Two vessels A and B of equal volume V_0 are joined by a narrow tube...

Initially, both vessels contain an ideal gas at pressure $p_0$, volume $V_0$, and temperature $T_0$.

When the valve is closed, the gases in vessels $A$ and $B$ are isolated from each other.

For vessel $A$, the walls are diathermic, which means it can exchange heat with the surroundings. Since the piston is pulled out slowly, the process is isothermal. The temperature remains constant at the atmospheric temperature $T_0$.

Using Boyle's law for the isothermal process in vessel $A$:

$p_A V_A = p_0 V_0$

Given that the final volume $V_A = 2V_0$:

$p_A (2V_0) = p_0 V_0 \Rightarrow p_A = \dfrac{p_0}{2}$

Thus, in vessel $A$, the temperature is $T_0$ and the pressure is $\dfrac{p_0}{2}$.

For vessel $B$, the walls are adiabatic, meaning no heat is exchanged with the surroundings. Since the expansion is slow, the process is reversible and adiabatic.

Using the adiabatic relations $T V^{\gamma - 1} = \text{constant}$ and $p V^\gamma = \text{constant}$:

$T_B V_B^{\gamma - 1} = T_0 V_0^{\gamma - 1}$

Given that the final volume $V_B = 2V_0$:

$T_B (2V_0)^{\gamma - 1} = T_0 V_0^{\gamma - 1} \Rightarrow T_B = \dfrac{T_0}{2^{\gamma - 1}}$

For pressure:

$p_B V_B^\gamma = p_0 V_0^\gamma$

$p_B (2V_0)^\gamma = p_0 V_0^\gamma \Rightarrow p_B = \dfrac{p_0}{2^\gamma}$

Thus, in vessel $B$, the temperature is $\dfrac{T_0}{2^{\gamma - 1}}$ and the pressure is $\dfrac{p_0}{2^\gamma}$.