Let’s approach this definition of change in entropy by looking again at a process that we described in Sections 18-11 and 19-11: the free expansion of an ideal gas. Figure 20-1a shows the gas in its initial equilibrium state i, confined by a closed stopcock to the left half of a thermally insulated container. If we open the stopcock, the gas rushes to fill the entire container, eventually reaching the final equilibrium state f shown in Fig. 20-1b. This is an irreversible process; all the molecules of the gas will never return to the left half of the container.
The p–V plot of the process, in Fig. 20-2, shows the pressure and volume of the gas in its initial state i and final state f. Pressure and volume are state properties, properties that depend only on the state of the gas and not on how it reached that state. Other state properties are temperature and energy. We now assume that the gas has still another state property—its entropy. Furthermore, we define the change in entropy Sf − Si of a system during a process that takes the system from an initial state i to a final state f as
Here Q is the energy transferred as heat to or from the system during the process, and T is the temperature of the system in kelvins. Thus, an entropy change depends not only on the energy transferred as heat but also on the temperature at which the transfer takes place. Because T is always positive, the sign of ΔS is the same as that of Q. We see from Eq. 20-1 that the SI unit for entropy and entropy change is the joule per kelvin.
There is a problem, however, in applying Eq. 20-1 to the free expansion of Fig. 20-1. As the gas rushes to fill the entire container, the pressure, temperature, and volume of the gas fluctuate unpredictably. In other words, they do not have a sequence of well-defined equilibrium values during the intermediate stages of the change from initial equilibrium state i to final equilibrium state f. Thus, we cannot trace a pressure–volume path for the free expansion on the p–V plot of Fig. 20-2 and, more important, we cannot find a relation between Q and T that allows us to integrate as Eq. 20-1 requires.
However, if entropy is truly a state property, the difference in entropy between states i and f must depend only on those states and not at all on the way the system went from one state to the other. Suppose, then, that we replace the irreversible free expansion of Fig. 20-1 with a reversible process that connects states i and f. With a reversible process we can trace a pressure–volume path on a p–V plot, and we can find a relation between Q and T that allows us to use Eq. 20-1 to obtain the entropy change.
We saw in Section 19-11 that the temperature of an ideal gas does not change during a free expansion: Ti = Tf = T. Thus, points i and f in Fig. 20-2 must be on the same isotherm. A convenient replacement process is then a reversible isothermal expansion from state i to state f, which actually proceeds along that isotherm. Furthermore, because T is constant throughout a reversible isothermal expansion, the integral of Eq. 20-1 is greatly simplified.
Figure 20-3 shows how to produce such a reversible isothermal expansion. We confine the gas to an insulated cylinder that rests on a thermal reservoir maintained at the temperature T. We begin by placing just enough lead shot on the movable piston so that the pressure and volume of the gas are those of the initial state i of Fig. 20-1a. We then remove shot slowly (piece by piece) until the pressure and volume of the gas are those of the final state f of Fig. 20-1b. The temperature of the gas does not change because the gas remains in thermal contact with the reservoir throughout the process.
Fig. 20-2 A p–V diagram showing the initial state i and the final state f of the free expansion of Fig. 20-1. The intermediate states of the gas cannot be shown because they are not equilibrium states.
The reversible isothermal expansion of Fig. 20-3 is physically quite different from the irreversible free expansion of Fig. 20-1. However, both processes have the same initial state and the same final state and thus must have the same change in entropy. Because we removed the lead shot slowly, the intermediate states of the gas are equilibrium states, so we can plot them on a p–V diagram (Fig. 20-4).
To apply Eq. 20-1 to the isothermal expansion, we take the constant temperature T outside the integral, obtaining
Because ∫ dQ = Q, where Q is the total energy transferred as heat during the process, we have
To keep the temperature T of the gas constant during the isothermal expansion of Fig. 20-3, heat Q must have been energy transferred from the reservoir to the gas. Thus, Q is positive and the entropy of the gas increases during the isothermal process and during the free expansion of Fig. 20-1.
