A cylindrical tube of volume V having adiabatic walls contains an...

Let the total volume of the cylindrical tube be $V$. Since the diathermic wall divides the tube into two equal parts, the volume of each part is $\dfrac{V}{2}$.

The number of moles of gas in the left and right parts initially are:
$n_1 = \dfrac{p_1 (V/2)}{R T_1}$
$n_2 = \dfrac{p_2 (V/2)}{R T_2}$

Since the outer walls are adiabatic and the inner wall is fixed, there is no heat exchange with the surroundings ($Q = 0$) and no work is done by the system ($W = 0$). Thus, the total internal energy of the system remains conserved.

Initial internal energy of the system:
$U_i = U_1 + U_2 = 1.5 n_1 R T_1 + 1.5 n_2 R T_2$
$U_i = 1.5 \left( \dfrac{p_1 V}{2 R T_1} \right) R T_1 + 1.5 \left( \dfrac{p_2 V}{2 R T_2} \right) R T_2$
$U_i = \dfrac{1.5 V}{2} (p_1 + p_2)$

Let the final equilibrium temperature be $T_f$. The final internal energy of the system is:
$U_f = 1.5 (n_1 + n_2) R T_f$
$U_f = 1.5 \left( \dfrac{p_1 V}{2 R T_1} + \dfrac{p_2 V}{2 R T_2} \right) R T_f$
$U_f = \dfrac{1.5 V}{2} \left( \dfrac{p_1}{T_1} + \dfrac{p_2}{T_2} \right) T_f$
$U_f = \dfrac{1.5 V}{2} \left( \dfrac{p_1 T_2 + p_2 T_1}{T_1 T_2} \right) T_f$

Equating the initial and final internal energies ($U_i = U_f$):
$\dfrac{1.5 V}{2} (p_1 + p_2) = \dfrac{1.5 V}{2} \left( \dfrac{p_1 T_2 + p_2 T_1}{T_1 T_2} \right) T_f$

$p_1 + p_2 = \left( \dfrac{p_1 T_2 + p_2 T_1}{T_1 T_2} \right) T_f$

$T_f = \dfrac{T_1 T_2 (p_1 + p_2)}{p_1 T_2 + p_2 T_1}$

Given $\lambda = p_1 T_2 + p_2 T_1$, we get:
$T_f = \dfrac{T_1 T_2 (p_1 + p_2)}{\lambda}$