Semipermeable membrane: which quantities become equal?

Question 1. The diagram distinguishes the two species and shows that only A can cross. Set $P_*=N_0k_BT_0/V_0$. Each compartment initially contains $3N_0$ particles:

$$P_{1,i}=\frac{(2N_0+N_0)k_BT_0}{V_0}=3P_*,\qquad P_{2,i}=\frac{(N_0+2N_0)k_BT_0}{V_0}=3P_*.$$

For A's partial pressure, count only A particles: $2N_0$ on the left and $N_0$ on the right. Initially $P_{1,i}=P_{2,i}=3P_*$, but $p_{A,1,i}=2P_*$ and $p_{A,2,i}=P_*$. Thus $\mu_{A,1}-\mu_{A,2}=k_BT_0\ln2>0$: A initially moves from left to right. For $\mathrm dN_{A,1}<0$ at equal temperatures, the total entropy change $(\mu_{A,2}-\mu_{A,1})\,\mathrm dN_{A,1}/T_0$ is positive. Equal total pressures do not suffice to stop diffusion. In the chemical-potential difference, the two $\mu_A^\circ(T_0)$ terms cancel; combining the logarithms gives $k_BT_0\ln(p_{A,1,i}/p_{A,2,i})=k_BT_0\ln2$.

Question 2. Total energy and the total number of A particles are conserved; the numbers of B particles on each side are constant. Using $\mathrm dU_2=-\mathrm dU_1$, $\mathrm dN_{A,2}=-\mathrm dN_{A,1}$ and $\mathrm dN_{B,1}=\mathrm dN_{B,2}=0$ gives

$$\mathrm dS_j=\frac{\mathrm dU_j}{T_j} -\frac{\mu_{A,j}}{T_j}\,\mathrm dN_{A,j} \quad (j=1,2).$$

The volume term vanishes because the membrane is fixed. Adding gives

$$\mathrm dS_{\mathrm{tot}} =\frac{\mathrm dU_1}{T_1}-\frac{\mathrm dU_1}{T_2} -\frac{\mu_{A,1}}{T_1}\mathrm dN_{A,1} +\frac{\mu_{A,2}}{T_2}\mathrm dN_{A,1}.$$

Collecting the coefficients gives

$$\mathrm dS_{\mathrm{tot}}= \left(\frac1{T_1}-\frac1{T_2}\right)\mathrm dU_1 +\left(\frac{\mu_{A,2}}{T_2}-\frac{\mu_{A,1}}{T_1}\right) \mathrm dN_{A,1}.$$

Heat conduction and A transfer are allowed: their variations can be tested independently. Stationarity therefore requires $1/T_1-1/T_2=0$, or $T_1=T_2$, followed by $(\mu_{A,2}-\mu_{A,1})/T=0$, or $\mu_{A,1}=\mu_{A,2}$. It requires neither $P_1=P_2$ (the membrane is fixed) nor $\mu_{B,1}=\mu_{B,2}$ (B cannot cross).

Question 3. Initially the total energy is $9N_0k_BT_0$ and the total number of particles remains $6N_0$. Indeed, $U_i=(3/2)(3N_0+3N_0)k_BT_0=9N_0k_BT_0$. At a common final temperature $T$, $U_f=(3/2)(6N_0)k_BT=9N_0k_BT$. Equality $U_f=U_i$ therefore gives $T=T_0$. Equality of A's chemical potentials becomes

$$k_BT_0\ln\frac{p_{A,1}}{p^\circ} =k_BT_0\ln\frac{p_{A,2}}{p^\circ}.$$

Dividing by $k_BT_0$ and exponentiating gives $p_{A,1}=p_{A,2}$. Since the volumes are equal, the factors $k_BT_0/V_0$ cancel: $N_{A,1}=N_{A,2}$. Since $N_{A,1}+N_{A,2}=3N_0$, each equals $3N_0/2$. Adding the B particles gives

$$P_{1,f}=\frac{(3N_0/2+N_0)k_BT_0}{V_0},\qquad P_{2,f}=\frac{(3N_0/2+2N_0)k_BT_0}{V_0}.$$

Thus,

$$P_{1,f}=\frac52P_*,\qquad P_{2,f}=\frac72P_*.$$

The membrane's support balances this pressure difference. Setting $a=N_{A,1}$ and $N_{A,2}=3N_0-a$, at fixed temperature and volumes, question 2 gives

$$\frac{\mathrm dS_{\mathrm{tot}}}{\mathrm da} =\frac{\mu_{A,2}-\mu_{A,1}}{T_0} =k_{\mathrm{B}}\ln\frac{p_{A,2}}{p_{A,1}} =k_{\mathrm{B}}\ln\frac{3N_0-a}{a}.$$

The energy term cancels because the temperatures are equal and $\mathrm dU_1+\mathrm dU_2=0$. To differentiate again, write $\ln(3N_0-a)-\ln a$: its derivative is $-1/(3N_0-a)-1/a$. Thus,

$$\frac{\mathrm dS_{\mathrm{tot}}}{\mathrm da} =k_{\mathrm{B}}\ln\frac{3N_0-a}{a},\qquad \frac{\mathrm d^2S_{\mathrm{tot}}}{\mathrm da^2} =-k_{\mathrm{B}}\left(\frac1a+\frac1{3N_0-a}\right)<0.$$

The equal distribution is the unique maximum on $0<a<3N_0$.

