Six moles of an ideal gas performs a cycle shown in figure. If the temperature are $T_A=600K,T_B=800K,T_C=2200K\text{\hspace{0.17em}and\hspace{0.17em}}{T}_D=1200K,$ the work done per cycle (in KJ) is

 

Processes $AtoB$ and $CtoD$ are parts of straight line graphs of the form $y=mx$  
Also $P=\frac{\mu R}{V}T(\mu =6)$     
$\Rightarrow P\alpha T.$ So volume remains constant for the graphs $AB$ and $CD$ 

 So no work is done during processes for $AtoB$ and $CtoD$  
i.e., $W_{AB}=W_{CD}=0and{W}_{BC}=P_2(V_C–V_B)=\mu R(T_C –T_B)$  
$=6R(2200–800)=6R\times 1400J$              
Also $W_{DA}=P_1(V_A–V_D)=\mu R(T_A–T_B)$  
$=6R(600–1200)=–6R\times 600J$               
Hence work done in complete cycle 
$W=W_{AB}+W_{BC}+W_{CD}+W_{DA}$  
$=0+6R\times 1400+0–6R\times 600$      

$=6R\times 900=6\times 8.3\times 800\approx 40kJ$