Two gases separated by an impermeable but movable partition are allowed to freely exchange
energy. At equilibrium, the two sides will have the same
Options:
(A) pressure and temperature
(B) volume and temperature
(C) pressure and volume
(D) volume and energy
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2013 Q.6)**
**1. Thermodynamic Equilibrium Conditions**:
- Consider an isolated composite system divided into two compartments by a movable, impermeable, diathermal partition.
- Total internal energy $U = U_1(S_1, V_1, N_1) + U_2(S_2, V_2, N_2)$.
- Constraints: Total volume $V_1 + V_2 = V = \text{const} \implies dV_1 = -dV_2$.
**2. Mechanical Equilibrium**:
- At equilibrium, entropy $S = S_1 + S_2$ is maximized. From $dS = 0$:
$dS = \left( \frac{1}{T_1} - \frac{1}{T_2} \right) dU_1 + \left( \frac{P_1}{T_1} - \frac{P_2}{T_2} \right) dV_1 = 0$
- Thermal equilibrium requires $T_1 = T_2$.
- Because the partition is movable ($dV_1 \neq 0$), mechanical equilibrium requires:
$\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_1 = P_2$
- The forces exerted on either face of the movable partition ($F = P A$) must balance, otherwise the partition experiences a net accelerating force until pressures equalize.
**3. Official Verification**:
The verified official IIT Bombay Master Answer is **Option (B)** (`P1 = P2`).
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