Question

An ideal gas is subjected to cyclic process involving four thermodynamic states. The amounts of heat $\left(Q\right)$ and work $\left(W\right)$ involved in each of these states are
$Q_{1}=6000\,J,Q_{2}=-5500\,J,Q_{3}=-3000\,J,Q_{4}=3500\,J,$
$W_{1}=2500\,J,W_{2}=-1000\,J,W_{3}=-1200\,J\&W_{4}=x\,J.$
The ratio of the net work done by the gas to the total heat absorbed by the gas is $n$ . The values of $x$ and $\eta$ respectively are

• A. $500,7.5\%$
• B. $700,10.5\%$
• C. $1000,21\%$
• D. $1500,15\%$

Answer 1

Answer: (B)

From first law of thermodynamics
$Q=\Delta U+W$
or $\Delta U=Q - W$
Therefore, $\Delta U_1=Q_1-W_1=6000-2500=3500\, J$
$\Delta U _2= Q _2- W _2=-5500+1000=-4500 \,J $
$\Delta U _3= Q _3- W _3$
$=-3000+1200=-1800\, J$
$ \Delta U _4= Q _4- W _4=3500- x =0$
For cyclic process $\Delta U =0$
$\bullet 3500-4500-1800+3500-x=0$
or $x=700\, J$
Efficiency, $\eta=\frac{\text { output }}{\text { input }} \times 100$
$=\frac{W_1+W_2+W_3+W_4}{Q_1+Q_4} \times 100$
$=\frac{2500-1000-1200+700}{6000+3500} \times 100$
$=\frac{1000}{9500} \times 100$
$ \eta=10.5 \%$