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

Initially, the separator is in the middle, so the volume of each gas is $V_{1i} = V_{2i} = \dfrac{V_0}{2}$.

Since the separator is slid slowly and both the tube and separator are adiabatic, the process is reversible and adiabatic for both gases. For an ideal gas undergoing a reversible adiabatic process, $pV^\gamma = \text{constant}$.

Let the final common pressure in equilibrium be $p$. Let the final volumes of the left and right parts be $V_{1f}$ and $V_{2f}$ respectively.

For the gas in the left part:

$p_1 \left(\dfrac{V_0}{2}\right)^\gamma = p V_{1f}^\gamma \Rightarrow V_{1f} = \dfrac{V_0}{2} \left(\dfrac{p_1}{p}\right)^{1/\gamma}$

For the gas in the right part:

$p_2 \left(\dfrac{V_0}{2}\right)^\gamma = p V_{2f}^\gamma \Rightarrow V_{2f} = \dfrac{V_0}{2} \left(\dfrac{p_2}{p}\right)^{1/\gamma}$

The total volume of the tube remains constant, so $V_{1f} + V_{2f} = V_0$.

Substituting the expressions for $V_{1f}$ and $V_{2f}$:

$\dfrac{V_0}{2} \left(\dfrac{p_1}{p}\right)^{1/\gamma} + \dfrac{V_0}{2} \left(\dfrac{p_2}{p}\right)^{1/\gamma} = V_0$

Dividing by $V_0 / 2$:

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