Q. An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states $Q_{1} = 6000\ J,\ Q_{2} = - 5500J,\ Q_{3} = - 3000J,\ Q_{4} = 3500J$ $W_{1} = 2500\ J,\ W_{2} = - 1000J,\ W_{3} = - 1200J,\ W_{4} = xJ$ The ratio of the net work done by the gas to the total heat absorbed by the gas is $\eta$. The values of $x$ and $\eta$ respectively are
A) 500; 7.5%
B) 700; 10.5%
C) 1000; 21%
D) 1500; 15%
Explanation
(b) From first law of thermodynamics,
$Q = \mathrm{\Delta}U + W \Rightarrow \mathrm{\Delta}U = Q - W$
$\therefore\mathrm{\Delta}U_{1} = Q_{1} - W_{1} = 6000 - 2500 = 3500J$
$\mathrm{\Delta}U_{2} = Q_{2} - W_{2} = - 5500 + 1000 = - 4500J$
$\mathrm{\Delta}U_{3} = Q_{3} - W_{3} = - 3000 + 1200 = - 1800J$
$\mathrm{\Delta}U_{4} = Q_{4} - W_{4} = 3500 - x$
For cyclic process, $\mathrm{\Delta}U = 0$
$\therefore 3500 - 4500 - 1800 + 3500 - x = 0 \Rightarrow x = 700J$
Efficiency, $\eta = \frac{output}{input} \times 100$
$= \frac{W_{1} + W_{2} + W_{3} + W_{4}}{Q_{1} + Q_{4}} \times 100 = \frac{1000}{9500} \times 100 \Rightarrow \eta = 10.5\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.