Adiabatic piston: mechanical equilibrium without thermal equilibrium

Question 1. The diagram shows the resistor in compartment 1, the insulated piston and electrical wires crossing the outer boundary. Gas 2 receives mechanical work through the piston and no heat. For “two gases + resistor”, no external heat is exchanged and the outer walls are rigid: only electrical work enters. Work exchanged across the piston is internal to this combined system.

Question 2. Gas 2 undergoes reversible adiabatic compression, so $P_0V_0^\gamma=P_fV_{2,f}^\gamma$. With $P_f=xP_0$,

$$\left(\frac{V_{2,f}}{V_0}\right)^\gamma=\frac1x, \qquad V_{2,f}=x^{-1/\gamma}V_0.$$

Set $a=x^{-1/\gamma}$. The equation of state at the initial and final states gives

$$\frac{T_{2,f}}{T_0}=\frac{P_fV_{2,f}}{P_0V_0}=xa, \qquad T_{2,f}=xaT_0.$$

The total volume is $2V_0$ and the piston imposes the same pressure in compartments 1 and 2:

$$V_{1,f}=2V_0-V_{2,f}=(2-a)V_0, \qquad T_{1,f}=T_0\frac{P_fV_{1,f}}{P_0V_0}=x(2-a)T_0.$$

Since $x>1$ and $\gamma>1$, $0<a<1$, hence $2-a>a$ and $T_{1,f}>T_{2,f}$. The partition allows work but no direct heat transfer. After the current is switched off, this state is a mechanical equilibrium subject to the adiabatic constraint in the adopted macroscopic model.

Question 3. For each gas, $\Delta U=nc_V(T_f-T_0)$ and $nc_VT_0=nRT_0/(\gamma-1)=P_0V_0/(\gamma-1)$. Substituting the calculated temperatures gives

$$\Delta U_1=\frac{P_0V_0}{\gamma-1}[x(2-a)-1],\qquad \Delta U_2=\frac{P_0V_0}{\gamma-1}(xa-1).$$

The energy stored in the resistor is negligible. The first law applied to the combined system therefore gives

$$\begin{aligned} W_{\mathrm{elec}}&=\Delta U_1+\Delta U_2\\ &=\frac{P_0V_0}{\gamma-1}[x(2-a)-1+xa-1]\\ &=\frac{P_0V_0}{\gamma-1}(2x-2) =\frac{2P_0V_0}{\gamma-1}(x-1). \end{aligned}$$

For gas 2, $Q_2=0$, so the work done by the piston is $W_2=\Delta U_2$. Since the piston stores no energy, its work on gas 1 is $-W_2$. These two work contributions cancel in the combined system's balance.

Question 4. For gas 2, the entropy formula gives

$$\begin{aligned} \Delta S_2&=nc_V\ln(xa)+nR\ln a\\ &=nc_V[\ln x+\ln a+(\gamma-1)\ln a]\\ &=nc_V\ln(xa^\gamma)=0, \end{aligned}$$

because $a^\gamma=1/x$. This recovers the isentropic nature of reversible compression. For gas 1,

$$\begin{aligned} \Delta S_1&=nc_V\ln[x(2-a)]+nR\ln(2-a)\\ &=nc_V[\ln x+\gamma\ln(2-a)] =nc_V\ln[x(2-a)^\gamma]. \end{aligned}$$

We have $x>1$ and $2-a>1$, so the logarithm's argument exceeds 1: $\Delta S_1>0$. The combined system receives no entropy from outside: electrical work carries none. Thus $S_{\mathrm{i}}=\Delta S_1+\Delta S_2=\Delta S_1$. Joule heating produces entropy even when heating is slow. The adiabatic relation therefore does not apply to gas 1.

Question 5. Once the current is switched off, the combined system is isolated. The now diathermal, movable partition allows final temperatures and pressures to become equal. Since both gases contain the same $n$, $P_*V_{1,*}=nRT_*=P_*V_{2,*}$ gives $V_{1,*}=V_{2,*}=V_0$. Conservation of energy requires

$$nc_V(T_*-T_{1,f})+nc_V(T_*-T_{2,f})=0.$$

Dividing by $nc_V$ and substituting the temperatures gives

$$2T_*=T_{1,f}+T_{2,f}=xT_0[(2-a)+a]=2xT_0,$$

hence $T_*=xT_0$ and $P_*=nRxT_0/V_0=xP_0$. For the entropy changes during this new process,

$$\begin{aligned} \Delta S_{1,\mathrm{add}} &=nc_V\ln\frac{xT_0}{x(2-a)T_0}+nR\ln\frac{V_0}{(2-a)V_0} =-nc_P\ln(2-a),\\ \Delta S_{2,\mathrm{add}} &=nc_V\ln\frac{xT_0}{xaT_0}+nR\ln\frac{V_0}{aV_0} =-nc_P\ln a. \end{aligned}$$

Their sum is the additional entropy produced, since the combined system is isolated:

$$S_{\mathrm{i},\mathrm{add}}=-nc_P[\ln(2-a)+\ln a]=-nc_P\ln[a(2-a)].$$

For $0<a<1$, $0<a(2-a)=1-(1-a)^2<1$, so its logarithm is negative. The minus sign makes $S_{\mathrm{i},\mathrm{add}}>0$.