An adiabatic cylindrical tube having a volume V_0 is separated into...

Since the separator is adiabatic and the process is slow, the expansion or compression of both gases is reversible and adiabatic.

For an adiabatic process, $pV^{\gamma} = \text{constant}$.

Let the final pressure of both gases in equilibrium be $p$. The initial volume of each part is $\dfrac{V_0}{2}$.

For the left part:
$p_1 \left(\dfrac{V_0}{2}\right)^{\gamma} = p V_1^{\gamma} \Rightarrow V_1 = \dfrac{V_0}{2} \left(\dfrac{p_1}{p}\right)^{1/\gamma}$

For the right part:
$p_2 \left(\dfrac{V_0}{2}\right)^{\gamma} = p V_2^{\gamma} \Rightarrow V_2 = \dfrac{V_0}{2} \left(\dfrac{p_2}{p}\right)^{1/\gamma}$

The total volume of the tube remains constant, so $V_1 + V_2 = V_0$.

Substituting the expressions for $V_1$ and $V_2$:
$\dfrac{V_0}{2} \left(\dfrac{p_1}{p}\right)^{1/\gamma} + \dfrac{V_0}{2} \left(\dfrac{p_2}{p}\right)^{1/\gamma} = V_0$

$\dfrac{1}{2 p^{1/\gamma}} \left( p_1^{1/\gamma} + p_2^{1/\gamma} \right) = 1$

\left(\dfrac{1}{p}\right)^{1/\gamma} = \dfrac{2}{p_1^{1/\gamma} + p_2^{1/\gamma}}$<br><br>Substituting this back into the equations for$V_1$and$V_2$:<br><br>$V_1 = \dfrac{V_0}{2} p_1^{1/\gamma} \left( \dfrac{2}{p_1^{1/\gamma} + p_2^{1/\gamma}} \right) = \dfrac{p_1^{1/\gamma} V_0}{p_1^{1/\gamma} + p_2^{1/\gamma}}$<br><br>$V_2 = \dfrac{V_0}{2} p_2^{1/\gamma} \left( \dfrac{2}{p_1^{1/\gamma} + p_2^{1/\gamma}} \right) = \dfrac{p_2^{1/\gamma} V_0}{p_1^{1/\gamma} + p_2^{1/\gamma}}