One mole of monoatomic gas is taken through cyclic process shown below. T_A = 300 K. Process AB is defined as PT = constant. 

(a) Process AB :

    $\begin{array}{l} \mathrm{PT}=\text{ constant }\\ \Rightarrow \frac{{\mathrm{nRT}}^2}{V}=\text{ constant }\\ \mathrm{W}={\int }_{\mathrm{A}}^{\mathrm{B}} \mathrm{PdV}={\int }_{\mathrm{A}}^{\mathrm{B}} \frac{\text{Constant }}{\mathrm{T}}\mathrm{dV}\\ \Rightarrow \frac{\mathrm{dV}}{\mathrm{dT}}=\frac{2\mathrm{nRT}}{\text{ Constant }}\\ \mathrm{W}={\int }_{300}^{100} \frac{\text{Constant }}{\mathrm{T}}\cdot \frac{2\mathrm{nRT}}{\text{ Constant }}\mathrm{dt}\\ \Rightarrow \frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{B}}}=\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{T}}_{\mathrm{A}}}\\ \Rightarrow \frac{1}{3}=\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{T}}_{\mathrm{A}}}\\ \Rightarrow {\mathrm{T}}_{\mathrm{B}}=\frac{300}{3}=100\\ \Rightarrow \mathrm{W}=2\mathrm{nR}(100-300)\\ \Rightarrow {\mathrm{W}}_{\mathrm{AB}}=-400\mathrm{nR}\end{array}$

(b) Process CA :

$\text{ Isochoric, so }\frac{\mathrm{P}}{\mathrm{T}}\text{ is constant : }$

      $\begin{array}{l} \frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{T}}_{\mathrm{C}}}=\frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{C}}}\\ \frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{T}}_{\mathrm{C}}}=\frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{B}}}\\ \Rightarrow \frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{T}}_{\mathrm{C}}}=\frac{1}{3}\\ \Rightarrow {\mathrm{T}}_{\mathrm{C}}=3{\mathrm{T}}_{\mathrm{A}}\\ {\mathrm{T}}_{\mathrm{C}}=900\mathrm{R}\\ \Rightarrow \mathrm{\Delta U}={\mathrm{nC}}_{\mathrm{V}}\mathrm{\Delta T}\\ =\left(1\right)\frac{3}{2}\mathrm{R}\times \left({\mathrm{T}}_{\mathrm{A}}-{\mathrm{T}}_{\mathrm{C}}\right)\\ =\frac{3}{2}\mathrm{R}\times (300-900)\\ =\frac{3}{2}\mathrm{R}\times -600=-900\mathrm{R}\\ \Rightarrow \left|\mathrm{\Delta U}\right|=\left(900\mathrm{R}\right)\end{array}$

(c) Process BC:

  $\mathrm{Isobaric}$

      $\begin{array}{l} \mathrm{Q}={\mathrm{nC}}_{\mathrm{P}}\mathrm{\Delta T}\\ \Rightarrow \mathrm{Q}=\left(1\right)\frac{5}{2}\mathrm{R}\times \left({\mathrm{T}}_{\mathrm{C}}-{\mathrm{T}}_{\mathrm{B}}\right)\\ \mathrm{Q}=\frac{5}{2}\mathrm{R}\times (900-100)\\ \Rightarrow \mathrm{Q}=\frac{5}{2}\mathrm{R}\times 800\\ \Rightarrow \mathrm{Q}=2000\mathrm{R}\end{array}$