Question 4. Stationarity with respect to volume adds $P_1=P_2$: the additional term in $\mathrm dS_{\mathrm{tot}}$ is $(P_1/T_1-P_2/T_2)\,\mathrm dV_1$. Its coefficient must vanish; with $T_1=T_2$, this requires equal total pressures. The final temperature is again $T_0$, from the same energy balance as in question 3. Since $p_{A,1}=p_{A,2}$, equal total pressures imply $p_{B,1}=p_{B,2}$. B cannot cross:

$$\frac{N_0}{V_1}=\frac{2N_0}{V_2},\qquad V_1+V_2=2V_0,$$

so $V_2=2V_1$, then $3V_1=2V_0$: $V_1=2V_0/3$ and $V_2=4V_0/3$. For A, equal partial pressures give $N_{A,1}/V_1=N_{A,2}/V_2$, hence $N_{A,2}=2N_{A,1}$. Since the sum is $3N_0$, we obtain $N_{A,1}=N_0$ and $N_{A,2}=2N_0$. Finally, $P_f=(N_0+N_0)k_BT_0/(2V_0/3)=3P_*$. The results are therefore

$$V_1=\frac23V_0,\quad V_2=\frac43V_0,\quad N_{A,1}=N_0,\quad N_{A,2}=2N_0,\quad P_f=3P_*.$$

Equal B partial pressures follow here from the other conditions and the ideal-mixture model, rather than from an allowed exchange of B. The amounts of A are no longer equal: their concentrations, and hence their partial pressures, become equal.

Question 5. Convert the volumes to SI units to calculate pressures: $V_0=10\ \mathrm L=10^{-2}\ \mathrm{m^3}$. The constant $\mathcal N_A$ converts an amount of substance to a particle number: $N_0=\mathcal N_A n_0$ with $n_0=1{,}0\ \mathrm{mol}$. First calculate

$$R=\mathcal N_A k_{\mathrm{B}} =(6{,}022\times10^{23})(1{,}380649\times10^{-23}) \simeq8{,}314\ \mathrm{J\,mol^{-1}\,K^{-1}}.$$

With $N_0k_B=n_0R$, the pressure scale introduced in question 1 is

$$P_* =\frac{N_0k_BT_0}{V_0} =\frac{n_0RT_0}{V_0} \simeq\frac{1{,}0\times8{,}314\times300}{10^{-2}} =249420\ \mathrm{Pa}\simeq2{,}494\ \mathrm{bar},$$

because $1\ \mathrm{bar}=10^5\ \mathrm{Pa}$. This pressure is a calculation scale; it is not itself a compartment's total pressure.

Fixed membrane. Each volume remains $10\ \mathrm L$. Question 3 gives $N_{A,1}=N_{A,2}=3N_0/2$. To convert particles to moles, divide by $\mathcal N_A$:

$$n_{A,1}=n_{A,2}=\frac{3N_0}{2\mathcal N_A} =\frac32\times1\ \mathrm{mol}=1{,}5\ \mathrm{mol}.$$

B cannot cross the membrane: $1\ \mathrm{mol}$ of B remains on the left and $2\ \mathrm{mol}$ on the right. The total amounts are therefore $2{,}5\ \mathrm{mol}$ and $3{,}5\ \mathrm{mol}$. At $T_0=300\ \mathrm K$, the equation $P=nRT/V$ gives separately

$$P_{1,f}=\frac{(1{,}5+1)\times8{,}314\times300}{10^{-2}} =623550\ \mathrm{Pa}\simeq6{,}24\ \mathrm{bar}, P_{2,f}=\frac{(1{,}5+2)\times8{,}314\times300}{10^{-2}} =872970\ \mathrm{Pa}\simeq8{,}73\ \mathrm{bar}.$$

The total pressures differ; fixing the membrane maintains this state. A's partial pressures are equal because A occupies equal volumes with equal amounts and temperatures.

Movable membrane. From question 4,

$$V_{1,f}=\frac23V_0=\frac23\times10\simeq6{,}67\ \mathrm L, \qquad V_{2,f}=\frac43V_0=\frac43\times10\simeq13{,}33\ \mathrm L.$$

The sum remains $20\ \mathrm L$, the vessel's volume. The A populations are now $N_{A,1}=N_0$ and $N_{A,2}=2N_0$:

$$n_{A,1}=\frac{N_0}{\mathcal N_A}=1\ \mathrm{mol},\qquad n_{A,2}=\frac{2N_0}{\mathcal N_A}=2\ \mathrm{mol}.$$

The total is indeed $3\ \mathrm{mol}$ of A. Adding B, the compartments contain $2$ and $4\ \mathrm{mol}$ of gas respectively. Use the exact volumes before rounding:

$$P_{1,f}=\frac{(1+1)\times8{,}314\times300}{(2/3)\times10^{-2}} =748260\ \mathrm{Pa}\simeq7{,}48\ \mathrm{bar}, P_{2,f}=\frac{(2+2)\times8{,}314\times300}{(4/3)\times10^{-2}} =748260\ \mathrm{Pa}\simeq7{,}48\ \mathrm{bar}.$$

Both calculations give the same total pressure, as required for mechanical equilibrium of the released membrane.