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

Let the initial volume of each part be $\dfrac{V}{2}$.

The initial internal energy of a gas is given by $U = 1.5 nRT = 1.5 pV_{part}$.

For the left part:
$U_{L,i} = 1.5 p \left(\dfrac{V}{2}\right) = \dfrac{3pV}{4}$

For the right part:
$U_{R,i} = 1.5 p \left(\dfrac{V}{2}\right) = \dfrac{3pV}{4}$

Total initial internal energy of the system:
$U_{i} = U_{L,i} + U_{R,i} = \dfrac{3pV}{4} + \dfrac{3pV}{4} = \dfrac{3pV}{2}$

Since the vessel is adiabatic and the separator is fixed, the total internal energy of the system remains conserved. The conducting separator allows heat exchange until a common final temperature $T_f$ is reached.

Total final internal energy:
$U_{f} = 1.5 (n_L + n_R) R T_f = 1.5 (1 + 2) R T_f = 4.5 R T_f$

Equating initial and final internal energies:
$4.5 R T_f = \dfrac{3pV}{2} \Rightarrow R T_f = \dfrac{pV}{3}$

The final internal energy of the left part is:
$U_{L,f} = 1.5 n_L R T_f = 1.5 (1) \left(\dfrac{pV}{3}\right) = \dfrac{pV}{2}$

The increase in the internal energy of the gas in the left part is:
$\Delta U_L = U_{L,f} - U_{L,i} = \dfrac{pV}{2} - \dfrac{3pV}{4} = -\dfrac{pV}{4}$