Showing That the Entropy Reaches a Maximum at Equilibrium in a Closed, Isolated System (In Sec. 4.1 we established that the entropy function will be a maximum at equilibrium in an isolated system. This is illustrated by example for the system shown here.) Figure 4.5-3 shows a well-insulated box of

Question

Showing That the Entropy Reaches a Maximum at Equilibrium in a Closed, Isolated System

(In Sec. 4.1 we established that the entropy function will be a maximum at equilibrium in an isolated system. This is illustrated by example for the system shown here.)

Figure 4.5-3 shows a well-insulated box of volume 6 [latex]m ^{3}[/latex] divided into two equal volumes. The left-hand cell is initially filled with air at [latex]100^{\circ}[/latex]C and 2 bar, and the right-hand cell is initially evacuated. The valve connecting the two cells will be opened so that gas will slowly pass from cell 1 to cell 2. The wall connecting the two cells conducts heat sufficiently well that the temperature of the gas in the two cells will always be the same. Plot on the same graph (1) the pressure in the second tank versus the pressure in the first tank, and (2) the change in the total entropy of the system versus the pressure in tank 1. At these temperatures and pressures, air can be considered to be an ideal gas of constant heat capacity.

Accepted Answer

For this system

Total mass [latex]=N=N_{1}+N_{2}[/latex]

Total energy [latex]=U=U_{1}+U_{2}[/latex]

Total entropy [latex]=S=S_{1}+S_{2}=N_{1} \underline{S}_{1}+N_{2} \underline{S}_{2}[/latex]

From the ideal gas equation of state and the fact that [latex]V=N \underline{V}[/latex] , we have

[latex]N_{1}^{i}=\frac{P V}{R T}=\frac{2 bar imes 3 m ^{3}}{8.314 imes 10^{-5} \frac{ bar m ^{3}}{ mol K } imes 373.15 K }=193.4 mol =0.1934 kmol[/latex]

Now since U = constant, [latex]T_{1}=T_{2}, ext { and for the ideal gas } \underline{U}[/latex] is a function of temperature only, we conclude that [latex]T_{1}=T_{2}=100^{\circ} C[/latex] at all times. This result greatly simplifies the computation. Suppose that the pressure in cell 1 is decreased from 2 bar to 1.9 bar by transferring some gas from cell 1 to cell 2. Since the temperature in cell 1 is constant, we have, from the ideal gas law,

[latex]N_{1}=0.95 N_{1}^{i}[/latex]

and by mass conservation, [latex]N_{2}=0.05 N_{1}^{i} ext {. Applying the ideal gas relation, we obtain } P_{2}=[/latex] 0.1 bar.
For any element of gas, we have, from Eq. 4.4-3,

[latex]\underline{S}\left(T_{2}, P_{2} ight)-\underline{S}\left(T_{1}, P_{1} ight)=C_{ P }^{*} \ln \left(\frac{T_{2}}{T_{1}} ight)-R \ln \left(\frac{P_{2}}{P_{1}} ight)[/latex] ( 4.4-3 )

[latex]\underline{S}^{f}-\underline{S}^{i}=C_{ P }^{*} \ln \frac{T^{f}}{T^{i}}-R \ln \frac{P^{f}}{P^{i}}=-R \ln \frac{P^{f}}{P^{i}}[/latex]

Figure 4.5-4 The system entropy change and the pressure in cell 2 as a function of the pressure in cell 1.

since temperature is constant. Therefore, to compute the change in entropy of the system, we visualize the process of transferring 0.05N1i moles of gas from cell 1 to cell 2 as having two effects:

1. To decrease the pressure of the [latex]0.95 N_{1}^{i}[/latex] moles of gas remaining in cell 1 from 2 bar to 1.9 ba

2. To decrease the pressure of the [latex]0.95 N_{1}^{i}[/latex] moles of gas that have been transferred from cell 1 to cell 2 from 2 bar to 0.1 bar

Thus

[latex]S^{f}-S^{i}=-0.95 N_{1}^{i} R \ln (1.9 / 2)-0.05 N_{1}^{i} R \ln (0.1 / 2)[/latex]

or

[latex]\begin{aligned}\frac{\Delta S}{N_{1}^{i} R} &=-0.95 \ln 0.95-0.05 \ln 0.05 \&=0.199\end{aligned} \quad\left\{\begin{array}{l}P_{1}=1.9 bar \P_{2}=0.1 bar\end{array} ight.[/latex]

Similarly, if [latex]P_{1}[/latex] = 1.8 bar,

[latex]\begin{aligned}\frac{\Delta S}{N_{1}^{i} R} &=-0.9 \ln 0.9-0.1 \ln 0.1 \&=0.325\end{aligned}\left\{\begin{array}{l}P_{1}=1.8 bar \P_{2}=0.2 bar\end{array} ight.[/latex]

and so forth. The results are plotted in Fig. 4.5-4.
From this figure it is clear that ΔS, the change in entropy from the initial state, and therefore the total entropy of the system, reaches a maximum value when [latex]P_{1}=P_{2}=1[/latex] bar. Consequently, the equilibrium state of the system under consideration is the state in which the pressure in both cells is the same, as one would expect. (Since the use of the entropy function leads to a solution that agrees with one’s intuition, this example should reinforce confidence in the use of the entropy function as a criterion for equilibrium in an isolated constant-volume system.)

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