One mole of an ideal gas $\left(C_{v,m}=\frac{5}{2}R\right)$ at 300K and 5 atm is expanded adiabatically to a final pressure of 2 atm against a constant pressure of 2 atm. Final temperature of the gas is :

Given,

$\begin{array}{l}{P}_{\text{ext }}=2\mathrm{atm}\\ P_1=5\mathrm{atm}\\ {\mathrm{P}}_2=2\mathrm{atm}\\ {\mathrm{T}}_1=300\mathrm{K}\\ {\mathrm{T}}_2=\mathrm{T}\left(\mathrm{say}\right)=?\\ \mathrm{n}=1\\ {\mathrm{C}}_{\mathrm{V}}=\frac{5}{2}\mathrm{R}\end{array}$

For an adiabatic process
$\begin{array}{l}\mathrm{W}={\mathrm{nC}}_{\mathrm{V}}\mathrm{dT}=-\mathrm{P\Delta V}\\ \therefore {\mathrm{nC}}_{\mathrm{V}}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=-{\mathrm{P}}_{\mathrm{axar}}\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)\\ \Rightarrow 1\times \frac{5}{2}\mathrm{R}(\mathrm{T}-300)=-2\left(\frac{{\mathrm{nRTT}}_2}{{\mathrm{P}}_2}-\frac{\mathrm{nRT}}{{\mathrm{P}}_1}\right)\\ \Rightarrow \frac{5}{2}\mathrm{R}(\mathrm{T}-300)=-2\mathrm{R}\left(\frac{\mathrm{T}}{2}-\frac{300}{5}\right)\\ \Rightarrow 5\mathrm{T}-1500=-2\mathrm{T}+240\\ \Rightarrow T\mathrm{T}=1740\Rightarrow \mathrm{T}=248.57\mathrm{K}\\ \therefore {\mathrm{T}}_2=\mathrm{T}=248.57\mathrm{K}\end{array}$