Step-by-Step Analysis

1. Initial Moles Calculation

Using the Ideal Gas Law \(PV = nRT\), we determine the initial number of moles in each cylinder:

• Cylinder A: \(n_A = \frac{2p_0 V_0}{RT_0}\)
• Cylinder B: \(n_B = \frac{p_0 V_0}{RT_0}\)

The total number of moles in the system is:

$$n_{total} = n_A + n_B = \frac{3p_0 V_0}{RT_0}$$

2. Conservation of Energy

Consider the system as the gas in both cylinders combined. Since the walls are not specified as conducting, and this is a rapid filling process involving boundary work, we treat the system as adiabatic but with work done on it by the piston in A.

First Law of Thermodynamics:

$$\Delta U = Q + W_{on}$$

Since \(Q = 0\) (adiabatic) and work is done by the piston maintaining constant pressure \(2p_0\):

$$W_{on} = P_{ext} (V_{initial} – V_{final}) = 2p_0 (V_0 – V_A’)$$

The change in internal energy for a mono-atomic gas (\(C_v = \frac{3}{2}R\)) is:

$$\Delta U = n_{total} C_v (T_f – T_0) = \frac{3p_0 V_0}{RT_0} \cdot \frac{3}{2}R \cdot (T_f – T_0)$$

3. Solving for Final Temperature (\(T_f\))

Equating the Work and Internal Energy change:

$$2p_0 (V_0 – V_A’) = \frac{9}{2} \frac{p_0 V_0}{T_0} (T_f – T_0)$$

We also need the Equation of State for the final condition. The final pressure is \(2p_0\) everywhere, cylinder B has volume \(V_0\), and cylinder A has volume \(V_A’\).

$$n_{total} = \frac{P_f V_{total}}{R T_f} \implies \frac{3p_0 V_0}{RT_0} = \frac{2p_0(V_A’ + V_0)}{R T_f}$$

From this mole balance, we isolate \(V_A’\):

$$\frac{3}{T_0} = \frac{2(V_A’ + V_0)}{V_0 T_f} \implies \frac{V_A’}{V_0} = \frac{3}{2}\frac{T_f}{T_0} – 1$$

Substitute \(\frac{V_A’}{V_0}\) back into the energy equation:

$$2p_0 V_0 \left( 1 – \left[ \frac{3}{2}\frac{T_f}{T_0} – 1 \right] \right) = \frac{9}{2} \frac{p_0 V_0}{T_0} (T_f – T_0)$$

Dividing by \(p_0 V_0\) and simplifying:

$$2 \left( 2 – \frac{3}{2}\frac{T_f}{T_0} \right) = \frac{9}{2} \left( \frac{T_f}{T_0} – 1 \right)$$ $$4 – 3\frac{T_f}{T_0} = 4.5\frac{T_f}{T_0} – 4.5$$ $$8.5 = 7.5 \frac{T_f}{T_0}$$

Thus:

$$\frac{T_f}{T_0} = \frac{8.5}{7.5} = \frac{17}{15}$$

4. Solving for Final Volume of A

Using the relation derived earlier: \(\frac{V_A’}{V_0} = \frac{3}{2}\frac{T_f}{T_0} – 1\)

$$V_A’ = V_0 \left( \frac{3}{2} \cdot \frac{17}{15} – 1 \right)$$ $$V_A’ = V_0 \left( \frac{17}{10} – 1 \right) = \frac{7}{10} V_0$$