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

Let the volume of each part be $V_1 = V_2 = \dfrac{V}{2}$.

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

The internal energy of the ideal gas is given as $U = 1.5 nRT = \dfrac{3}{2} nRT$. This implies the molar heat capacity at constant volume is $C_v = \dfrac{3}{2} R$.

Since the outer walls are adiabatic and the total volume is fixed, the total internal energy of the system remains conserved.
$U_1 + U_2 = U_{1f} + U_{2f}$
$n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T_f$

Substituting the values of $n_1 and$n_2$:<br>$\dfrac{p_1 V}{2 R T_1} T_1 + \dfrac{p_2 V}{2 R T_2} T_2 = \left( \dfrac{p_1 V}{2 R T_1} + \dfrac{p_2 V}{2 R T_2} \right) T_f$<br>$p_1 + p_2 = \left( \dfrac{p_1 T_2 + p_2 T_1}{T_1 T_2} \right) T_f$<br><br>Given$\lambda = p_1 T_2 + p_2 T_1$, we can write:<br>$T_f = \dfrac{(p_1 + p_2) T_1 T_2}{\lambda}$<br><br>The heat flown from the right gas to the left gas is equal to the change in internal energy of the left gas (since the volume of each part is fixed, work done$W = 0$).<br>$\Delta Q = \Delta U_1 = n_1 C_v (T_f - T_1)$<br>$\Delta Q = \dfrac{p_1 V}{2 R T_1} \left( \dfrac{3}{2} R \right) (T_f - T_1) = \dfrac{3 p_1 V}{4 T_1} (T_f - T_1)$<br><br>Now, substituting the value of$T_f$:<br>$T_f - T_1 = \dfrac{(p_1 + p_2) T_1 T_2}{\lambda} - T_1 = T_1 \left(\dfrac{p_1 T_2 + p_2 T_2 - (p_1 T_2 + p_2 T_1)}{\lambda} \right)
T_f - T_1 = T_1 \left(\dfrac{p_2 T_2 - p_2 T_1}{\lambda} \right) = \dfrac{p_2 T_1 (T_2 - T_1)}{\lambda}

Substituting this back into the equation for $\Delta Q$:
$\Delta Q =$\dfrac{3 p_1 V}{4 T_1} \times \dfrac{p_2 T_1 (T_2 - T_1)}{\lambda} = \dfrac{3 p_1 p_2 (T_2 - T_1) V}{4 \lambda}

Answer: \dfrac{3 p_1 p_2 (T_2 - T_1) V}{4 \lambda}