SSC JE Mechanical Engineering 11 Oct 2020 Shift 1 Papers

${\mathit{P}}_1=1$ bar, ${\mathit{P}}_2=8$ bar, ${\mathit{T}}_1=295\mathit{K},\mathrm{\gamma }=1.4$
we can find the intial volume ${\mathit{V}}_1$ by ideal gas equation, ${\mathit{P}}_1{\mathit{V}}_1={\mathit{m}\mathit{R}\mathit{T}}_1$
$1\times {10}^5\times {\mathit{V}}_1=1\times 287\times 295$
Work done using isentropic process $\mathit{W}$
$\mathit{W}=\frac{1.4}{1.4-1}\times 1\times {10}^5\mathrm{}$ $\times 0.846\left[{\left(\frac{8}{1}\right)}^{\frac{1.4-1}{1.4}}-1\right]$
$\mathit{W}=240.269\mathit{k}\mathit{W}$
Work done using isothermal process W
$\mathit{W}=1\times {10}^5\times 0.846\times \ln \frac{8}{1}$
$\mathit{W}=175.920\mathit{k}\mathit{W}$
Therefore, the difference between the isentropic and isothermal work is,