Q.
An ideal gas is taken through the cycle $A \rightarrow B \rightarrow C \rightarrow A$ as shown in figure. If the net heat supplied to the gas in cycle is 5J, work done by the gas in the process $C \rightarrow A$
A) $- 5$ J
B) $- 10$ J
C) $- 15$ J
D) $- 20$ J
Explanation
(a) $\mathrm{\Delta}W_{AB} = p\mathrm{\Delta}V = 10(2 - 1) = 10J$
$\mathrm{\Delta}W_{BC} = 0$, becuase $V$ is constant,
From first law of thermodynamics,
$\mathrm{\Delta}Q = \mathrm{\Delta}W + \mathrm{\Delta}U$
As $ABCA$ is a cyclic process, therefore,
$\mathrm{\Delta}U = 0$
$\therefore\ \mathrm{\Delta}Q = \mathrm{\Delta}W_{AB} + \mathrm{\Delta}W_{BC} + \mathrm{\Delta}W_{CA}$
$= \mathrm{\Delta}W_{AB} + \mathrm{\Delta}W_{CA}$
or $\mathrm{\Delta}W_{CA} = \mathrm{\Delta}Q - \mathrm{\Delta}W_{AB} = 5 - 10 = - 5\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.