Chemistry - Thermodynamics - I Question with Solution | TestHub

${\mathrm{T}}_{\mathrm{A}}=\frac{{\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}}{\mathrm{nR}}=\frac{3(41)}{(1)(0.082)}=1500\mathrm{K}$

${\mathrm{T}}_{\mathrm{C}}=1500\mathrm{K}$ (as process AC is isothermal)

${\mathrm{T}}_{\mathrm{B}}=\frac{{\mathrm{P}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}}{\mathrm{nR}}=\frac{1(82)}{(1)(0.82)}=1000\mathrm{K}$

Work in polytropic process AB

$\begin{array}{rlrl}{W}_{AB}& =\frac{nR\left(T_2-T_1\right)}{\chi -1}& & \end{array}$

$=\frac{\left(1\right)R(1000-1500)}{0.6}=-833.33R$

work in isothermal process ( ${\mathrm{W}}_{\mathrm{CA}}$ )

$=-nRT\ln \left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)=-1(R)(1500)\ln \left(\frac{41}{82}\right)$

Total work $=-833.33R+1(R)(1500)\ln 2\Rightarrow 17.8$ litre-atm