on a pool table. This process amounts to transferring energy from “hot” molecules to “cold” ones. Collisional transfer of energy does not require intimate mixing and it can take place when two solids are placed into contact, or when two fluids come into contact with a common solid wall. Molecules in the solid phase, though restricted in the range of their motion, move vigorously about their equilibrium position and collide with their neighbors, thus absorbing energy from higher-energy molecules and transferring it to those with lower energy. The rate of this transfer varies among materials, but this affects only how long it takes to reach equilibrium, not the final equilibrium itself. A thermal insulator, we may note here, is a material that offers a resistance to heat transfer so that the amount transferred over the duration of a typical application is negligibly small. Over time, however, heat will escape, even through an insulator, and thermal equilibrium will be established across all systems that are in contact with each other.
Heat shares some important characteristics with work:
1. It is a transient form of energy that is observed during a change of state (process); once the system is in equilibrium with its surroundings there is no net heat transfer because both system and surroundings are at the same temperature.3
3. On an instantaneous basis, individual collisions may transfer energy in either direction, from the system to the bath or the other way around, even at equilibrium. Overall, however, there is no net transfer between system and surroundings if the conditions of thermal equilibrium are met.
2. As a transient form, heat is not a storable mode of energy: once it enters a system, it is stored as internal energy. It is incorrect to say that a system “contains heat,” or to speak of energy that “is converted into heat.”
3. It has a direction, from the system to the surroundings, or vice versa.
4. It is a path function whose value is determined by the entire path of the process. This property of heat is not obvious at the moment but will become so in the next section.
Sign convention for heat: Heat is positive if it is transferred to the system from the surroundings, negative if it is transferred from the system to the surroundings.
A useful and accurate method for the calculation of liquid molar volumes at saturation is the Rackett equation. In its modified form, it gives the molar volume of saturated liquid as
where VL is the molar volume at saturation, Tc is the critical temperature, Pc is the critical pressure, Tr = T/Tc is the reduced temperature and ZR is a parameter specific to the fluid. Perry’s Chemical Engineers’ Handbook [2] lists the values for some common fluids10. If ZR is not available, it can be replaced by the compressibility factor at the critical point, Zc = PcVc/RTc, which is usually tabulated. If we use the critical compressibility factor in eq. (2.56), the Rackett equation is simplified and takes the form
where Vc is the critical volume. The Rackett equation is empirical but quite accurate, even in the above simplified form, which contains no adjustable parameters.
Note
Molar Volume of Compressed Liquid
At temperatures well below critical, liquids are practically incompressible and we may assume κ ≈ 0. In this approximation, an isotherm on the PV graph is almost perpendicular to the volume axis. Accordingly, the molar volume of a compressed liquid is essentially the same as that of the saturated liquid at the same temperature. The volume of saturated liquid is often tabulated, or it can be calculated from empirical equations such as the Rackett equation. Therefore, the volume of compressed liquid can be estimated quite accurately from the volume of the saturated liquid at the same temperature. This approximation breaks down close to the critical point where the liquid phase becomes quite compressible.
Example 2.13: Thermal Expansion of Acetone
Perry’s Chemical Engineers’ Handbook provides the following equation for the thermal expansion of liquids (p. 2-131 in Ref. [2]):
V = V0 (1+ a1t + a2t2 + a3t3)
where V0 is the specific volume at 0 °C, V is the volume at temperature t, and t is in °C. For acetone, the values of the parameters a1, a2, a3, are
Calculate the volumetric coefficient of thermal expansion of acetone at 20 °C and the percent increase in volume upon a temperature increase from 20 °Cto30 °C at constant pressure.
Solution We apply eq. (2.52) to the equation for V given in the problem statement, noting that the derivative must be taken in terms of absolute temperature whereas the equation is given with temperature in °C. Using T = t + 273.15, the required derivative is
In this equation, t is in °C, and the result is in K−1 or in °C−1 (with respect to temperature differences, as in dT, one degree in the Celsius scale is equal to one kelvin). Using the values of the parameters given in the problem statement at t = 20 °C, we find
β20 °C = 1.426 × 10−3 K−1.
The volume change upon increasing temperature from t1 to t2 is found from eq. (2.54). Keeping pressure constant (dP = 0), integration with respect to temperature gives
The coefficient β is a function of temperature and is given by eq. [A]. A quick approximation is obtained by assuming β to be constant and equal to some average value between t1 and t2. The above integral then becomes
For the average β we will use the simple average between the values at 20 °C and at 30 °C. The coefficient at 30 °C is calculated from eq. [A] and we find β30° C = 1.466 × 10−3 K−1. The mean β is
The volume change is calculated from eq. [C] using t2 − t1 = 10 °C= 10 K:
The volume increases by V2/V1 − 1 = 1.45%. If instead of using an average β we perform the integration in eq. [B] using the full form of β from eq. [A], we find V2/V1 = 1.01457. The approximate calculation in this case is essentially the same as the exact result.
Example 2.14: Constant-Volume Heating of Liquid
A glass container is filled with acetone at 25 °C and sealed, leaving no air inside. Determine the pressure that develops in the container when it is heated to 35 °C. The isothermal compressibility of acetone is (see Table 2-188 in Perry’s Chemical Engineers’ Handbook [2]),
κ = 52 × 10−6 bar−1
You may ignore the expansion of the container.
Solution If the expansion of the container is neglected, the volume of the acetone (total, specific, or molar) remains constant. Using dV = 0 in eq. (2.55) we obtain a relationship between temperature and pressure:
Treating β and κ as constants, integration gives
Using the average value of β between temperatures 20 °C and 30 °C, calculated in Example 2.13, and the value for κ given in the problem statement, the pressure increase is
The glass container will most likely break.
Comments This result may seem counterintuitive: if volume increases only by 1.45% when temperature increases by 10 °C, why is it that the pressure generated at fixed volume is so high? This behavior is a consequence of the steepness of the isotherms in the liquid region. Figure 2-14 shows a graph of these isotherms calculated from eq. (2.54) with the values of β and κ for acetone. These lines decrease very steeply, reflecting the fact that volume changes very little with pressure. During constant-volume heating, the state moves along a vertical line from the initial state at 1 bar, 20 °C, to final state at 30 °C. Because isotherms are very steep, pressure increases substantially along a constant-volume path. This behavior is common to all substances that are fairly incompressible. This is, the reason that frozen water pipes burst in winter and water bottles in the freezer break. The physical reason is that molecules in liquids and solids are at close distances from each other, and very close to the point that the intermolecular potential becomes very strongly repulsive (see Figure 1-3). In this region, very small changes to the position of molecules can create enormous repulsion. When we increase temperature, molecules have enough energy for such small changes in position. Normally, the resulting repulsion causes the system to expand but if the volume is constrained, the resulting pressure can be enormous.
Figure 2-14: Pressure-volume isotherms of acetone (see Example 2.14).
Example 2.15: Rackett Equation
Use the simplified Rackett equation to estimate the density of saturated water at 100 °C.
Solution We collect the following data for water:
Vc = 56 × 10−6 m3/mol, Zc = 0.229, Tc = 647.3K.
At T = 100 °C = 373.15 we have Tr = 0.5765. Using the Rackett equation we find
The value from the steam tables is 0.001043 m3/kg = 1.043 cm3/g, a difference of −2.4%.
Example 2.16: Density of Compressed Liquid
Estimate the density of water at 100 °C, 5 bar.
Solution Under the given conditions water is compressed liquid. The pressure is low compared to the critical point; therefore, we may take the volume to be equal to that of saturated liquid at 100 °C. This value is 0.001043 m3/kg.
2.12 Summary
The equation of state is a fundamental property of fluids that describes the relationship between molar volume, temperature, and pressure. This relationship is expressed mathematically as an equation between V, P, and T, or alternatively, between Z, P, and T. The PV graph is a graphical representation of this relationship and a convenient way to present the phase behavior of a pure component. It is a good idea to draw a qualitative PV graph when solving problems involving heating, cooling, compression, or expansion of pure fluids, as a way of becoming oriented in thermodynamic space. Simple processes on this graph is represented by simple paths and help visualize any phase transformations that may occur.
The accurate prediction of the molar volume, or, equivalently, of the compressibility factor, is a requirement in engineering calculations. Several methodologies were discussed in this chapter:
• Tabulated values. Tabulations are available for a large number of pure components. The steam tables is one such example. Usually, however, property tabulations are not as detailed or extensive as the steam tables. Tables are generally the most accurate source for properties. They are not convenient, however, for large-scale calculations.
• Generalized graphs. These graphs are correlations based on corresponding states. Using just three physical constants, critical pressure and temperature, and acentric factor, one can estimate the molar volume of a pure fluid over a very wide range of conditions. This is a very important advantage but it comes with certain limitations. The method is approximate and is based on graphs that have been tweaked to provide overall good accuracy for several different fluids. Necessarily, the agreement will be better for some and worse for others. The method should be used only for normal fluids that are not polar to a significant degree. This covers a large number of compounds but excludes many industrially important molecules, chiefly among them, water.
• Equations of state. An equation of state in mathematical form has the advantage that it can be used in repetitive calculations and is especially suited for computer-based calculations. Industrial software for chemical process design makes extensive use of such equations. The equations discussed in this chapter (van der Waals, Soave-Redlich-Kwong, and Peng-Robinson) incorporate the principle of corresponding states in their constants, which can be computed for any fluid given the critical pressure, critical temperature, and acentric factor. These equations must be used with the same caution as generalized graphs: they should be applied to nonpolar fluids only. The constants of these equations, specifically the part that involves the acentric factor, have been obtained from fitting data for several different fluids to obtain overall good agreement among them. It is possible to modify the constants of an equation of state to improve accuracy for a particular substance, however, the constants given here are specifically for normal fluids.
Of the many equations that were discussed here, two deserve a special note:
Ideal-gas law. The ideal-gas law is the theoretical limit of any equation of state at low pressure. It is a universal limit but its range of applicability is limited to the region we call ideal-gas state. You should resist the temptation to apply the ideal-gas law without justification.
Virial equation. The virial equation has rigorous basis on theory. However, its truncated form, it should be treated as an extrapolation to somewhat higher pressures that those in the ideal-gas state.
• Empirical equations. Empirical equations usually have no basis on theory but give nonetheless very accurate predictions. The Rackett equation is one such example. The limitation is that the range of applicability is typically narrow. The Rackett equation, for example, is only good along the line of saturated liquid.
The availability of several alternative methodologies is a toolbox that facilitates the job of the engineer. It is the responsibility of the engineer to choose the right tool from this toolbox for a given situation.
2.13 Problems
Problem 2.1: a) Use the steam tables to determine the phase of water (liquid or vapor) at following conditions: 25 °C, 1 bar; 80 °C, 10 bar; 120 °C; 50 bar.
b) The vapor pressure of bromobenzene at 40 °C is 10 mm Hg. Determine the phase of bromobenzene at 40 °C, 1 atm.
c) The normal boiling point of fluorobenzene is 84.7 °C. What is the phase of fluorobenzene at 25 °C, 1 atm?
