Two gases separated by an internal partition

Question 1. The diagram shows a rigid vessel of total volume $2V$ divided by a partition. On the left: $(n_1,T_1,V,P_0)$; on the right: $(3n_1,T_1/2,V,P_{2,i})$. The outer walls are adiabatic. With a fixed, impermeable, adiabatic internal partition, no exchange is allowed: each gas retains its state even if the temperatures and pressures differ. The partition's support sustains the pressure difference.

Question 2 — Algebraic calculation. The ideal-gas equation gives

$$P_{1,i}=\frac{n_1RT_1}{V}=P_0, \qquad P_{2,i}=\frac{3n_1R(T_1/2)}{V}=\frac32P_0.$$

Question 3 — Fixed diathermal partition. During a small energy transfer with $\mathrm{d}U_2=-\mathrm{d}U_1$, the volumes remain constant: for each gas, $\mathrm dS_j=\mathrm dU_j/T_j$. Adding gives

$$\mathrm dS_{\mathrm{tot}}=\frac{\mathrm dU_1}{T_1} +\frac{\mathrm dU_2}{T_2} =\frac{\mathrm dU_1}{T_1}-\frac{\mathrm dU_1}{T_2}.$$

The temperatures in this variation refer to the state being considered. Factoring gives

$$\mathrm{d}S_{\mathrm{tot}} =\left(\frac1{T_1}-\frac1{T_2}\right)\mathrm{d}U_1.$$

At equilibrium, entropy is maximal; the coefficient of $\mathrm{d}U_1$ must therefore vanish, requiring $T_{1,f}=T_{2,f}=T_f$. Conservation of energy gives

$$n_1\frac32R(T_f-T_1) +3n_1\frac32R\left(T_f-\frac{T_1}{2}\right)=0,$$

Dividing the energy balance by $(3/2)n_1R$ gives

$$(T_f-T_1)+3\left(T_f-\frac{T_1}{2}\right)=0, \quad 4T_f-\frac52T_1=0, \quad T_f=\frac58T_1.$$

Then $P_{1,f}=n_1RT_f/V=(T_f/T_1)P_0$ and $P_{2,f}=3n_1RT_f/V=3(T_f/T_1)P_0$, so

$$T_f=\frac58T_1, \qquad P_{1,f}=\frac58P_0, \qquad P_{2,f}=\frac{15}{8}P_0.$$

Because the partition is fixed, mechanical equilibrium between the gases is not required: their final pressures may differ.

Question 4 — Entropy balance. Gas 2 receives energy entirely as heat,

$$Q_{1\to2}=\Delta U_2 =3n_1\frac32R\left(\frac58T_1-\frac12T_1\right) =\frac9{16}n_1RT_1.$$

Gas 1 receives the opposite heat: $Q_1=-Q_{1\to2}<0$, whereas $Q_2=Q_{1\to2}>0$. The partition is fixed, so $W_1=W_2=0$. The combined system is isolated, so $S_{\mathrm{e}}=0$ and $S_{\mathrm{i}}=\Delta S_{\mathrm{tot}}$. Since the volumes are constant,

$$\Delta S_1=\frac32n_1R\ln\frac{T_f}{T_1} =\frac32n_1R\ln\frac58, \Delta S_2=\frac32(3n_1)R\ln\frac{T_f}{T_1/2} =\frac92n_1R\ln\frac54.$$

Gas 1 loses entropy and gas 2 gains it. Their sum is

$$S_{\mathrm{i}} =\frac32n_1R\left[ \ln\!\left(\frac58\right) +3\ln\!\left(\frac54\right) \right]>0.$$

This entropy production reflects spontaneous heat transfer between gases initially at different temperatures. To verify the sign without numerical values, combine the logarithms:

$$\ln\frac58+3\ln\frac54 =\ln\left[\frac58\left(\frac54\right)^3\right] =\ln\frac{625}{512}>0.$$

Question 5 — Movable diathermal partition. With $\mathrm{d}U_2=-\mathrm{d}U_1$ and $\mathrm{d}V_2=-\mathrm{d}V_1$,

$$\mathrm{d}S_{\mathrm{tot}} =\left(\frac1{T_1}-\frac1{T_2}\right)\mathrm{d}U_1 +\left(\frac{P_1}{T_1}-\frac{P_2}{T_2}\right)\mathrm{d}V_1.$$

The partition independently allows energy and volume to be redistributed. Stationarity must therefore hold for every $\mathrm dU_1$ and every $\mathrm dV_1$: both coefficients vanish. The first requires $T_{1,f}=T_{2,f}$; the second then gives $P_{1,f}/T_f=P_{2,f}/T_f$, or $P_{1,f}=P_{2,f}$. Conservation of energy again gives $T_f=5T_1/8$. At common pressure and temperature, volumes are proportional to amounts of substance. Since their sum is $2V$,

$$\frac{V_{2,f}}{V_{1,f}}=\frac{3n_1RT_f/P_f}{n_1RT_f/P_f}=3, \qquad V_{1,f}+3V_{1,f}=2V.$$

Thus $V_{1,f}=V/2$ and $V_{2,f}=3V/2$.

The common pressure is $P_f=n_1R(5T_1/8)/(V/2)=(5/4)P_0$. Hence

$$V_{1,f}=\frac V2, \qquad V_{2,f}=\frac{3V}{2}, \qquad P_f=\frac54P_0.$$

The internal-energy changes are the same as for the fixed partition, but energy is now exchanged as both heat and work. The entropy balance contains the same temperature terms as in question 4, with additional volume terms:

$$\Delta S_1=\frac32n_1R\ln\frac58+n_1R\ln\frac{V/2}{V}, \Delta S_2=\frac92n_1R\ln\frac54+3n_1R\ln\frac{3V/2}{V}.$$

Adding gives

$$S_{\mathrm{i}}=\Delta S_{\mathrm{tot}} =\frac32n_1R\left[ \ln\!\left(\frac58\right)+3\ln\!\left(\frac54\right) \right] +n_1R\left[ \ln\!\left(\frac12\right)+3\ln\!\left(\frac32\right) \right]>0.$$

The increase compared with the fixed-partition case is

$$n_1R\left[\ln\frac12+3\ln\frac32\right] =n_1R\ln\left[\frac12\left(\frac32\right)^3\right] =n_1R\ln\frac{27}{16}>0.$$

Allowing the partition to move therefore increases the final entropy accessible to the system.