Two cylinders are connected by a fixed diathermic partition A, and a removable adiabatic partition B placed adjacent to A as shown in figure Initially n moles of ideal mono-atomic gas is present in both cylinders at normal at pressure $P_0$ Both the gas occupy same volume $V_0$ initially. Now the piston of left cylinder is compressed in adiabatic manner so that volume of left portion becomes ${\mathrm{V}}_0/2$ and then left piston is clamped. Then the adiabatic slider B is removed so that the two cylinder come in thermal equation. Assume all other surfaces except A to be adiabatic, for this situation mark the correct statement (s).

As the gas of left chamber is compressed adiabatically to half of its initial volume

$P_0{V}_0=P_1{\left(V_0/2\right)}^{\gamma }\Rightarrow P_1=P_0\times 2^{\gamma }$

i.e. pressure in left chamber after the completion of adiabatic process is $P_1$ while pressure in right chamber remains same as it is isolated from surrounding or you say no energy transfer between right chamber & surrounding.

Due to this adiabatic compression, temperature will also change in left chamber,

${\mathrm{T}}_0{\mathrm{V}}_0^{\gamma -1}={\mathrm{T}}_1{\left({\mathrm{V}}_0/2\right)}^{\gamma -1}\Rightarrow {\mathrm{T}}_1=2^{\gamma -1}{\mathrm{T}}_0\text{ where }{\mathrm{T}}_0=\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{\mathrm{nR}}$

when adiabatic seperator is removed temperature on both sides are different, temperature on left is  $T_1$ while temperature on right is ${\mathrm{T}}_0\text{ as }{\mathrm{T}}_1>{\mathrm{T}}_0.$ So heat transfer takes place from left chamber to right chamber and the gas in right chamber starts expanding, as piston (right side) is free to move the pressure remains constant as $P_0$ So expansion in right chamber is taking place at constant temperature. During this expansion, of gas in right chamber , the volume of gas in left chamber remains constant. but temperature & press changes.

For the instant just after the removval of adiabatic seperator we can take the pressure of gas in left chamber as same just before the removal i.e. ${\mathrm{P}}_0\times 2^{\gamma }$ Let final temperature on both sides is T, then the energy transferred from left to right is used to increase the temperature of right part & to expand the gas. i.e.

$\begin{array}{r}{\mathrm{nC}}_v\left({\mathrm{T}}_1-\mathrm{T}\right)={\mathrm{nC}}_{\mathrm{p}}\left(\mathrm{T}-{\mathrm{T}}_0\right)={\mathrm{nC}}_v\left(\mathrm{T}-{\mathrm{T}}_0\right)+\mathrm{W}\\ \Rightarrow \mathrm{T}=\frac{5{\mathrm{T}}_0+3{\mathrm{T}}_1}{8}=\left(\frac{5+3\times 2^{\gamma -1}}{8}\right){\mathrm{T}}_0=1.22{\mathrm{T}}_0\end{array}$

So energy transferred is, 

$\mathrm{Q}=n{\mathrm{C}}_v\left({\mathrm{T}}_1-\mathrm{T}\right)=\frac{3\mathrm{nR}}{2}\left(2^{\gamma -1}-1.22\right){\mathrm{T}}_0=0.55{\mathrm{P}}_0{\mathrm{V}}_0$

Work done, 

$\begin{array}{l}W=P_0\left(V_f-V_i\right)=nR\left(T-T_0\right)\\ =0.22nR{T}_0=0.22P_0{V}_0\end{array}$