Problem 2.2: Determine the temperature and phase of water from the following information. If the phase is a vapor-liquid mixture, report the fraction of vapor and liquid.
a) The specific volume of water is 100 cm3/g and the pressure 40 bar.
b) The specific volume of water is 100 cm3/g and the pressure 6 bar.
Problem 2.3: Use the steam tables to do the following:
a) A drum 3.5 m3 in volume contains steam at 1 bar, 210 °C. Determine the mass of steam in the drum.
b) Wet steam with quality 15% vapor is to be stored under pressure at 20 bar in a thermally insulated vessel. What is the temperature?
c) If the total mass to be stored is 525 kg, what is the required volume of the vessel?
The conditions are the same as in part (b).
Problem 2.4: A closed tank contains a vapor-liquid mixture of steam at 45 bar with liquid content 25% by mass.
a) Heat is added until the pressure becomes 80 bar. What is the temperature?
b) The pressure is changed until the contents become 100% saturated vapor. What is that pressure?
c) With pressure held constant at 45 bar, steam is added to or removed from the tank, as needed, until the contents are 25% liquid by volume. Calculate the mass of steam that must be added or removed and report it as a percentage of the total mass originally in the tank.
Problem 2.5: The following data are available for a gas:
To calculate the molar volume at 25 °C, 12 bar, our team came up with two different suggestions: (a) linear interpolation for V between the given pressure, or (b) linear interpolation for the density ρ = 1/V between the two pressures. Which method do you recommend and why? Under what conditions is your recommendation accurate?
Problem 2.6: A 12 m3 tank contains a liquid/vapor mixture of steam at 15 bar. The volume of the liquid in the tank is 0.5 m3.
a) What is the temperature?
b) What is the total mass in the tank?
c) What is quality of the steam?
d) 87% of the mass in the tank is removed while keeping pressure constant at 15 bar. What is the final temperature in the tank?
Problem 2.7: An eight-liter pressure cooker contains a mixture of steam and liquid water at 2 bar. Through a level indicator we can see that the liquid occupies 25% of the volume inside the cooker.
a) What is the temperature inside the cooker?
b) What is the total mass of water (liquid plus vapor) in the cooker?
c) What is the mass fraction of the liquid?
d) The cooker, while it remains sealed, is placed under running water until its temperature cools to 25 °C. What is the pressure in the cooker?
e) What force does it take to unseal the cooker? The cover is circular with a radius of 20 cm.
Problem 2.8: a) A pressure cooker is filled to the brim with water at 80 °Cand the lid is locked. The temperature is then changed until the contents become saturated liquid. What is the temperature and pressure at that point?
b) A closed pressure cooker contains 50% by volume liquid and 50% water vapor at 1 bar. The pressure is then changed until the point where the contents become a single phase. Is that phase saturated liquid or saturated vapor?
c) A closed rigid vessel that contains a pure fluid is cooled until the contents become saturated vapor. Determine whether the initial state is superheated vapor, compressed liquid, or vapor/liquid.
d) A closed rigid vessel that contains a pure fluid is heated until the contents become saturated liquid. Determine whether the initial state is superheated vapor, compressed liquid, or vapor/liquid.
Problem 2.9: A tank whose volume is 12 m3 contains 6.2 kg of water at 1.4 bar.
a) What is the phase (liquid, vapor, liquid/vapor mixture)?
b) What is the temperature?
c) We add more steam to the tank while maintaining its temperature constant at the value calculated in part b. As a result, the pressure in the tank increases. Determine how much water (in kg) of steam must be added to bring the steam in the tank to the point of condensation.
d) Draw a qualitative PV graph and show the path of the process for part c.
Problem 2.10: a) Use data from the steam tables to construct the PV graph of water. Show the saturated liquid, the saturated vapor, the critical point. Include the isotherms at 100 °C, 200 °C, 300 °C and 400 °C. Make two plots, one using linear axes and one in which the pressure axis is linear but the volume axis is logarithmic.
b) Make a Z − P plot of water using data from the steam tables showing the same information as the PV graph above.
Problem 2.11: Determine whether the truncated virial equation is valid for ethane at the following states:
a) 10 bar, 25 °C.
b) Saturated vapor at 10 bar.
c) 10 bar, −35 °C.
Additional data: The boiling point of ethane at 10 bar is −29 °C.
Problem 2.12: The R&D division of your company has released the following limited data on proprietary compound X-23:
Using this incomplete information estimate as best as you can the following:
a) Phase of X-23 at 12 bar, 25 °C.
b) The molar mass of X-23.
c) The second virial coefficient at 25 °C.
d) The required volume of a tank that is needed to store 20 kg of X-23 at 12 bar, 25 C.
e) State clearly and justify as best as you can all your assumptions and the methods you use.
Problem 2.13: Methane is stored under pressure in a 1 m3 tank. The pressure in the tank is 20 bar and the temperature is 25 °C.
a) Calculate the compressibility factor of methane in the tank from the virial equation truncated after the second term.
b) What is the amount (moles) of methane in the tank?
c) You want to store twice as much methane in the tank at the same temperature. What will be the pressure in the tank?
d) Is it appropriate to use the virial equation for this problem? Explain.
At 25 °C the second virial coefficient of methane is −4.22 × 10−5 m3/mol.
Problem 2.14: Use the truncated virial equation to answer the questions below:
a) A 5 m3 tank contains nitrogen at 110 K, 7 bar. How many kg of nitrogen are in the tank?
b) The tank is cooled until the contents become saturated vapor. What is the pressure and temperature in the tank?
c) Is the use of the truncated virial equation justified in this problem? Additional data: The saturation pressure of nitrogen is given by the following empirical equation:
with Psat is in mm Hg and T is in kelvin.
Problem 2.15: a) Calculate the second virial coefficient of water at 200 °C using only data from the steam tables.
b) Use the truncated virial equation along with the second virial coefficient calculated above to estimate the volume of water at 200 °C, 14 bar. How does this value compare with the volume obtained from the steam tables? Discuss this comparison.
Problem 2.16: 1000 kg of methane is to be stored in a tank at 25 °C, 75 bar.
What is the required volume of the tank? (Hint: Use the Pitzer equation with the Lee-Kesler values.)
Problem 2.17: 200 kg of carbon dioxide are stored in a tank at 25 °C and 70 bar.
a) Is carbon dioxide an ideal gas under the conditions in the tank?
b) What is the volume of the tank?
c) How much carbon dioxide must be removed for the pressure of the tank to fall to1bar?
The molecular weight of CO2 is 44 g/mol.
Problem 2.18: A full cylinder of ethylene (C2H4) at 25 °C contains 50 kg of gas at 80 bar.
a) Is ethylene an ideal gas under these conditions? Explain.
b) What is the volume of the cylinder?
c) What is the pressure in the cylinder after 90% of the ethylene has been removed, if temperature is 25 °C?
The molecular weight of ethylene is 28 g/mol.
Problem 2.19: Use the Lee-Kesler method to answer the following: 2000 kg of krypton is to be stored under pressure in a tank at 110 bar, 20 °C. The tank is designed to withstand pressures up to 180 bar.
a) Determine the volume of the tank.
b) Is it safe to store 2500 kg in the tank at 25 °C?
c) Is the Lee-Kesler method appropriate?
Problem 2.20: A tank is divided by a rigid, thermally conducting partition into two equal parts, A and B, each 10 m3 in volume. Part A contains saturated liquid n-butane at 20 °C, 2.07 bar; part B contains saturated n-butane vapor, also at 20 °C, 2.07 bar. Each part is equipped with a safety alarm that will go off if pressure exceeds 40 bar.
a) How many moles of n-butane is in part A of the tank?
b) How many moles of n-butane is in part B of the tank?
c) The tank is heated slowly in such a way that the temperature in both parts rises at the same rate. As soon as the alarm goes off, the heating stops. Which alarm goes off, that of part A or part B?
d) What is the temperature when the alarm sounds?
Additional data: The volume expansivity and the isothermal compressibility of liquid n-butane are given below and may be assumed to be constant:
β = 2.54 × 10−3 K−1, κ = 3.4 × 10−4 bar−1.
Problem 2.21: a) A tank contains 10,000 kg of xenon at 132 °C, 82 bar. The plant supervisor asks you to remove xenon and fill the tank with 10,000 kg of steam at 200 °C. What is the pressure in the tank when it is filled with steam?
b) After the tank has been filled with steam, 5000 kg are withdrawn for use elsewhere in the plant. What is the pressure, if temperature remains at 200 °C?
Problem 2.22: The boiling point of o-xylene at 1 bar is 139 °C.
a) What is the state of o-xylene at 0.1 bar, 200 °C?
b) 100 moles of o-xylene are to be loaded in a tank at 0.1 bar, 200 °C. What is the required volume of the tank?
c) The tank can safely withstand pressures up to 44.9 bar. How much o-xylene can be stored in the tank under maximum pressure at 200 °C?
Problem 2.23: a) Determine the percent change in volume when olive oil is heated at constant pressure from 18 °C to 40 °C.
b) Olive oil is stored in a full container at 18 °C. Determine the pressure that will develop in the container if the temperature in the storage room rises to 40 °C.
Additional data: The volume expansion of olive oil is given by the empirical equation
V = V0(1 + a1t + a2t2 + a3t3)
where t is temperature in °C, V0 is the volume at 0 °C, and the coefficients in the equation are
The coefficient of isothermal compression is
κ = 52 × 10−6 bar−1.
(Data from Perry’s Chemical Engineers’ Handbook, 7th ed., Tables 2-147 and 2-188.)
Problem 2.24: A 0.5 m3 tank will be used to store CO2 at 20 °C. Using the SRK equation answer the following:
a) Determine the maximum amount (kg) of CO2 that can be stored safely if the tank can withstand a maximum pressure of 70 bar.
b) Repeat if the maximum pressure is 60 bar.
c) Repeat at 50 bar.
Problem 2.25: Use the SRK equation to answer the following:
a) 5000 kg of isobutane is to be stored in a tank at 60 psi, 70°F. What is the required tank volume?
b) Since the temperature in the summer can get as high as 95°F, determine the pressure that the tank must withstand to avoid rupture.
Problem 2.26: Use the SRK equation to calculate the molar volume of isobutane at the following states:
a) 30 °C, 1 bar.
b) 30 °C, 10 bar.
c) Saturated liquid at 30 °C.
d) Saturated vapor at 30 °C.
For parts c and d, compare with the saturated molar volumes reported in the NIST Web Book.
Additional information: The saturation pressure at 30 °C is 4.05 bar.
Problem 2.27: Use the SRK equation to perform the following calculations for isobutane to make a graph that shows three isotherms, one at 30 °C, one at the critical temperature, and one at 150 °C. Make the axis logarithmic in volume and linear in pressure, and select the range in the two axes so that the graph is not crowded and its important features are seen clearly. Report volume in m3/mol, pressure in bar and temperature in °C. Annotate the graph and label the axes properly.
Problem 2.28: The parameters β and κ of a substance are reported to be functions of pressure and temperature and are given below:
β = 1 /T, κ = 1 /P.
a) Determine the equation of state. Assume that at pressure P0 and temperature T0 the volume is known to be V0. Your final answer then should be in terms of P, V, T, P0, T0 and V0.
b) Is this equation appropriate for liquids?
