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
$Q_1=6000J,Q_2=-5500J,Q_3=-3000J,Q_4=3500J,$
$W_1=2500J,W_2=-1000J,W_3=-1200J\&W_4=xJ.$
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, $\Delta U_1=Q_1-W_1=6000-2500=3500J$
$\Delta U_2=Q_2-W_2=-5500+1000=-4500J$
$\Delta U_3=Q_3-W_3$
$=-3000+1200=-1800J$
$\Delta U_4=Q_4-W_4=3500-x=0$
For cyclic process $\Delta U=0$
$\bullet 3500-4500-1800+3500-x=0$
or $x=700J$
Efficiency, $\eta =\frac{\text{ output }}{\text{ input }}\times 100$
$=\frac{W_1+W_2+W_3+W_4}{Q_1+Q_4}\times 100$
$=\frac{2500-1000-1200+700}{6000+3500}\times 100$
$=\frac{1000}{9500}\times 100$
$\eta =10.5\%$