Q. A sample of an ideal gas is taken through a cycle a shown in figure. It absorbs $50J$ of energy during the process $AB$, no heat during $BC$, rejects $70J$ during $CA.\ 40J$ of work is done on the gas during $BC$. Internal energy of gas at $A$ is $1500J$, the internal energy at $C$ would be
A) $1590\ J$
B) $1620\ J$
C) $1540\ J$
D) $1570\ J$
Explanation
(a) $\mathrm{\Delta}W_{AB} = 0$ as $V =$ constant
$\therefore\mathrm{\Delta}Q_{AB} = \mathrm{\Delta}U_{AB} = 50J$ [Given]
$U_{A} = 1500J$
$\therefore U_{B} = (1500 + 50)J = 1550J$
$\mathrm{\Delta}W_{BC} = - \mathrm{\Delta}U_{BC} = - 40J$ [Given]
$\therefore\mathrm{\Delta}U_{BC} = 40J$
$\therefore U_{C} = (1550 + 40)J = 1590J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.