Problem 2.29: Calculate the coefficient of isothermal compressibility of isobutane as saturated liquid and saturated vapor at 30 °C, Psat = 4.05 bar, using the SRK equation. Report the result in bar−1.
Hint: Recall from calculus that
Problem 2.30: Use the steam tables to calculate the value of β of water at the following states:
a) 1 bar, 25 °C.
b) 20 bar, 25 °C.
c) 1 bar, 200 °C.
Problem 2.31: Use the Rackett equation to estimate the volume expansivity of liquid ethanol at 25 °C. State your assumptions clearly.
As a state property, the molar (or specific) volume can be determined once as a function of pressure and temperature, and tabulated for future use. Tabulations have been compiled for a large number of pure fluids. In very common use are the steam tables, which contain tabulations of the properties of water. Steam is a basic utility in chemical plants as a heat transfer fluid for cooling or heating, as well as for power generation (pressurized steam), and its properties are needed in many routine calculations. Thermodynamic tables for water are published by the American Society of Mechanical Engineers (ASME) and are available in various forms, printed and electronic. A copy is included in the appendix. We will use them not only because water is involved in many industrial processes but also as a demonstration of how to work with tabulated values in general.
Note
Interpolations
When working with tabulated values it is necessary to perform interpolations if the desired conditions lie between entries in the table. Suppose that a table contains the values of a function f (x) at x = x1 and x = x2 and we wish to obtain the value of f at an intermediate point x such that x1 < x < x2, we assume a linear relationship between f and x and write
where f1, f2 are the values of the function at points x1, x2, respectively. The procedure is shown graphically in Figure 2-5a. If the value of x is outside the interval (x1, x2), the same formula may be used and this calculation is called an extrapolation. Extrapolations should be avoided because they can be subject to large error. They may be used if the desired value is beyond the last tabulated entry, however, one cannot be certain about the accuracy of the result.
Figure 2-5: Linear interpolation: (a) simple interpolation; (b) double interpolation.
In thermodynamics we are usually dealing with functions of two variables, for example, f (x, y). If point (x, y) is such that the value y is found in the table but the value of x falls between tabulated values, the above equation may be used to interpolate with respect to x. If both x and y fall between tabulated entries, a double interpolation is necessary. The procedure involves three simple interpolations, and can be outlined as follows (Figure 2-5b): first interpolate between the tabulated values at (x1, y1) and (x1, y2) to obtain the value of f at point C; do the same between points (x2, y1) and (x2, y2) to obtain f at point C′. Finally, interpolate between C and C′ to obtain the value at the desired point B. Alternatively, interpolate to obtain points A and A′ followed by interpolation between A and A′ to obtain B—the result is the same. These steps can be combined into a single equation which takes the form:
with
Here, the notation fij refers to f(xi, yj). If both a and b are between 0 and 1, this calculation is indeed an interpolation and produces a result that is surrounded by the four tabulated values used in the calculation. This equation can be used for extrapolations outside these four values provided the distance is not large.
Example 2.4: Interpolation
Use the steam tables to determine the specific volume of water at 1.25 bar, 185 °C.
Solution From the superheated steam tables we find the following values for the specific volume (in m3/kg):
In this problem, pressure (1.25 bar) and temperature (185 °C) both lie between the tabulated values. Therefore, a double interpolation is required. We do the calculation first by successive interpolations, then by applying equation (2.14) for double interpolations.
Method 1—successive interpolations: We first obtain the volume at P = 1.25 bar, T1 = 150 °C by interpolation between the values listed in the first row of the above data:
This amounts to obtaining the volume at point C of Figure 2-5. Next, we obtain the volume at P = 1.25 bar, T1 = 200 °C by interpolation between the values listed in the second row:
This amounts to calculating the volume at point C′. Finally, we interpolate between V1 (point C) and V2 (C′) to obtain the volume at 1.25 bar, 185 °C:
Method II—double interpolation: To apply the double-interpolation formula, we take x to be temperature and y to be pressure, and fxy to be the specific volume at temperature x and pressure y. The factors a and b are calculated from eq. (2.15):
Both methods give the same answer, as they should.
Example 2.5: Locating the State in the Steam Tables
A 12m3 pressurized vessel contains 200 kg of steam at 40 bar. What is the temperature?
Solution The specific volume of the steam in the tank is
We know two intensive properties, pressure and specific volume; the state, therefore, is fully specified. We must locate a point in the steam tables with P = 40 bar, V = 0.06 m3/kg.
From the entries at 40 bar we obtain the following data:
Interpolating for temperature at V = 0.06 m3/kg we have
Therefore, the temperature in the tank is 307.2 °C.
Example 2.6: Lever Rule
An additional 170 kg of steam is added to the tank of the previous example. If the final pressure is 50 bar, determine the temperature and phase of the contents of the tank.
Solution The new specific volume is
This volume lies between the volume of the saturated liquid and saturated vapor at 50 bar:
P = 50 bar T = 263.94 °C: VL = 0.0012864 m3/kg, VV = 0.039446 m3/kg.
The state, therefore, is wet steam at 50 bar, 263.94 °C. To determine the mass fractions of each phase we use the lever rule in eq. (2.10):
Therefore, the quality of the steam in the tank is 81.6%.
2.3 Compressibility Factor and the ZP Graph
The compressibility factor, Z, is defined as the ratio
where V is the molar volume, P is pressure, and T is absolute temperature. It is a dimensionless quantity and a state function. In the ideal-gas state, Vig = RT/P, and the compressibility factor is unity:
More precisely, this is the limiting value of the compressibility factor of any real gas when pressure is reduced to zero under constant temperature:
This result states that the compressibility factor along a line of constant temperature goes to 1 as P is reduced to zero. In other words, on a graph of Z versus pressure, all isotherms at zero pressure must meet at Z = 1.
The compressibility factor represents an alternative way of presenting molar volume, since the molar volume can be obtained easily if the compressibility factor is known at a given pressure and temperature:
Mathematically, the equation of state can be represented as either a relationship between P, V, and T, or between Z, P, and T. The latter relationship is quite useful in presenting the volumetric behavior of fluids. Its graphical representation is given on the ZP graph, whose general form is shown in Figure 2-6. Here again we have the vapor-liquid boundary in the form of bell-shaped curve that is now seated on the vertical axis. The vapor region is at the top, the liquid at the bottom. Generally, compressibility factors in the liquid phase are smaller than those in the vapor phase because the molar volume of the liquid is small. All isotherms meet at P = 0, Z = 1, as anticipated on the basis of eq. (2.18). This point of convergence represents the ideal-gas state. The ZP graph illustrates the path to the ideal-gas state: it is approached by following an isotherm to zero pressure. Since all isotherms converge to this point, the ideal-gas state can be reached from any initial state. We can now see why there is no such a thing as an ideal-gas state: if such a gas existed, its ZP graph would consist of a single horizontal line at Z = 1, which would represent all isotherms. No substance exists that exhibits such behavior.
Figure 2-6: The ZP graph of pure fluid.
Although the mathematical definition places the ideal-gas state at a single point (P = 0, Z = 1), from a practical point of view we will consider a gas to be in the ideal-gas state if the compressibility factor is sufficiently close to 1. For calculations that do not require high accuracy we will assume a gas to be in the ideal-gas state if the compressibility factor is within ±5% of the theoretical value of 1. The pressure range over which this approximation is valid varies with temperature.
With reference to Figure 2-6, the isotherm at T4 remains closer to 1 over a wider interval of pressures, compared to the isotherm at T1, which has a larger negative slope and decreases faster. In general, to determine whether a gas at given pressure and temperature can be treated as ideal we must check with a ZP graph. We will return to this question in the next section.
2.4 Corresponding States
It is found experimentally that the ZP graphs of different fluids look very similar to each other as if they are scaled versions of a single, universal, graph. This underlying graph is revealed if pressure and temperature are rescaled by appropriate factors. We introduce a set of reduced (dimensionless) variables by scaling pressure and temperature with the corresponding values at the critical point:
Using the reduced coordinates it is possible to combine ZP data for several compounds on the same graph. Figure 2-7 shows the resulting graph for selected molecules. The remarkable feature of this graph is that the compressibility factor of the six compounds shown on this figure agree with each other quite well. Even though the dependence of the compressibility factor on temperature and pressure is different for each fluid, they all seem to arise from the same reduced function. This observation gives rise to the correlation of corresponding states, which is expressed as follows:
At the same reduced temperature and pressure, fluids have approximately the same compressibility factor.
Figure 2-7: The compressibility factor of selected molecules as a function of reduced pressure and temperature. Data compiled from E. W. Lemmon, M. O. McLinden and D. G. Friend, “Thermophysical Properties of Fluid Systems.” In NIST Chemistry WebBook, NIST Standard Reference Database No. 69, eds. P. J. Linstrom and W. G. Mallard, National Institute of Standards and Technology, Gaithersburg MD, 20899, http://webbook.nist.gov (retrieved December 11, 2010).
We express this mathematically by writing,
The practical implication is important: to the extent that eq. (2.20) is obeyed, the compressibility factor of any fluid can be described by a universal equation that is a function of reduced temperature and reduced pressure. Figure 2-7 is a graphical representation of this equation. Accordingly, the compressibility factor of a fluid can be determined at any pressure and temperature using just two parameters, the critical temperature and critical pressure.
The correlation of corresponding states is not an exact physical law and is not obeyed to the same degree of accuracy by all fluids. Although the agreement in Figure 2-7 between different fluids is impressive, it is not exact. The spread of the data points in Figure 2-7 is not due to experimental error (the values have been calculated from validated models) but reflect systematic deviations. These are more clearly seen near the saturation curve and in the liquid region. The correlation arises from similarities in the intermolecular potential. In general, agreement is very good in the gas phase, where interactions are unimportant. In the liquid region, interactions are important and the individual chemical character of molecules becomes more apparent. Even so, nonpolar molecules that are nearly spherical in shape (e.g., Ar, CH4) agree remarkably well. Polar molecules or molecules with more complex structures (e.g., normal heptane) show the largest deviations because their interactions are more complex and dissimilar. Molecules like water, which is both polar and associates strongly via hydrogen bonding, exhibit even more deviations from this correlation. The principle of corresponding states should be treated as a working hypothesis that can provide useful but not always highly accurate estimates of the compressibility factor.
Acentric Factor and the Pitzer Method
Eq. (2.20) is a two-parameter correlation because it requires two physical properties, critical temperature and critical pressure. To improve the predictive power of the principle of corresponding state while retaining its simplicity, a third parameter is introduced, the acentric factor. It is a dimensionless parameter that is defined according to the equation,
where is the reduced saturation pressure of the fluid at reduced temperature Tr = 0.7. It is a characteristic property of the fluid and is found in tables, usually along with the critical properties of the fluid. It was introduced by Pitzer as a measure of the sphericity of the molecule. More generally, it should be understood as a combined measure of the shape and polarity. Symmetric nonpolar molecules, such as Ar, have ω = 0. These are called simple fluids and are found to obey the two-parameter correlation of corresponding states quite well. Nonspherical or polar molecules have a larger acentric factor. For most fluids the acentric factor is positive and in the range 0 to 0.4, although small negative values of ω are also possible.
