Question

Six moles of an ideal gas undergo a cyclic process shown in the figure. If the temperatures are given as $T_A=600\text{K}$, $T_B=800\text{K}$, $T_C=2200\text{K}$ and $T_D=1200\text{K}$, then the work done per cycle is

• A. $20\text{kJ}$
• B. $30\text{kJ}$
• C. $40\text{kJ}$
• D. $60\text{kJ}$

Answer 1

The correct option is C $40\text{kJ}$

Given,
Number of moles of ideal gas, $\mu =6$


Processes $AB$ and $CD$ have straight line graphs of the form $Y=mX$
As, $P=\frac{\mu R}{V}T$, by comparision, we get
$P\propto T$.
So, $AB$ and $CD$ are isochoric processes. Thus, work done during processes $AB$ and $CD$ is equal to zero
i.e $W_{AB}=W_{CD}=0$.

Also from the figure, $BC$ and $DA$ are isobaric processes.
Work done during process $BC$
$W_{BC}=P_2(V_C-V_B)=\mu R(T_C-T_B)$
$=6R(2200-800)=6\times 8.3\times 1400=69.72\text{kJ}$
Similarly, $W_{DA}=P_1(V_A-V_D)=\mu R(T_A-T_B)$
$=6R(600-1200)=-6\times 8.3\times 600=-29.88\text{kJ}$

Hence, work done in complete cycle
$W=W_{AB}+W_{BC}+W_{CD}+W_{DA}$
$W=0+69.72\text{kJ}+0-29.88\text{kJ}$
$\Rightarrow W=39.84\text{kJ}\approx 40\text{kJ}$
Hence, option (c) is the correct answer.