Q. An ideal gas is made to go through a cyclic thermodynamical process in four steps. The amount of heat involved are $Q_{1} = 600\ J,\ Q_{2} = - 400\ J,\ Q_{3} = - 300\ J$ and $Q_{4} = 200\ J$ respectively. The corresponding work involved are $W_{1} = 300\ J,\ W_{2} = - 200\ J,\ W_{3} = - 150\ J$ and $W_{4}$. What is the value of $W_{4}$
A) $- 50J$
B) $100J$
C) $150J$
D) $50J$
Explanation
(c) From the first law of thermodynamics,
$\mathrm{\Delta}Q = \mathrm{\Delta}U + \mathrm{\Delta}W$
For a cyclic process, $\mathrm{\Delta}U = 0$
$\therefore\mathrm{\Delta}Q = \mathrm{\Delta}W$
$\mathrm{\Delta}Q = Q_{1} + Q_{2} + Q_{3} + Q_{4}$
$= 600J - 400J - 300J + 200J = 100J$
$\mathrm{\Delta}W = W_{1} + W_{2} + W_{3} + W_{4}$
$\mathrm{\Delta}W = 300J - 200J - 150J + W_{4}$
$= - 50J + W_{4}$
Substitute the value of $\mathrm{\Delta}Q$ and $\mathrm{\Delta}W$ in eqn. (i), we get
$100J = - 50J + W_{4}$ or $W_{4} = 150J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.