In the Pitzer method, the compressibility factor is expressed in the form:
where Z(0) and Z(1) are universal functions that depend on Tr and Pr. Function Z(0) represents the compressibility factor of simple fluids (ω = 0) and Z(1) represents a correction that is proportional to the acentric factor. The two functions Z(0) and Z(1) may be calculated once and tabulated against reduced pressure and temperature for future use. The resulting tables and charts are called generalized because they are not limited to a specific molecule.
Various methodologies have been developed for the calculation of the functions in eq. (2.22), but the most widely used is that of Lee and Kesler.5 The Lee-Kesler result for Z(0) is plotted in Figure 2-8. Recall that this term represents the compressibility factor of a simple fluid, therefore, it has the familiar appearance of the ZP graph. Figure 2-9 shows the same graph in semi-logarithmic scales that cover an expanded range of pressures. The Z(1) is shown in Figure 2-10. The correction factor is zero in the vicinity of the ideal-gas state. It increases in absolute value with increasing pressure and it may take positive or negative values. The sharp changes in subcritical temperatures correspond to a shift from the vapor branch to the liquid branch of the isotherm. The vapor/liquid transitions in Figures 2-8, 2-9, and 2-10 apply only to simple fluids (ω = 0). For nonsimple fluids the phase boundary also depends on the acentric factor ω and is generally shifted to lower reduced pressure compared to simple fluids.6
5. B. I. Lee and M. G. Kesler. A generalized thermodynamic correlation based on three-parameter corresponding states. AIChE J., 21(3):510, 527 1975. doi: 10.1002/aic.690210313.
6. The determination of the precise location of the phase boundary is discussed in Chapter 7.
Figure 2-8: Generalized graph of Z(0) based on the Lee-Kesler method.
Figure 2-9: Generalized graph of Z(0) based on the Lee-Kesler calculation (extended range of pressures).
Figure 2-10: Generalized graph of Z(1) based on the Lee-Kesler calculation.
The strength of molecular interactions is determined by the mean intermolecular distance and the property that most directly reflects this distance is molar density, or its reciprocal, molar volume. The approximate relationship between molar volume and mean intermolecular distance is given by (see Example 1.1),
where NA is Avogadro’s number. The packing density of molecules in given volume reflects the strength of the potential interaction. In gases (large V), distances are large and interactions weak. In liquids, the opposite is true. The volume that is occupied by a fixed number of molecules depends both on temperature and pressure. The relationship between volume, pressure, and temperature is of fundamental importance and its mathematical form is known as equation of state. In this chapter we examine this relationship in graphical and mathematical form. The learning objectives of this chapter are to develop the following skills:
1. Using the PVT graph to identify the phase of a pure fluid.
2. Working with tabulated values of P, V, T (steam tables).
3. Identify the region of applicability of the ideal-gas law and the truncated virial equation.
4. Working with cubic equations of state.
5. Working with generalized correlations for the compressibility factor.
6. Representing processes on the PV graph.
2.1 The PVT Behavior of Pure Fluid
The molar volume, V, is the volume occupied by 1 mol of the substance, and is the inverse of the molar density, ρ:1
1. We will use the same general symbol, ρ, for both the molar density and the mass density. We will annotate them differently only if they both appear in the same equation.
The specific volume is the volume occupied by 1 kg of substance. The specific volume, the mass density, and the molar densities are related to each other:
where ρ′ is the mass density (kg/m3) and Mm is molar mass (kg/mol). The volume occupied by a given amount of matter depends on temperature and pressure: it decreases under compression and (for most substances) increases upon heating. The relationship between P, V, T is characteristic of a substance but the general features of this relationship are common to all pure fluids and will be examined below.
The relationship between volume, pressure, and temperature is represented graphically by a three-dimensional surface whose general shape is shown in Figure 2-1. This graph has been rotated to show pressure in the vertical axis, with the mesh lines on the surface representing lines of constant temperature (left to right), and lines of constant volume (front to back). A point on the surface gives the molar volume at the indicated pressure and temperature. The bell-shaped curve facing the pressure-volume plane is the vapor liquid boundary. To its left is the liquid region (steep part of the surface), to its right the vapor region. At sufficiently low temperatures a system exhibits a phase transition to a solid; this region is not shown on this graph.2
2. The surface in Figure 2-1 was calculated using the Soave-Redlich-Kwong equation of state (see Section 2.6), which is appropriate for liquids and gases but not for solids.
Figure 2-1: The PVT surface of a pure fluid.
The three-dimensional representation is useful for the purposes of visualizing the PVT relationship but is impractical for routine use. We work, instead, with projections of the PVT surface on one of the three planes, PV, PT, or VT. A projection is a view of the three-dimensional surface from an angle perpendicular to the projection plane and reduces the graph into a two-dimensional plot. The most commonly used projections are those on the PV and the PT planes.
The PV Graph
The PV graph is the projection on the PV plane and is shown in Figure 2-2. The characteristic feature of this graph is the vapor-liquid region, represented by a bell-shaped curve that consists of two branches, saturated liquid (to the left) and saturated vapor (to the right). The two branches meet at the top and this point defines the critical point of the fluid. Temperature is indicated by contours of constant temperature (isotherms). These are lines with the general direction from the upper left corner to the lower right. The isotherm that passes through the critical point corresponds to the critical temperature, TC. This isotherm has an inflection point at the critical point, namely, its first and second derivatives are both zero:
Figure 2-2: The PV graph of pure fluid (the solid phase is not shown).
From there on, it decreases smoothly into the vapor region. The region to the left of the saturated liquid is called subcooled liquid to indicate that its temperature is below the boiling point that corresponds to its pressure. For example, at A the temperature is T1, lower than the saturation temperature (T2) that corresponds to pressure PA. Alternatively, it is called compressed liquid to indicate that pressure is higher than the saturation pressure that corresponds to its temperature (the terms subcooled and compressed liquid are used interchangeably). Vapor to the right of the saturated line is superheated because its temperature is higher than the boiling temperature that corresponds to its pressure. At B, for example, temperature (T4) is higher than the saturation temperature (T2) that corresponds to its pressure.
The organization of information on the PV graph can be better understood by conducting a heating or cooling process and following the path on the graph. Suppose we add heat under constant pressure starting with liquid at state A, which is at pressure PA and temperature TA = T1. The process is depicted by the line AB, drawn at constant pressure PA. Between states A and L, heating causes the volume to increase somewhat but the increase is relatively small because the thermal expansion of liquids is small. At point L the liquid is saturated and at the verge of boiling. Adding heat at this point causes liquid to evaporate and produce more vapor, moving the state along the line LV. During this process both pressure and temperature remain constant (line LV is both an isotherm and an isobar). At point V all the liquid has evaporated and the system is saturated vapor. Adding more heat causes temperature and molar volume to increase and moves the state along the line VB. If we start at state B and perform a constant-pressure cooling process, we will observe the reverse course of events. During BV, the vapor is cool. At point V, the state is saturated vapor at the verge of condensation. Removing heat at this point causes vapor to condense until the steam becomes 100% saturated liquid (state L). Upon further cooling, the system moves further into the subcooled region.
Note
Boiling in Open Air
If the heating/cooling process that is described here is conducted in an open container, for example, by heating water in an open flask at atmospheric pressure, the behavior will be somewhat different than the one described here. In an open container, water forms vapor at any temperature below boiling, not just at the boiling point. The important difference is that in an open container we are dealing with a multicomponent system that contains not only water, but also air. A vapor-liquid mixture with two or more components behaves differently from the pure components. Multicomponent phase equilibrium is treated in the second part of this book and until then, it should be understood that we are dealing with pure fluids. The process described by the path AB may be thought to take place inside a sealed cylinder fitted with a piston and initially filled with liquid containing no air at all.
The Critical Point
The critical point is an important state and its pressure and temperature have been tabulated for a large number of pure substances. In approaching the critical point from below, the distance between points L and V decreases, indicating that the molar volume of the saturated liquid and saturated vapor come closer together. At the critical point the two saturated phases coincide: vapor and liquid become indistinguishable and the phase boundary seizes to exist. The region of the phase diagram above Pc and Tc is referred to as supercritical fluid. No isotherms or isobars in this region intersect the vapor-liquid boundary. If point E is heated isobarically to final state F, one will observe a continuous transition from a dense, liquid-like state, to a dilute, gaslike state. In the supercritical region the notions of “liquid” and “vapor” are not helpful. These terms are meaningful when both phases can exist simultaneously and can be identified as distinct from each other. The term supercritical fluid avoids these ambiguities.
Properties near the critical point are quite different compared to states at lower temperatures and pressures. As the difference between vapor and liquid becomes less clear near the critical point, the liquid becomes substantially more compressible than typical liquids. This is indicated on the PV graph by the gentle slope of the isotherm as it approaches the critical point. Isotherms below but near the critical temperature (not shown in Figure 2-2) show similar behavior. The usual approximation that treats liquids as incompressible is acceptable only at temperatures well below the critical. In the supercritical region, the behavior of a fluid is somewhere between that of a liquid and a gas. The gentle slope of the isotherms indicates that the fluid is quite compressible, even at high, liquidlike densities (low molar volumes). Other properties, in particular, the solubility of various nonvolatile solutes, are often found to be quite enhanced compared to the subcritical region. As an example, the enhanced solubility of caffeine in supercritical carbon dioxide (Tc = 304.1 K, Pc = 73.8 bar) makes it possible to use carbon dioxide as a solvent to extract caffeine from coffee, thus avoiding the use of other solvents with potential toxic effects.
A Special Limit: The Ideal-Gas State
If the molar volume is increased sufficiently, the effect of molecular interactions decreases, and in the limit that it becomes infinite, it vanishes completely. When this condition is met we say that the system is in the ideal-gas state. The mathematical specification of the ideal-gas state is :
or, equivalently,
The stipulation of constant temperature is necessary. Without it, it would be possible to maintain the system in the liquid (or even solid) phase, even at very low pressures, thus never reaching a state where intermolecular distances are large. On the PV graph, the ideal-gas state is found near the lower-right corner.
In the ideal-gas state, the PVT relationship is universal for all gases, regardless of chemical composition and this relationship is given by the ideal-gas law:
where Vig is molar volume, T is absolute temperature, and R is a universal constant (ideal-gas constant), whose value in the SI system is
The superscript “ig” will be used to indicate results that are valid only in the ideal-gas state. The ideal-gas law should be viewed as the limiting form of the equation of state of any real fluid when pressure is reduced under constant temperature. Even though the ideal-gas state represents an idealization (infinite distance between molecules), in practice eq. (2.6) provides satisfactory results if the actual state of a gas is sufficiently close to the ideal-gas state. More often than not, engineering problems require calculations at conditions where the ideal-gas law is not valid. It is important to be aware of the limitations of the ideal-gas law and never use it without proper justification.