To summarize:
To find the entropy change for an irreversible process occurring in a closed system, replace that process with any reversible process that connects the same initial and final states. Calculate the entropy change for this reversible process with Eq. 20-1.
When the temperature change ΔT of a system is small relative to the temperature (in kelvins) before and after the process, the entropy change can be approximated as
where Tavg is the average temperature of the system in kelvins during the process.
CHECK POINT 1 Water is heated on a stove. Rank the entropy changes of the water as its temperature rises (a) from 20°C to 30°C, (b) from 30°C to 35°C, and (c) from 80°C to 85°C, greatest first.
Fig. 20-3 The isothermal expansion of an ideal gas, done in a reversible way. The gas has the same initial state i and same final state f as in the irreversible process of Figs. 20-1 and 20-2.
Fig. 20-4 A p–V diagram for the reversible isothermal expansion of Fig. 20-3. The intermediate states, which are now equilibrium states, are shown.
Suppose 1.0 mol of nitrogen gas is confined to the left side of the container of Fig. 20-1a. You open the stopcock, and the volume of the gas doubles. What is the entropy change of the gas for this irreversible process? Treat the gas as ideal.
Solution: We need two Key Ideas here. One is that we can determine the entropy change for the irreversible process by calculating it for a reversible process that provides the same change in volume. The other is that the temperature of the gas does not change in the free expansion. Thus, the reversible process should be an isothermal expansion—namely, the one of Figs. 20-3 and 20-4.
From Table 19-4, the energy Q added as heat to the gas as it expands isothermally at temperature T from an initial volume Vi to a final volume Vf is
in which n is the number of moles of gas present. From Eq. 20-2 the entropy change for this reversible process is
Substituting n = 1.00 mol and Vf/Vi = 2, we find
Thus, the entropy change for the free expansion (and for all other processes that connect the initial and final states shown in Fig. 20-2) is
Because ΔS is positive, the entropy increases, in accordance with the entropy postulate of Section 20-2.
CHECK POINT 2 An ideal gas has temperature T1 at the initial state i shown in the p–V diagram here. The gas has a higher temperature T2 at final states a and b, which it can reach along the paths shown. Is the entropy change along the path to state a larger than, smaller than, or the same as that along the path to state b?
Figure 20-5a shows two identical copper blocks of mass m = 1.5 kg: block L at temperature TiL = 60°C and block R at temperature TiR = 20°C. The blocks are in a thermally insulated box and are separated by an insulating shutter. When we lift the shutter, the blocks eventually come to the equilibrium temperature Tf = 40°C (Fig. 20-5b). What is the net entropy change of the two-block system during this irreversible process? The specific heat of copper is 386 J/kg · K.
Solution: The Key Idea here is that to calculate the entropy change, we must find a reversible process that takes the system from the initial state of Fig. 20-5a to the final state of Fig. 20-5b. We can calculate the net entropy change ΔSrev of the reversible process using Eq. 20-1, and then the entropy change for the irreversible process is equal to ΔSrev. For such a reversible process we need a thermal reservoir whose temperature can be changed slowly (say, by turning a knob). We then take the blocks through the following two steps, illustrated in Fig. 20-6.
Step 1 With the reservoir’s temperature set at 60°C, put block L on the reservoir. (Since block and reservoir are at the same temperature, they are already in thermal equilibrium.) Then slowly lower the temperature of the reservoir and the block to 40°C. As the block’s temperature changes by each increment dT during this process, energy dQ is transferred as heat from the block to the reservoir. Using Eq. 18-14, we can write this transferred energy as dQ = mc dT, where c is the specific heat of copper. According to Eq. 20-1, the entropy change ΔSL of block L during the full temperature change from initial temperature TiL (= 60°C = 333 K) to final temperature Tf (= 40°C = 313 K) is
Fig. 20-5 (a) In the initial state, two copper blocks L and R, identical except for their temperatures, are in an insulating box and are separated by an insulating shutter. (b) When the shutter is removed, the blocks exchange energy as heat and come to a final state, both with the same temperature Tf.
