Calculate the work done by the system in an irreversible (single step) adiabatic expansion of 2 mole of a polyatomic gas $(\mathrm{\gamma }=4/3)$ from 300 K and pressure 10 atm to 1 atm : (in KJ)

In irreversible adiabatic process work done

$\mathrm{w}=-{\mathrm{p}}_{\text{ext }}\left({\mathrm{v}}_2-{\mathrm{v}}_1\right)\text{ (or) }\mathrm{w}=\frac{\mathrm{nR}}{\mathrm{\gamma }-1}\left[{\mathrm{T}}_2-{\mathrm{T}}_1\right]$

For irreversible process${\mathrm{pv}}^{\mathrm{r}}$ is not valid and final temperature is to be obtained by equating the above two expressions.

$\begin{array}{rr} & \text{ So, }-{\mathrm{p}}_2\left({\mathrm{v}}_2-{\mathrm{v}}_1\right)=\frac{\mathrm{nR}}{\mathrm{\gamma }-1}\left[{\mathrm{T}}_2-{\mathrm{T}}_1\right]\\ & {\mathrm{P}}_2\left(\frac{{\mathrm{nT}}_1\mathrm{R}}{{\mathrm{P}}_1}-\frac{{\mathrm{nT}}_2\mathrm{R}}{{\mathrm{P}}_2}\right)=\frac{\mathrm{nR}}{\mathrm{\gamma }-1}\left[{\mathrm{T}}_2-{\mathrm{T}}_1\right]\\ & \frac{{\mathrm{P}}_2{\mathrm{T}}_1}{\mathrm{P}}-{\mathrm{T}}_2=\frac{1}{\mathrm{\gamma }-1}\left[{\mathrm{T}}_2-{\mathrm{T}}_1\right]\\ & \text{ so, }{\mathrm{T}}_1\left[\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}+\frac{1}{\mathrm{\gamma }-1}\right],{\mathrm{T}}_2\left[1+\frac{1}{\mathrm{\gamma }-1}\right]\\ & {\mathrm{T}}_2={\mathrm{T}}_1\left(\frac{\mathrm{\gamma }-1}{\mathrm{\gamma }}\right)\left[\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}+\frac{1}{\mathrm{\gamma }-1}\right]\\ & \mathrm{ }=300\left[\frac{1/3}{4/3}\right]\left[\frac{1}{10}+\frac{1}{1/3}\right]\\ & \mathrm{ }=300\times \frac{1}{4}\times \left[\frac{31}{10}\right]=232.5\\ & \mathrm{w}=\frac{2\times 8.314\times {10}^{-3}}{1/3}[232.5-300]\text{; so, ans }3\end{array}$