Historically, a large part of thermodynamics was developed before the emergence of atomic and molecular theories of matter. This part has come to be known as classical thermodynamics and makes no reference to molecular concepts. It is based on two basic principles (“laws”) and produces a rigorous mathematical formalism that provides exact relationships between properties and forms the basis for numerical calculations. It is a credit to the ingenuity of the early developers of thermodynamics that they were capable of developing a correct theory without the benefit of molecular concepts to provide them with physical insight and guidance. The limitation is that classical thermodynamics cannot explain why a property has the value it does, nor can it provide a convincing physical explanation for the various mathematical relationships. This missing part is provided by statistical thermodynamics. The distinction between classical and statistical thermodynamics is partly artificial, partly pedagogical. Artificial, because thermodynamics makes physical sense only when we consider the molecular phenomena that produce the observed behaviors. From a pedagogical perspective, however, a proper statistical treatment requires more time to develop, which leaves less time to devote to important engineering applications. It is beyond the scope of this book to provide a bottom-up development of thermodynamics from the molecular level to the macroscopic. Instead, our goal is to develop the knowledge, skills, and confidence to perform thermodynamic calculations in chemical engineering settings. We will use molecular concepts throughout the book to shed light to new concepts but the overall development will remain under the general umbrella of classical thermodynamics. Those who wish to pursue the connection between the microscopic and the macroscopic in more detail, a subject that fascinated some of the greatest scientific minds, including Einstein, should plan to take an upper-level course in statistical mechanics from a chemical engineering, physics, or chemistry program.
The Laws of Classical Thermodynamics
Thermodynamics is built on a small number of axiomatic statements, propositions that we hold to be true on the basis of our experience with the physical world. Statistical and classical thermodynamics make use of different axiomatic statements; the axioms of statistical thermodynamics have their basis on statistical concepts; those of classical thermodynamics are based on behavior that we observe macroscopically. There are two fundamental principles in classical thermodynamics, commonly known as the first and second law.3 The first law expresses the principle that matter has the ability to store energy within. Within the context of classical thermodynamics, this is an axiomatic statement since its physical explanation is inherently molecular. The second law of thermodynamics expresses the principle that all systems, if left undisturbed, will move towards equilibrium –never away from it. This is taken as an axiomatic principle because we cannot prove it without appealing to other axiomatic statements. Nonetheless, contact with the physical world convinces us that this principle has the force a universal physical law.
3. The term law comes to us from the early days of science, a time during which scientists began to recognize mathematical order behind what had seemed up until then to be a complicated physical world that defies prediction. Many of the early scientific findings were known as “laws,” often associated with the name of the scientist who reported them, for example, Dalton’s law, Ohm’s law, Mendel’s law, etc. This practice is no longer followed. For instance, no one refers to Einstein’s famous result, E = mc2, as Einstein’s law.
Other laws of thermodynamics are often mentioned. The “zeroth” law states that, if two systems are in thermal equilibrium with a third system, they are in equilibrium with each other. The third law makes statements about the thermodynamic state at absolute zero temperature. For the purposes of our development, the first and second law are the only two principles needed in order to construct the entire mathematical theory of thermodynamics. Indeed, these are the only two equations that one must memorize in thermodynamics; all else is a matter of definitions and standard mathematical manipulations.
The “How” and the “Why” in Thermodynamics
Engineers must be skilled in the art of how to perform the required calculations, but to build confidence in the use of theoretical tools it is also important to have a sense why our methods work. The “why” in thermodynamics comes from two sources. One is physical: the molecular picture that gives meaning to “invisible” quantities such as heat, temperature, entropy, equilibrium. The other is mathematical and is expressed through exact relationships that connect the various quantities. The typical development of thermodynamics goes like this:
(a) Use physical principles to establish fundamental relationships between key properties. These relationships are obtained by applying the first and second law to the problem at hand.
(b) Use calculus to convert the fundamental relationships from step (a) into useful expressions that can be used to compute the desired quantities.
Physical intuition is needed in order to justify the fundamental relationships in step (a). Once the physical problem is converted into a mathematical one (step [b]), physical intuition is no longer needed and the gear must shift to mastering the “how.” At this point, a good handle of calculus becomes indispensable, in fact, a prerequisite for the successful completion of this material. Especially important is familiarity with functions of multiple variables, partial derivatives and path integrations.
1.3 Definitions
System
The system is the part of the physical world that is the object of a thermodynamic calculation. It may be a fixed amount of material inside a tank, a gas compressor with the associated inlet and outlet streams, or an entire chemical plant. Once the system is defined, anything that lies outside the system boundaries belongs to the surroundings. Together system and surroundings constitute the universe. A system can interact with its surroundings by exchanging mass, heat, and work. It is possible to construct the system in such way that some exchanges are allowed while others are not. If the system can exchange mass with the surroundings it called open, otherwise it is called closed. If it can exchange heat with the surroundings it is called diathermal, otherwise it is called adiabatic. A system that is prevented from exchanging either mass, heat, or work is called isolated. The universe is an isolated system.
A simple system is one that has no internal boundaries and thus allows all of its parts to be in contact with each other with respect to the exchange of mass, work, and heat. An example would be a mole of a substance inside a container. A composite system consists of simple systems separated by boundaries. An example would be a box divided into two parts by a firm wall. The construction of the wall would determine whether the two parts can exchange mass, heat, and work. For example, a permeable wall would allow mass transfer, a diathermal wall would allow heat transfer, and so on.
Figure 1-5: Examples of systems (see Example 1.2). The system is indicated by the dashed line. (a) Closed tank that contains some liquid and some gas. (b) The liquid portion in a closed, thermally insulated tank that also contains some gas. (c) Thermally insulated condenser of a laboratory-scale distillation unit.
Solution In (a) we have a tank that contains a liquid and a vapor. This system is inhomogeneous, because it consists of two phases; closed, because it cannot exchange mass with the surroundings; and simple, because it does not contain any internal walls. Although the liquid can exchange mass with the vapor, the exchange is internal to the system. There is no mention of insulation. We may assume, therefore, that the system is diathermal.
In (b) we have the same setup but the system is now defined to be just the liquid portion of the contents. This system is simple, open, and diathermal. Simple, because there are no internal walls; open, because the liquid can exchange mass with the vapor by evaporation or condensation; and diathermal, because it can exchange heat with the vapor. In this case, the insulation around the tank is not sufficient to render the system adiabatic because of the open interface between the liquid and vapor.
In (c) we have a condenser similar to those found in chemistry labs. Usually, hot vapor flows through the center of the condenser while cold water flows on the outside, causing the vapor to condense. This system is open because it allows mass flow through its boundaries. It is composite because of the wall that separates the two fluids. It is adiabatic because it is insulated from the surroundings. Even though heat is transferred between the inner and outer tube, this transfer is internal to the system (it does not cross the system bounds) and does not make the system diathermal.
Comments In the condenser of part (c), we determined the system to be open and adiabatic. Is it not possible for heat to enter through the flow streams, making the system diathermal? Streams carry energy with them and, as we will learn in Chapter 6, this is in the form of enthalpy. It is possible for heat to cross the boundary of the system inside the flow stream through conduction, due to different temperatures between the fluid stream just outside the system and the fluid just inside it. This heat flows slowly and represents a negligible amount compared to the energy carried by the flow. The main mode heat transfer is through the external surface of the system. If this is insulated, the system may be considered adiabatic.
Equilibrium
It is an empirical observation that a simple system left undisturbed, in isolation of its surroundings, must eventually reach an ultimate state that does not change with time. Suppose we take a rigid, insulated cylinder, fill half of it with liquid nitrogen at atmospheric pressure and the other half with hot, pressurized nitrogen, and place a wall between the two parts to keep them separate. Then, we rupture the wall between the two parts and allow the system to evolve without any disturbance from the outside. For some time the system will undergo changes as the two parts mix. During this time, pressure and temperature will vary, and so will the amounts of liquid and vapor. Ultimately, however, the system will reach a state in which no more changes are observed. This is the equilibrium state.
Equilibrium in a simple system requires the fulfillment of three separate conditions:
1.Mechanical equilibrium: demands uniformity of pressure throughout the system and ensures that there is no net work exchanged due to pressure differences.
2.Thermal equilibrium: demands uniformity of temperature and ensures no net transfer of heat between any two points of the system.
3.Chemical equilibrium: demands uniformity of the chemical potential and ensures that there is no net mass transfer from one phase to another, or net conversion of one chemical species into another by chemical reaction.
The chemical potential will be defined in Chapter 7.
Although equilibrium appears to be a static state of no change, at the molecular level it is a dynamic process. When a liquid is in equilibrium with a vapor, there is continuous transfer of molecules between the two phases. On an instantaneous basis the number of molecules in each phase fluctuates; overall, however, the molecular rates to and from each phase are equal so that, on average, there is no net transfer of mass from one phase to the other.
Constrained Equilibrium
If we place two systems into contact with each other via a wall and isolate them from the rest of their surroundings, the overall system is isolated and composite. At equilibrium, each of the two parts is in mechanical, thermal, and chemical equilibrium at its own pressure and temperature. Whether the two parts establish equilibrium with each other will depend on the nature of the wall that separates them. A diathermal wall allows heat transfer and the equilibration of temperature. A movable wall (for example, a piston) allows the equilibration of pressure. A selectively permeable wall allows the chemical equilibration of the species that are allowed to move between the two parts. If a wall allows certain exchanges but not others, equilibrium is established only with respect to those exchanges that are possible. For example, a fixed conducting wall allows equilibration of temperature but not of pressure. If the wall is fixed, adiabatic, and impermeable, there is no exchange of any kind. In this case, each part establishes its own equilibrium independently of the other.
Extensive and Intensive Properties
In thermodynamics we encounter various properties, for example, density, volume, heat capacity, and others that will be defined later. In general, property is any quantity that can be measured in a system at equilibrium. Certain properties depend on the actual amount of matter (size or extent of the system) that is used in the measurement. For example, the volume occupied by a substance, or the kinetic energy of a moving object, are directly proportional to the mass. Such properties will be called extensive. Extensive properties are additive: if an amount of a substance is divided into two parts, one of volume V1 and one of volume V2, the total volume is the sum of the parts, V1 + V2. In general, the total value of an extensive property in a system composed of several parts is the sum of its parts. If a property is independent of the size of the system, it will be called intensive. Some examples are pressure, temperature, density. Intensive properties are independent of the amount of matter and are not additive.
As a result of the proportionality that exists between extensive properties and amount of material, the ratio of an extensive property to the amount of material forms an intensive property. If the amount is expressed as mass (in kg or lb), this ratio will be called a specific property; if the amount is expressed in mole, it will be called a molar property. For example, if the volume of 2 kg of water at 25 °C, 1 bar, is measured to be 2002 cm3, the specific volume is
and the molar volume is
In general for any extensive property F we have a corresponding intensive (specific or molar) property:
The relationship between specific and molar property is
where Mm is the molar mass (kg/mol).
Note
Nomenclature
We will refer to properties like volume as extensive, with the understanding that they have an intensive variant. The symbol V will be used for the intensive variant, whether molar or specific. The total volume occupied by n mole (or m mass) of material will be written as Vtot, nV, or mV. No separate notation will be used to distinguish molar from specific properties. This distinction will be made clear by the context of the calculation.
