1. Given data

• Each side contains 1 mole of ideal gas.
• $C_V = 2R$.
• Initially:
  • Left gas: $V_0, T_0$
  • Right gas: $V_0, T_0$
• A small-power heater supplies heat $Q$ to the left gas.
• The partition is frictionless and movement is slow, so the process is quasi-static.
• Final volume of right gas is $V_R = \dfrac{V_0}{2}$.

Since total volume is constant, $V_L + V_R = 2V_0$ so finally, $V_L = 2V_0 - \frac{V_0}{2} = \frac{3V_0}{2}.$

2. Behavior of the right side gas

The right side is thermally insulated, so for the right gas, $Q_R = 0.$ Since the partition moves slowly, the compression is quasi-static and hence reversible adiabatic.

For 1 mole ideal gas, $\gamma = \frac{C_P}{C_V} = \frac{C_V + R}{C_V} = \frac{2R + R}{2R} = \frac{3}{2}.$

Using $TV^{\gamma-1} = \text{constant},$ we get $T_R V_R^{\gamma-1} = T_0 V_0^{\gamma-1}.$

Here, $\gamma - 1 = \frac{1}{2}.$ So, $T_R \left(\frac{V_0}{2}\right)^{1/2} = T_0 V_0^{1/2}.$ Hence, $T_R = T_0 \sqrt{2}.$

3. Mechanical equilibrium of partition

At every stage, pressures on both sides are equal. Therefore in the final state, $P_L = P_R.$

Using ideal gas law for 1 mole each, $\frac{RT_L}{V_L} = \frac{RT_R}{V_R}.$ So, $\frac{T_L}{V_L} = \frac{T_R}{V_R}.$

Substitute $V_L = \dfrac{3V_0}{2}$ and $V_R = \dfrac{V_0}{2}$: $\frac{T_L}{3V_0/2} = \frac{T_R}{V_0/2}.$ Thus, $T_L = 3T_R = 3\sqrt{2}\,T_0.$

4. Use first law for the whole system

Take both gases together as the system.

• Cylinder is thermally insulated externally.
• The only energy entering the system is the electrical heating $Q$ supplied to the left gas.
• Total volume of the two-gas system is constant, so there is no external work done by the system.

Therefore, $Q = \Delta U_L + \Delta U_R.$

For 1 mole ideal gas, $\Delta U = nC_V \Delta T = 2R\Delta T.$

Hence, $Q = 2R(T_L - T_0) + 2R(T_R - T_0).$

Substitute $T_L = 3\sqrt{2}T_0$ and $T_R = \sqrt{2}T_0$: $Q = 2R\left(3\sqrt{2}T_0 - T_0\right) + 2R\left(\sqrt{2}T_0 - T_0\right).$

$Q = 2RT_0\left(3\sqrt{2} - 1 + \sqrt{2} - 1\right)$ $Q = 2RT_0(4\sqrt{2} - 2)$ $Q = 4RT_0(2\sqrt{2} - 1).$

Therefore, $\frac{Q}{RT_0} = 4(2\sqrt{2} - 1).$

5. Option check

This matches:

Option B: $4(2\sqrt 2 - 1)$

6. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.