1. Given data

• Each side contains 1 mole of an ideal gas.
• Partition is thermally insulating and frictionless.
• Process is slow, so pressures on both sides remain equal at every instant.
• Initially: \qquad T_L=T_R=T_0$$
• Finally, right side volume becomes $V_R=\frac{V_0}{2}$
• Since total volume is constant, $V_L=2V_0-\frac{V_0}{2}=\frac{3V_0}{2}$
• For each gas, $C_V=2R$.

We need to find $\dfrac{T_R}{T_0}$.

2. Nature of process for the right gas

The partition is thermally insulating, so the right-side gas does not receive heat from the left side. Thus for the right gas, $dQ_R=0$ So the right gas undergoes a quasi-static adiabatic compression.

For an ideal gas in a reversible adiabatic process, $PV^\gamma=\text{constant}$ with $\gamma=\frac{C_P}{C_V}$ Now, $C_P=C_V+R=2R+R=3R$ Hence, $\gamma=\frac{3R}{2R}=\frac{3}{2}$

3. Use adiabatic relation between $T$ and $V$

For a reversible adiabatic process of an ideal gas, $TV^{\gamma-1}=\text{constant}$ Therefore, for the right gas, $T_0 V_0^{\gamma-1}=T_R\left(\frac{V_0}{2}\right)^{\gamma-1}$ Since $\gamma-1=\frac{3}{2}-1=\frac{1}{2}$ we get $T_0 V_0^{1/2}=T_R\left(\frac{V_0}{2}\right)^{1/2}$ So, $\frac{T_R}{T_0}=\frac{V_0^{1/2}}{(V_0/2)^{1/2}}=\sqrt{2}$

Thus, $\boxed{\frac{T_R}{T_0}=\sqrt{2}}$

4. Check options

• A: $\sqrt{2}$ ✅
• B: $\sqrt{3}$ ❌
• C: $2$ ❌
• D: $3$ ❌

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer is also A. Hence, they agree.