State of Pure Component
Experience teaches that if we fix temperature and pressure, all other intensive properties of a pure component (density, heat capacity, dielectric constant, etc.) are fixed. We express this by saying that the state of a pure substance is fully specified by temperature and pressure. For the molar volume V, for example, we write
which reads “V is a function of T and P.” The term state function will be used as a synonym for “thermodynamic property.” If eq. (1.5) is solved for temperature, we obtain an equation of the form
which reads “T is a function of P and V.” It is possible then to define the state using pressure and molar volume as the defining variables, since knowing pressure and volume allows us to calculate temperature. Because all properties are related to pressure and temperature, the state may be defined by any combination of two intensive variables, not necessarily T and P. Temperature and pressure are the preferred choice, as both variables are easy to measure and control in the laboratory and in an industrial setting. Nonetheless, we will occasionally consider different sets of variables, if this proves convenient.
Note
Fixing the State
If two intensive properties are known, the state of single-phase pure fluid is fixed, i.e., all other intensive properties have fixed values and can be obtained either from tables or by calculation.
State of Multicomponent Mixture
The state of a multicomponent mixture requires the specification of composition in addition to temperature and pressure. Mixtures will be introduced in Chapter 8. Until then the focus will be on single components.
Process and Path
The thermodynamic plane of pure substance is represented by two axes, T and P. A point on this plane represents a state, its coordinates corresponding to the temperature and pressure of the system. The typical problem in thermodynamics involves a system undergoing a change of state: a fixed amount of material at temperature TA and pressure PA is subjected to heating/cooling, compression/expansion, or other treatments to final state (TC, PC). A change of state is called a process. On the thermodynamic plane, a process is depicted by a path, namely, a line of successive states that connect the initial and final state (see Figure 1-6). Conversely, any line on this plane represents a process that can be realized experimentally. Two processes that are represented by simple paths on the TP plane are the constant-pressure (or isobaric) process, and the constant-temperature (or isothermal) process. The constant-pressure process is a straight line drawn at constant pressure (path AB in Figure 1-6); the constant-temperature process is drawn at constant temperature (path BC). Any two points on the TP plane can be connected using a sequence of isothermal and isobaric paths.
Figure 1-6: Illustration of two different paths (ABC, AB′C) between the same initial (A) and final (C) states. Paths can be visualized as processes (heating/cooling, compression/expansion) that take place inside a cylinder fitted with a piston.
Processes such as the constant-pressure, constant-temperature, and constant-volume process are called elementary. These are represented by simple paths during which one state variable (pressure, temperature, volume) is held constant. They are also simple to conduct experimentally. One way to do this is using a cylinder fitted with a piston. By fitting the piston with enough weights we can exert any pressure on the contents of the cylinder, and by making the piston movable we allow changes of volume due to heating/cooling to take place while keeping the pressure inside the cylinder constant. To conduct an isothermal process we employ the notion of a heat bath, or heat reservoir. Normally, when a hot system is used to supply heat to a colder one, its temperature drops as a result of the transfer of heat. If we imagine the size of the hot system to approach infinity, any finite transfer of heat to (or from) another system represents an infinitesimal change for the large system and does not change its temperature by any appreciable amount. The ambient air is a practical example of a heat bath with respect to small exchanges of heat. A campfire, for example, though locally hot, has negligible effect on the temperature of the air above the campsite. The rising sun, on the other hand, changes the air temperature appreciably. Therefore, the notion of an “infinite” bath must be understood as relative to the amount of heat that is exchanged. A constant-temperature process may be conducted by placing the system into contact with a heat bath. Additionally, the process must be conducted in small steps to allow for continuous thermal equilibration. The constant-volume process requires that the volume occupied by the system remain constant. This can be easily accomplished by confining an amount of substance in a rigid vessel that is completely filled. Finally, the adiabatic process may be conducted by placing thermal insulation around the system to prevent the exchange of heat.
We will employ cylinder-and-piston arrangement primarily as a mental device that allows us to visualize the mathematical abstraction of a path as a physical process that we could conduct in the laboratory.
Quasi-Static Process
At equilibrium, pressure and temperature are uniform throughout the system. This ensures a well-defined state in which, the system is characterized by a single temperature and single pressure, and represented by a single point on the PT plane. If we subject the system to a process, for example, heating by placing it into contact with a hot source, the system will be temporarily moved away from equilibrium and will develop a temperature gradient that induces the necessary transfer of heat. If the process involves compression or expansion, a pressure gradient develops that moves the system and its boundaries in the desired direction. During a process the system is not in equilibrium and the presence of gradients implies that its state cannot be characterized by a single temperature and pressure. This introduces an inconsistency in our depiction of processes as paths on the TP plane, since points on this plane represent equilibrium states of well-defined pressure and temperature. We resolve this difficulty by requiring the process to take place in a special way, such that the displacement of the system from equilibrium is infinitesimally small. A process conducted in such manner is called quasi static. Suppose we want to increase the temperature of the system from T1 to T2. Rather than contacting the system with a bath at temperature T2, we use a bath at temperature T1 + δT, where δT is a small number, and let the system equilibrate with the bath. This ensures that the temperature of the system is nearly uniform (Fig. 1-7). Once the system is equilibrated to temperature T1 + δT, we place it into thermal contact with another bath at temperature T1 + 2δT, and repeat the process until the final desired temperature is reached. Changes in pressure are conducted in the same manner. In general, in a quasi-static process we apply small changes at a time and wait between changes for the system to equilibrate. The name derives from the Latin quasi (“almost”) and implies that the process occurs as if the system remained at a stationary equilibrium state.
Figure 1-7: (a) Typical temperature gradient in heat transfer. (b) Heat transfer under small temperature difference. (c) Quasi-static idealization: temperatures in each system are nearly uniform and almost equal to each other.
All macroscopic behavior of matter is the result of phenomena that take place at the microscopic level and arise from force interactions among molecules. Molecules exert a variety of forces: direct electrostatic forces between ions or permanent dipoles; induction forces between a permanent dipole and an induced dipole; forces of attraction between nonpolar molecules, known as van der Waals (or dispersion) forces; other specific chemical forces such as hydrogen bonding. The type of interaction (attraction or repulsion) and the strength of the force that develops between two molecules depends on the distance between them. At far distances the force is zero. When the distance is of the order of several Å, the force is generally attractive. At shorter distances, short enough for the electron clouds of the individual atoms to begin to overlap, the interaction becomes very strongly repulsive. It is this strong repulsion that prevents two atoms from occupying the same point in space and makes them appear as if they possess a solid core. It is also the reason that the density of solids and liquids is very nearly independent of pressure: molecules are so close to each other that adding pressure by any normal amounts (say 10s of atmospheres) is insufficient to overcome repulsion and cause atoms to pack much closer.
Intermolecular Potential
The force between two molecules is a function of the distance between them. This force is quantified by intermolecular potential energy, Φ(r), or simply intermolecular potential, which is defined as the work required to bring two molecules from infinite distance to distance r. Figure 1-3 shows the approximate intermolecular potential for CO2. Carbon dioxide is a linear molecule and its potential depends not only on the distance between the molecules but also on their relative orientation. This angular dependence has been averaged out for simplicity. To interpret Figure 1-3, we recall from mechanics that force is equal to the negative derivative of the potential with respect to distance:
Figure 1-3: Approximate interaction potential between two CO2 molecules as a function of their separation distance. The potential is given in kelvin; to convert to joule multiply by the Boltzmann constant, kB = 1.38 × 10−23 J/K. The arrows show the direction of the force on the test molecule in the regions to the left and to the right of r*.
That is, the magnitude of the force is equal to the slope of the potential with a negative sign that indicates that the force vector points in the direction in which the potential decreases. To visualize the force, we place one molecule at the origin and a test molecule at distance r. The magnitude of the force on the test molecule is equal to the derivative of the potential at that point (the force on the first molecule is equal in magnitude and opposite in direction). If the direction of force is towards the origin, the force is attractive, otherwise it is repulsive. The potential in Figure 1-3 has a minimum at separation distance r* = 4.47 Å. In the region r > r* the slope is positive and the force is attractive. The attraction is weaker at longer distances and for r larger than about 9 Å the potential is practically flat and the force is zero. In the region r < r* the potential is repulsive and its steep slope indicates a very strong force that arises from the repulsive interaction of the electrons surrounding the molecules. Since the molecules cannot be pushed much closer than about r ≈ r*, we may regard the distance r* to be the effective diameter of the molecule.1 Of course, even simple molecules like argon are not solid spheres; therefore, the notion of a molecular diameter should not be taken literally.
1. The closest center-to-center distance we can bring two solid spheres is equal to the sum of their radii. For equal spheres, this distance is equal to their diameter.
The details of the potential vary among different molecules but the general features are always the same: Interaction is strongly repulsive at very short distance, weakly attractive at distance of the order of several Å, and zero at much larger distances. These features help to explain many aspects of the macroscopic behavior of matter.
Temperature and Pressure
In the classical view of molecular phenomena, molecules are small material objects that move according to Newton’s laws of motion, under the action of forces they exert on each other through the potential interaction. Molecules that collide with the container walls are reflected back, and the force of this collision gives rise to pressure. Molecules also collide among themselves,2 and during these collisions they exchange kinetic energy. In a thermally equilibrated system, a molecule has different energies at different times, but the distribution of energies is overall stationary and the same for all molecules. Temperature is a parameter that characterizes the distribution of energies inside a system that is in equilibrium with its surroundings. With increasing temperature, the energy content of matter increases. Temperature, therefore, can be treated as a measure of the amount of energy stored inside matter.
2. Molecular collisions do not require solid contact as macroscopic objects do. If two molecules come close enough in distance, the steepness of the potential produces a strong repulsive force that causes their trajectories to deflect.
Note
Maxwell-Boltzmann Distribution
The distribution of molecular velocities in equilibrium is given by the Maxwell-Boltzmann equation:
where m is the mass of the molecule, v is the magnitude of the velocity, T is absolute temperature, and kB is the Boltzmann constant. The fraction of molecules with velocities between any two values v1 to v2 is equal to the area under the curve between the two velocities (the total area under the curve is 1). The velocity vmax that corresponds to the maximum of the distribution, the mean velocity s, and the mean of the square of the velocity are all given in terms of temperature:
The Maxwell-Boltzmann distribution is a result of remarkable generality: it is independent of pressure and applies to any material, regardless of composition or phase. Figure 1-4 shows this distribution for water at three temperatures. At the triple point, the solid, liquid, and vapor, all have the same distribution of velocities.
Figure 1-4: Maxwell-Boltzmann distribution of molecular velocities in water.
Phase Transitions
The minimum of the potential represents a stable equilibrium point. At this distance, the force between two molecules is zero and any small deviations to the left or to the right produce a force that points back to the minimum. A pair of molecules trapped at this distance r* would form a stable pair if it were not for their kinetic energy, which allows them to move and eventually escape from the minimum. The lifetime of a trapped pair depends on temperature. At high temperature, energies are higher, and the probability that a pair will remain trapped is low. At low temperature a pair can survive long enough to trap additional molecules and form a small cluster of closely packed molecules. This cluster is a nucleus of the liquid phase and can grow by further collection to form a macroscopic liquid phase. Thus we have a molecular view of vapor-liquid equilibrium. This picture highlights the fact that to observe a vapor-liquid transition, the molecular potential must exhibit a combination of strong repulsion at short distances with weak attraction at longer distances. Without strong repulsion, nothing would prevent matter from collapsing into a single point; without attraction, nothing would hold a liquid together in the presence of a vapor. We can also surmise that molecules that are characterized by a deeper minimum (stronger attraction) in their potential are easier to condense, whereas a shallower minimum requires lower temperature to produce a liquid. For this reason, water, which associates via hydrogen bonding (attraction) is much easier to condense than say, argon, which is fairly inert and interacts only through weak van der Waals attraction.
