Question

Six moles of an ideal gas undergoes a thermodynamic cycle as shown in figure. If the temperatures are $T_A=600\text{K},T_B=800\text{K},T_C=2200\text{K}$ and $T_D=1200\text{K}$, the work done per cycle is approximately:
(Take universal gas constant, $R=\frac{25}{3}\text{J/mole. K}$)

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

Answer 1

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

The processes $(A\to B)$ and $(C\to D)$ are parts of straight line passing through origin.
$\Rightarrow P\propto T$
$\Rightarrow \frac{P}{T}=\text{constant}$
This indicates isochoric process.
Hence, the volume remains constant for the processes $\text{AB}$ and $\text{CD}$.
So no work is done during processes for (A to B) and (C to D)
$\Rightarrow W_{AB}=W_{CD}=0$
Processes $BC$ and $DA$ are isobaric processes,
$\Rightarrow W_{BC}=P\Delta V=nR\Delta T$
Or, $W_{BC}=nR(T_C-T_B)$
Or,$W_{BC}=6R(2200-800)=(6R\times 1400)\text{J}$
Also $W_{DA}=nR\Delta T=nR(T_A-T_D)$
$W_{DA}=6R(600-1200)=(-6R\times 600)\text{J}$
Hence, work done in complete cycle is:
$W=W_{AB}+W_{BC}+W_{CD}+W_{DA}$
$W=0+(6R\times 1400)+0+(-6R\times 600)$
$W=6R\times 800$
$\therefore W=6\times \frac{25}{3}\times 800=40000\text{J}=40\text{k J}$