Question

A gas is expanded from volume $V_{0}$ to $2V_{0}$ under three different processes. In figure process $1$ is an isobaric process, process $2$ is isothermal and process $3$ is adiabatic. Let $\Delta U_{1}, \, \Delta U_{2}$ and $\Delta U_{3}$ be the change in internal energy of the gas in these three processes. Then

• A. $\Delta U_{1}>\Delta U_{2}>\Delta U_{3}$
• B. $\Delta U_{1} < \Delta U_{2} < \Delta U_{3}$
• C. $\Delta U_{2} < \Delta U_{1} < \Delta U_{3}$
• D. $\Delta U_{2} < \Delta U_{3} < \Delta U_{1}$

Answer 1

Answer: (A)

Process $1$ is isobaric $\left(\right.p=$ constant) expansion
Hence, temperature of gas will increase
$\therefore \, \Delta U_{1}=$ positive
Process $2$ is an isothermal expansion
$\therefore \Delta U_{2}=0$
Process $3$ is an adiabatic expansion
Hence, temperature of gas will fall
$\therefore \Delta U_{3}=$ negative
$\therefore \Delta U_{1}>\Delta U_{2}>\Delta U_{3}$