Fig. 20-6 The blocks of Fig. 20-5 can proceed from their initial state to their final state in a reversible way if we use a reservoir with a controllable temperature (a) to extract heat reversibly from block L and (b) to add heat reversibly to block R.
Inserting the given data yields
Step 2 With the reservoir’s temperature now set at 20°C, put block R on the reservoir. Then slowly raise the temperature of the reservoir and the block to 40°C. With the same reasoning used to find ΔSL, you can show that the entropy change ΔSR of block R during this process is
The net entropy change ΔSrev of the two-block system undergoing this two-step reversible process is then
Thus, the net entropy change ΔSirrev for the two-block system undergoing the actual irreversible process is
This result is positive, in accordance with the entropy postulate of Section 20-2.
The pericarp part of a popcorn kernel is a small container of liquid water. When the popcorn is heated to about 180°C, the pericarp walls burst, and the liquid vaporizes so rapidly that the vapor explodes the rest of the kernel to many times its original volume. The explosion produces the audible pop of the popcorn. If the water in a pericarp has a mass of about 4 mg, what is the change in entropy of the water due to the vaporization and expansion?
Solution: The vaporization and expansion of the water in the pericarp constitute an irreversible process in which values for temperature and pressure vary in an unknown manner. A Key Idea here is that we can calculate the change ΔS in the water’s entropy in the irreversible process by calculating ΔS for two reversible processes: (1) vaporization at 180°C and (2) adiabatic expansion of the vapor. (We assume the expansion is adiabatic because it takes place so rapidly that the vapor cannot exchange energy as heat with its environment.)
Process 1: Vaporization. Because we assume the vaporization occurs reversibly at a constant temperature, we can calculate the change in entropy from Eq. 20-2 (ΔS = Q/T). Substituting for Q with Eq. 18-16 (Q = Lm), using the latent heat of vaporization LV = 2256 kJ/kg from Eq. 18-17, and converting 180°C to 453 K, we find
Process 2: Adiabatic expansion. Because we assume the expansion does not allow any exchange of energy as heat Q between the water vapor and its environment, Q is zero. From Eq. 20-1, we see that change in the vapor’s entropy is also zero. Thus, ΔS2 = 0.
The total change in entropy for the vaporization and expansion is
Each audible pop in popping popcorn signals an increase in the water’s entropy of about 0.02 J/K.
Entropy as a State Function
We have assumed that entropy, like pressure, energy, and temperature, is a property of the state of a system and is independent of how that state is reached. That entropy is indeed a state function (as state properties are usually called) can be deduced only by experiment. However, we can prove it is a state function for the special and important case in which an ideal gas is taken through a reversible process.
To make the process reversible, it is done slowly in a series of small steps, with the gas in an equilibrium state at the end of each step. For each small step, the energy transferred as heat to or from the gas is dQ, the work done by the gas is dW, and the change in internal energy is dEint. These are related by the first law of thermodynamics in differential form (Eq. 18-27):
Because the steps are reversible, with the gas in equilibrium states, we can use Eq. 18-24 to replace dW with p dV and Eq. 19-45 to replace dEint with nCV dT. Solving for dQ then leads to
Using the ideal gas law, we replace p in this equation with nRT/V. Then we divide each term in the resulting equation by T, obtaining
Now let us integrate each term of this equation between an arbitrary initial state i and an arbitrary final state f to get
The quantity on the left is the entropy change ΔS (= Sf − Si) defined by Eq. 20-1. Substituting this and integrating the quantities on the right yield
Note that we did not have to specify a particular reversible process when we integrated. Therefore, the integration must hold for all reversible processes that take the gas from state i to state f. Thus, the change in entropy ΔS between the initial and final states of an ideal gas depends only on properties of the initial state (Vi and Ti) and properties of the final state (Vf and Tf); ΔS does not depend on how the gas changes between the two states.
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