Work In Thermodynamic Process

Let us call the region on the left $A$ and the region on the left $B$ . Let the area of the piston be $A$ .

Without loss of generality, let the piston move to the left by a distance $x$ . Let us call the pressure in $A$ at that moment $P_{1}$ and the pressure in $B$ , $P_{2}$ .

Now as the piston is being moved very slowly,

$\begin{aligned} F_{ext}+P_1A &=P_2A\\ F_{ext} &=(P_2-P_1)A\\ F_{ext}\mathrm{d}x=\mathrm{d}W&=(P_2-P_1)A\mathrm{d}x \end{aligned}$

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Now as the processes of compression in $B$ and expansion in $A$ are both adiabatic, we can use the fact that $PV^{\gamma}=$ constant .

$\begin{cases}PV^{\gamma}=P_{1} (V+Ax)^{\gamma} \\ PV^{\gamma}=P_{2}(V-Ax)^{\gamma} \end{cases}$ $\implies P_2-P_1=PV^{\gamma} \left(\dfrac{1}{(V-Ax)^{\gamma}}-\dfrac{1}{(V+Ax)^{\gamma}}\right)$

Now, the piston is moved until $V+Ax=3(V-Ax) \implies x=\dfrac{V}{2A}$ .

Finally, $\begin{aligned}W&=\int_{0}^{V/2A} (P_2-P_1)A\mathrm{d}x\\ &=\int_{0}^{V/2A} PV^{\gamma} \left(\dfrac{1}{(V-Ax)^{\gamma}}-\dfrac{1}{(V+Ax)^{\gamma}}\right) A \mathrm{d}x \end{aligned}$

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Now substitute $\gamma=\dfrac{5}{3}$ and integrate.

After doing that you should get $W=PV \left(\dfrac{\sqrt[3]{12}-3\sqrt[3]{4}-6}{2}\right) \implies \boxed{k\approx 0.526}$

Try using energy conservation. Final temperature of both the chamber will be easily calculated and the change of internal energy change is work done by the external force.