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

Let the volume of each part be $V_0 = \dfrac{V}{2}$.

The number of moles of gas in the left part is:
$n_1 = \dfrac{p_1 V_0}{R T_1}$

The number of moles of gas in the right part is:
$n_2 = \dfrac{p_2 V_0}{R T_2}$

Since the outer walls are adiabatic and the total volume is constant, the total internal energy of the system remains conserved. Let the final common temperature be $T_f$.

$U_{initial} = U_{final}$
$n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T_f$

Solving for $T_f$:
$T_f = \dfrac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$

Substituting the expressions for $n_1$ and $n_2$:
$n_1 T_1 + n_2 T_2 = \dfrac{p_1 V_0}{R} + \dfrac{p_2 V_0}{R} = \dfrac{V_0}{R}(p_1 + p_2)$

n_1 + n_2 = \dfrac{V_0}{R} \left( \dfrac{p_1}{T_1} + \dfrac{p_2}{T_2} \right) = \dfrac{V_0}{R} \left(\dfrac{p_1 T_2 + p_2 T_1}{T_1 T_2} \right)

Therefore, the final temperature is:
$T_f =$\dfrac{\dfrac{V_0}{R}(p_1 + p_2)}{\dfrac{V_0}{R} \left( \dfrac{p_1 T_2 + p_2 T_1}{T_1 T_2} \right)} = \dfrac{(p_1 + p_2) T_1 T_2}{p_1 T_2 + p_2 T_1}

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

Now, applying the ideal gas law to find the final pressure on the left side ($p_{1f}$) and right side ($p_{2f}$), keeping in mind that the volume of each part remains $V_0$:

For the left side:
\dfrac{p_1}{T_1} = \dfrac{p_{1f}}{T_f} \Rightarrow$p_{1f} =$\dfrac{p_1}{T_1}$T_f =$\dfrac{p_1}{T_1} \left[ \dfrac{(p_1 + p_2) T_1 T_2}{\lambda} \right] = \dfrac{p_1 T_2 (p_1 + p_2)}{\lambda}

For the right side:
\dfrac{p_2}{T_2} = \dfrac{p_{2f}}{T_f} \Rightarrow$p_{2f} =$\dfrac{p_2}{T_2}$T_f =$\dfrac{p_2}{T_2} \left[ \dfrac{(p_1 + p_2) T_1 T_2}{\lambda} \right] = \dfrac{p_2 T_1 (p_1 + p_2)}{\lambda}

Answer: \dfrac{p_1 T_2 (p_1 + p_2)}{\lambda} and \dfrac{p_2 T_1 (p_1 + p_2)}{\lambda}