An adiabatic vessel of total volume V is partitioned into two equal...

Let the initial temperatures of the left and right parts be $T_1 and$T_2$respectively.<br><br>Using the ideal gas equation$pV = nRT$:<br>For the left part:$p \left(\dfrac{V}{2}\right) = 1 \cdot R \cdot T_1 \Rightarrow T_1 = \dfrac{pV}{2R}$<br>For the right part:$p \left(\dfrac{V}{2}\right) = 2 \cdot R \cdot T_2 \Rightarrow T_2 = \dfrac{pV}{4R}$<br><br>Since the vessel is adiabatic, the total internal energy of the system remains constant. Let the final equilibrium temperature be$T_f$.<br><br>$n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T_f$<br><br>$1 \cdot \dfrac{pV}{2R} + 2 \cdot \dfrac{pV}{4R} = (1 + 2) T_f$<br><br>$\dfrac{pV}{R} = 3 T_f \Rightarrow T_f = \dfrac{pV}{3R}$<br><br>The separator is fixed, so the volume of the right part is constant, which means the work done is zero ($W = 0$). By the first law of thermodynamics, the heat given to the gas in the right part is equal to its change in internal energy:<br><br>$Q = \Delta U_2 = n_2 C_v (T_f - T_2)$<br><br>Given$U = 1.5 nRT$, we have$C_v = 1.5 R = \dfrac{3}{2} R$.<br><br>$Q = 2 \cdot \left(\dfrac{3}{2}R\right)\left(\dfrac{pV}{3R} - \dfrac{pV}{4R}\right)

$Q = 3R \left(\dfrac{4pV - 3pV}{12R}\right) =$3R \left(\dfrac{pV}{12R}\right) = \dfrac{pV}{4}