Question

Six moles of an ideal gas performs a cycle as shown in figure. If the temperatures are $T_A=600K,T_B=800K,T_C=2200K$ and $T_D=1200K$
, then the work done for complete cycle is (in kJ)

Answer 1

Processes $A$ to $B$ and $C$ to $D$ are parts of straight line graphs of the form $y=mx$
Also $P=\frac{nR}{V}T(n=6)$
⇒ $P\propto T$. So volume remains constant for the graphs $AB$ and $CD$


So no work is done during processes for $A$ to $B$ and $C$ to $D$ i.e.
$W_{AB}=W_{CD}=0\mathrm{and}{W}_{BC}=P_2(V_c-V_B)=nR(T_c-T_B)=6R(2200-800)=6R\times 1400J$
$\mathrm{Also}{W}_{DA}=P_1(V_A-V_D)=nR(T_A-T_B)=6R(600-1200)=-6R\times 600J$
Hence, work done in complete cycle
$W=W_{AB}+W_{BC}+W_{CD}+W_{DA}=0+(6R\times 1400)+0-(6R\times 600)=6R\times 800=6\times 8.3\times 800=40kJ$