Note
Condensed Phases
The properties of liquids depend on both temperature and pressure, but the effect of pressure is generally weak. Molecules in a liquid (or in a solid) phase are fairly closely packed so that increasing pressure does little to change molecular distances by any appreciable amount. As a result, most properties of liquids are quite insensitive to pressure and can be approximately taken to be functions of temperature only.
Example 1.1: Density of Liquid CO2
Estimate the density of liquid carbon dioxide based on Figure 1-3.
Solution The mean distance between molecules in the liquid is approximately equal to r*, the distance where the potential has a minimum. If we imagine molecules to be arranged in a regular cubic lattice at distance r* from each other, the volume of NA molecules would be . The density of this arrangement is
where Mm is the molar mass. Using r* = 4.47 Å = 4.47 × 10−10 m, Mm = 44.01 × 10−3 kg/mol,
Perry’s Handbook (7th ed., Table 2-242) lists the following densities of saturated liquid CO2 at various temperatures:
According to this table, density varies with temperature from 1130.7 kg/m3 at 216.6 K to 466.2 kg/m3 at 304.2 K. The calculated value corresponds approximately to 285 K.
Comments The calculation based on the intermolecular potential is an estimation. It does not account for the effect of temperature (assumes that the mean distance between molecules is r* regardless of temperature) and that molecules are arranged in a regular cubic lattice. Nonetheless, the final result is of the correct order of magnitude, a quite impressive result given the minimal information used in the calculation.
Ideal-Gas State
Figure 1-3 shows that at distances larger than about 10 Å the potential of carbon dioxide is fairly flat and the molecular force nearly zero. If carbon dioxide is brought to a state such that the mean distance between molecules is more than 10 Å we expect that molecules would hardly register the presence of each other and would largely move independently of each other, except for brief close encounters. This state can be reproduced experimentally by decreasing pressure (increasing volume) while keeping temperature the same. This is called the ideal-gas state. It is a state—not a gas—and is reached by any gas when pressure is reduced sufficiently. In the ideal-gas state molecules move independently of each other and without the influence of the intermolecular potential. Certain properties in this state become universal for all gases regardless of the chemical identity of their molecules. The most important example is the ideal-gas law, which describes the pressure-volume-temperature relationship of any gas at low pressures.
1.2 Statistical versus Classical Thermodynamics
Historically, a large part of thermodynamics was developed before the emergence of atomic and molecular theories of matter. This part has come to be known as classical thermodynamics and makes no reference to molecular concepts. It is based on two basic principles (“laws”) and produces a rigorous mathematical formalism that provides exact relationships between properties and forms the basis for numerical calculations. It is a credit to the ingenuity of the early developers of thermodynamics that they were capable of developing a correct theory without the benefit of molecular concepts to provide them with physical insight and guidance. The limitation is that classical thermodynamics cannot explain why a property has the value it does, nor can it provide a convincing physical explanation for the various mathematical relationships. This missing part is provided by statistical thermodynamics. The distinction between classical and statistical thermodynamics is partly artificial, partly pedagogical. Artificial, because thermodynamics makes physical sense only when we consider the molecular phenomena that produce the observed behaviors. From a pedagogical perspective, however, a proper statistical treatment requires more time to develop, which leaves less time to devote to important engineering applications. It is beyond the scope of this book to provide a bottom-up development of thermodynamics from the molecular level to the macroscopic. Instead, our goal is to develop the knowledge, skills, and confidence to perform thermodynamic calculations in chemical engineering settings. We will use molecular concepts throughout the book to shed light to new concepts but the overall development will remain under the general umbrella of classical thermodynamics. Those who wish to pursue the connection between the microscopic and the macroscopic in more detail, a subject that fascinated some of the greatest scientific minds, including Einstein, should plan to take an upper-level course in statistical mechanics from a chemical engineering, physics, or chemistry program.
The Laws of Classical Thermodynamics
Thermodynamics is built on a small number of axiomatic statements, propositions that we hold to be true on the basis of our experience with the physical world. Statistical and classical thermodynamics make use of different axiomatic statements; the axioms of statistical thermodynamics have their basis on statistical concepts; those of classical thermodynamics are based on behavior that we observe macroscopically. There are two fundamental principles in classical thermodynamics, commonly known as the first and second law.3 The first law expresses the principle that matter has the ability to store energy within. Within the context of classical thermodynamics, this is an axiomatic statement since its physical explanation is inherently molecular. The second law of thermodynamics expresses the principle that all systems, if left undisturbed, will move towards equilibrium –never away from it. This is taken as an axiomatic principle because we cannot prove it without appealing to other axiomatic statements. Nonetheless, contact with the physical world convinces us that this principle has the force a universal physical law.
3. The term law comes to us from the early days of science, a time during which scientists began to recognize mathematical order behind what had seemed up until then to be a complicated physical world that defies prediction. Many of the early scientific findings were known as “laws,” often associated with the name of the scientist who reported them, for example, Dalton’s law, Ohm’s law, Mendel’s law, etc. This practice is no longer followed. For instance, no one refers to Einstein’s famous result, E = mc2, as Einstein’s law.
Other laws of thermodynamics are often mentioned. The “zeroth” law states that, if two systems are in thermal equilibrium with a third system, they are in equilibrium with each other. The third law makes statements about the thermodynamic state at absolute zero temperature. For the purposes of our development, the first and second law are the only two principles needed in order to construct the entire mathematical theory of thermodynamics. Indeed, these are the only two equations that one must memorize in thermodynamics; all else is a matter of definitions and standard mathematical manipulations.
The “How” and the “Why” in Thermodynamics
Engineers must be skilled in the art of how to perform the required calculations, but to build confidence in the use of theoretical tools it is also important to have a sense why our methods work. The “why” in thermodynamics comes from two sources. One is physical: the molecular picture that gives meaning to “invisible” quantities such as heat, temperature, entropy, equilibrium. The other is mathematical and is expressed through exact relationships that connect the various quantities. The typical development of thermodynamics goes like this:
(a) Use physical principles to establish fundamental relationships between key properties. These relationships are obtained by applying the first and second law to the problem at hand.
(b) Use calculus to convert the fundamental relationships from step (a) into useful expressions that can be used to compute the desired quantities.
Physical intuition is needed in order to justify the fundamental relationships in step (a). Once the physical problem is converted into a mathematical one (step [b]), physical intuition is no longer needed and the gear must shift to mastering the “how.” At this point, a good handle of calculus becomes indispensable, in fact, a prerequisite for the successful completion of this material. Especially important is familiarity with functions of multiple variables, partial derivatives and path integrations.
The system is the part of the physical world that is the object of a thermodynamic calculation. It may be a fixed amount of material inside a tank, a gas compressor with the associated inlet and outlet streams, or an entire chemical plant. Once the system is defined, anything that lies outside the system boundaries belongs to the surroundings. Together system and surroundings constitute the universe. A system can interact with its surroundings by exchanging mass, heat, and work. It is possible to construct the system in such way that some exchanges are allowed while others are not. If the system can exchange mass with the surroundings it called open, otherwise it is called closed. If it can exchange heat with the surroundings it is called diathermal, otherwise it is called adiabatic. A system that is prevented from exchanging either mass, heat, or work is called isolated. The universe is an isolated system.
A simple system is one that has no internal boundaries and thus allows all of its parts to be in contact with each other with respect to the exchange of mass, work, and heat. An example would be a mole of a substance inside a container. A composite system consists of simple systems separated by boundaries. An example would be a box divided into two parts by a firm wall. The construction of the wall would determine whether the two parts can exchange mass, heat, and work. For example, a permeable wall would allow mass transfer, a diathermal wall would allow heat transfer, and so on.
Figure 1-5: Examples of systems (see Example 1.2). The system is indicated by the dashed line. (a) Closed tank that contains some liquid and some gas. (b) The liquid portion in a closed, thermally insulated tank that also contains some gas. (c) Thermally insulated condenser of a laboratory-scale distillation unit.
Solution In (a) we have a tank that contains a liquid and a vapor. This system is inhomogeneous, because it consists of two phases; closed, because it cannot exchange mass with the surroundings; and simple, because it does not contain any internal walls. Although the liquid can exchange mass with the vapor, the exchange is internal to the system. There is no mention of insulation. We may assume, therefore, that the system is diathermal.
In (b) we have the same setup but the system is now defined to be just the liquid portion of the contents. This system is simple, open, and diathermal. Simple, because there are no internal walls; open, because the liquid can exchange mass with the vapor by evaporation or condensation; and diathermal, because it can exchange heat with the vapor. In this case, the insulation around the tank is not sufficient to render the system adiabatic because of the open interface between the liquid and vapor.
In (c) we have a condenser similar to those found in chemistry labs. Usually, hot vapor flows through the center of the condenser while cold water flows on the outside, causing the vapor to condense. This system is open because it allows mass flow through its boundaries. It is composite because of the wall that separates the two fluids. It is adiabatic because it is insulated from the surroundings. Even though heat is transferred between the inner and outer tube, this transfer is internal to the system (it does not cross the system bounds) and does not make the system diathermal.
Comments In the condenser of part (c), we determined the system to be open and adiabatic. Is it not possible for heat to enter through the flow streams, making the system diathermal? Streams carry energy with them and, as we will learn in Chapter 6, this is in the form of enthalpy. It is possible for heat to cross the boundary of the system inside the flow stream through conduction, due to different temperatures between the fluid stream just outside the system and the fluid just inside it. This heat flows slowly and represents a negligible amount compared to the energy carried by the flow. The main mode heat transfer is through the external surface of the system. If this is insulated, the system may be considered adiabatic.
Equilibrium
It is an empirical observation that a simple system left undisturbed, in isolation of its surroundings, must eventually reach an ultimate state that does not change with time. Suppose we take a rigid, insulated cylinder, fill half of it with liquid nitrogen at atmospheric pressure and the other half with hot, pressurized nitrogen, and place a wall between the two parts to keep them separate. Then, we rupture the wall between the two parts and allow the system to evolve without any disturbance from the outside. For some time the system will undergo changes as the two parts mix. During this time, pressure and temperature will vary, and so will the amounts of liquid and vapor. Ultimately, however, the system will reach a state in which no more changes are observed. This is the equilibrium state.
Equilibrium in a simple system requires the fulfillment of three separate conditions:
1.Mechanical equilibrium: demands uniformity of pressure throughout the system and ensures that there is no net work exchanged due to pressure differences.
2.Thermal equilibrium: demands uniformity of temperature and ensures no net transfer of heat between any two points of the system.
3.Chemical equilibrium: demands uniformity of the chemical potential and ensures that there is no net mass transfer from one phase to another, or net conversion of one chemical species into another by chemical reaction.
The chemical potential will be defined in Chapter 7.
