Q.
A gas is expanded from volume $V_{0}$ to $2V_{0}$ under three different processes, in figure process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let $\mathrm{\Delta}U_{1},\ \mathrm{\Delta}U_{2}$ and $\mathrm{\Delta}U_{3}$ be the change in internal energy of the gas in these three processes. Then
A) $\mathrm{\Delta}U_{1} > \mathrm{\Delta}U_{2} > \mathrm{\Delta}U_{3}$
B) $\mathrm{\Delta}U_{1} < \mathrm{\Delta}U_{2} < \mathrm{\Delta}U_{3}$
C) $\mathrm{\Delta}U_{2} < \mathrm{\Delta}U_{1} > \mathrm{\Delta}U_{3}$
D) $\mathrm{\Delta}U_{2} < \mathrm{\Delta}U_{3} < \mathrm{\Delta}U_{1}$
Explanation
(a) Process 1 is isobaric $(p =$ constant) expansion
Hence, temperature of gas will increase
$\therefore\ \mathrm{\Delta}U_{1} =$ negative
Process 2 is an adiabatic expansion
$\therefore\ \mathrm{\Delta}U_{2} = 0$
Process 3 is an adiabatic expansion
Hence, temperature of gas will fall
$\therefore\ \ \mathrm{\Delta}U_{3} =$ constant
$\therefore\ \ \mathrm{\Delta}U_{1} > \mathrm{\Delta}U_{2} > \mathrm{\Delta}U_{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.