Although equilibrium appears to be a static state of no change, at the molecular level it is a dynamic process. When a liquid is in equilibrium with a vapor, there is continuous transfer of molecules between the two phases. On an instantaneous basis the number of molecules in each phase fluctuates; overall, however, the molecular rates to and from each phase are equal so that, on average, there is no net transfer of mass from one phase to the other.
Constrained Equilibrium
If we place two systems into contact with each other via a wall and isolate them from the rest of their surroundings, the overall system is isolated and composite. At equilibrium, each of the two parts is in mechanical, thermal, and chemical equilibrium at its own pressure and temperature. Whether the two parts establish equilibrium with each other will depend on the nature of the wall that separates them. A diathermal wall allows heat transfer and the equilibration of temperature. A movable wall (for example, a piston) allows the equilibration of pressure. A selectively permeable wall allows the chemical equilibration of the species that are allowed to move between the two parts. If a wall allows certain exchanges but not others, equilibrium is established only with respect to those exchanges that are possible. For example, a fixed conducting wall allows equilibration of temperature but not of pressure. If the wall is fixed, adiabatic, and impermeable, there is no exchange of any kind. In this case, each part establishes its own equilibrium independently of the other.
Extensive and Intensive Properties
In thermodynamics we encounter various properties, for example, density, volume, heat capacity, and others that will be defined later. In general, property is any quantity that can be measured in a system at equilibrium. Certain properties depend on the actual amount of matter (size or extent of the system) that is used in the measurement. For example, the volume occupied by a substance, or the kinetic energy of a moving object, are directly proportional to the mass. Such properties will be called extensive. Extensive properties are additive: if an amount of a substance is divided into two parts, one of volume V1 and one of volume V2, the total volume is the sum of the parts, V1 + V2. In general, the total value of an extensive property in a system composed of several parts is the sum of its parts. If a property is independent of the size of the system, it will be called intensive. Some examples are pressure, temperature, density. Intensive properties are independent of the amount of matter and are not additive.
As a result of the proportionality that exists between extensive properties and amount of material, the ratio of an extensive property to the amount of material forms an intensive property. If the amount is expressed as mass (in kg or lb), this ratio will be called a specific property; if the amount is expressed in mole, it will be called a molar property. For example, if the volume of 2 kg of water at 25 °C, 1 bar, is measured to be 2002 cm3, the specific volume is
and the molar volume is
In general for any extensive property F we have a corresponding intensive (specific or molar) property:
The relationship between specific and molar property is
where Mm is the molar mass (kg/mol).
Note
Nomenclature
We will refer to properties like volume as extensive, with the understanding that they have an intensive variant. The symbol V will be used for the intensive variant, whether molar or specific. The total volume occupied by n mole (or m mass) of material will be written as Vtot, nV, or mV. No separate notation will be used to distinguish molar from specific properties. This distinction will be made clear by the context of the calculation.
State of Pure Component
Experience teaches that if we fix temperature and pressure, all other intensive properties of a pure component (density, heat capacity, dielectric constant, etc.) are fixed. We express this by saying that the state of a pure substance is fully specified by temperature and pressure. For the molar volume V, for example, we write
which reads “V is a function of T and P.” The term state function will be used as a synonym for “thermodynamic property.” If eq. (1.5) is solved for temperature, we obtain an equation of the form
which reads “T is a function of P and V.” It is possible then to define the state using pressure and molar volume as the defining variables, since knowing pressure and volume allows us to calculate temperature. Because all properties are related to pressure and temperature, the state may be defined by any combination of two intensive variables, not necessarily T and P. Temperature and pressure are the preferred choice, as both variables are easy to measure and control in the laboratory and in an industrial setting. Nonetheless, we will occasionally consider different sets of variables, if this proves convenient.
Note
Fixing the State
If two intensive properties are known, the state of single-phase pure fluid is fixed, i.e., all other intensive properties have fixed values and can be obtained either from tables or by calculation.
State of Multicomponent Mixture
The state of a multicomponent mixture requires the specification of composition in addition to temperature and pressure. Mixtures will be introduced in Chapter 8. Until then the focus will be on single components.
Process and Path
The thermodynamic plane of pure substance is represented by two axes, T and P. A point on this plane represents a state, its coordinates corresponding to the temperature and pressure of the system. The typical problem in thermodynamics involves a system undergoing a change of state: a fixed amount of material at temperature TA and pressure PA is subjected to heating/cooling, compression/expansion, or other treatments to final state (TC, PC). A change of state is called a process. On the thermodynamic plane, a process is depicted by a path, namely, a line of successive states that connect the initial and final state (see Figure 1-6). Conversely, any line on this plane represents a process that can be realized experimentally. Two processes that are represented by simple paths on the TP plane are the constant-pressure (or isobaric) process, and the constant-temperature (or isothermal) process. The constant-pressure process is a straight line drawn at constant pressure (path AB in Figure 1-6); the constant-temperature process is drawn at constant temperature (path BC). Any two points on the TP plane can be connected using a sequence of isothermal and isobaric paths.
Figure 1-6: Illustration of two different paths (ABC, AB′C) between the same initial (A) and final (C) states. Paths can be visualized as processes (heating/cooling, compression/expansion) that take place inside a cylinder fitted with a piston.
Processes such as the constant-pressure, constant-temperature, and constant-volume process are called elementary. These are represented by simple paths during which one state variable (pressure, temperature, volume) is held constant. They are also simple to conduct experimentally. One way to do this is using a cylinder fitted with a piston. By fitting the piston with enough weights we can exert any pressure on the contents of the cylinder, and by making the piston movable we allow changes of volume due to heating/cooling to take place while keeping the pressure inside the cylinder constant. To conduct an isothermal process we employ the notion of a heat bath, or heat reservoir. Normally, when a hot system is used to supply heat to a colder one, its temperature drops as a result of the transfer of heat. If we imagine the size of the hot system to approach infinity, any finite transfer of heat to (or from) another system represents an infinitesimal change for the large system and does not change its temperature by any appreciable amount. The ambient air is a practical example of a heat bath with respect to small exchanges of heat. A campfire, for example, though locally hot, has negligible effect on the temperature of the air above the campsite. The rising sun, on the other hand, changes the air temperature appreciably. Therefore, the notion of an “infinite” bath must be understood as relative to the amount of heat that is exchanged. A constant-temperature process may be conducted by placing the system into contact with a heat bath. Additionally, the process must be conducted in small steps to allow for continuous thermal equilibration. The constant-volume process requires that the volume occupied by the system remain constant. This can be easily accomplished by confining an amount of substance in a rigid vessel that is completely filled. Finally, the adiabatic process may be conducted by placing thermal insulation around the system to prevent the exchange of heat.
We will employ cylinder-and-piston arrangement primarily as a mental device that allows us to visualize the mathematical abstraction of a path as a physical process that we could conduct in the laboratory.
Quasi-Static Process
At equilibrium, pressure and temperature are uniform throughout the system. This ensures a well-defined state in which, the system is characterized by a single temperature and single pressure, and represented by a single point on the PT plane. If we subject the system to a process, for example, heating by placing it into contact with a hot source, the system will be temporarily moved away from equilibrium and will develop a temperature gradient that induces the necessary transfer of heat. If the process involves compression or expansion, a pressure gradient develops that moves the system and its boundaries in the desired direction. During a process the system is not in equilibrium and the presence of gradients implies that its state cannot be characterized by a single temperature and pressure. This introduces an inconsistency in our depiction of processes as paths on the TP plane, since points on this plane represent equilibrium states of well-defined pressure and temperature. We resolve this difficulty by requiring the process to take place in a special way, such that the displacement of the system from equilibrium is infinitesimally small. A process conducted in such manner is called quasi static. Suppose we want to increase the temperature of the system from T1 to T2. Rather than contacting the system with a bath at temperature T2, we use a bath at temperature T1 + δT, where δT is a small number, and let the system equilibrate with the bath. This ensures that the temperature of the system is nearly uniform (Fig. 1-7). Once the system is equilibrated to temperature T1 + δT, we place it into thermal contact with another bath at temperature T1 + 2δT, and repeat the process until the final desired temperature is reached. Changes in pressure are conducted in the same manner. In general, in a quasi-static process we apply small changes at a time and wait between changes for the system to equilibrate. The name derives from the Latin quasi (“almost”) and implies that the process occurs as if the system remained at a stationary equilibrium state.
Figure 1-7: (a) Typical temperature gradient in heat transfer. (b) Heat transfer under small temperature difference. (c) Quasi-static idealization: temperatures in each system are nearly uniform and almost equal to each other.
Quasi Static is Reversible
A process that is conducted in quasi-static manner is essentially at equilibrium at every step along the way. This implies that the system can retrace its path if all inputs (temperature and pressure differences) reverse sign. For this reason, the quasi-static process is also a reversible process. If a process is conducted under large gradients of pressure and temperature, it is neither quasi static nor reversible. Here is an exaggerated example that demonstrates this fact. If an inflated balloon is punctured with a sharp needle, the air in the balloon will escape and expand to the conditions of the ambient air. This process is not quasi static because expansion occurs under a nonzero pressure difference between the air in the ballon and the air outside. It is not reversible either: we cannot bring the deflated balloon back to the inflated state by reversing the action that led to the expansion, i.e., by “de-puncturing” it. We can certainly restore the initial state by patching the balloon and blowing air into it, but this amounts to performing an entirely different process. The same is true in heat transfer. If two systems exchange heat under a finite (nonzero) temperature difference, as in Figure 1-7(a), reversing ΔT is not sufficient to cause heat to flow in the reverse direction because the temperature gradient inside system 1 continues to transfer heat in the original direction. For a certain period of time the left side of system 1 will continue to receive heat until the gradient adjusts to the new temperature of system 2. Only when a process is conducted reversibly is it possible to recover the initial state by exactly retracing the forward path in the reverse direction. The quasi-static way to expand the gas is to perform the process against an external pressure that resists the expansion and absorbs all of the work done by the expanding gas. To move in the forward direction, the external pressure would have to be slightly lower than that of the gas; to move in the reverse direction, it would have to be slightly higher. In this manner the process, whether expansion or compression, is reversible. The terms quasi static and reversible are equivalent but not synonymous. Quasi static refers to how the process is conducted (under infinitesimal gradients); reversible refers to the characteristic property that such process can retrace its path exactly. The two terms are equivalent in the sense that if we determine that a process is conducted in a quasi-static manner we may conclude that it is reversible, and vice versa. In practice, therefore, the two terms may be used interchangeably.
is the reduced saturation pressure of the fluid at reduced temperature Tr = 0.7. It is a characteristic property of the fluid and is found in tables, usually along with the critical properties of the fluid. It was introduced by Pitzer as a measure of the sphericity of the molecule. More generally, it should be understood as a combined measure of the shape and polarity. Symmetric nonpolar molecules, such as Ar, have ω = 0. These are called simple fluids and are found to obey the two-parameter correlation of corresponding states quite well. Nonspherical or polar molecules have a larger acentric factor. For most fluids the acentric factor is positive and in the range 0 to 0.4, although small negative values of ω are also possible.
, and the mean of the square of the velocity are all given in terms of temperature:
. The density of this arrangement is
