Question

A gas is expanded from volume $V_0$ to $2V_0$ under three different processes, as shown in the figure. Process $1$ is 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 expansion ($P =$ constant)
Hence temperature of gas will increase.
$\therefore \Delta U_1 =$ positive
Process $2$ is an isothermal process
$\therefore \Delta U_2 = 0 $
Process $3$ is an adiabatic expansion
Hence temperature of gas will fall.
$\therefore \Delta U_3 =$ negative
$\therefore \Delta U_1 > \Delta U_2 > \Delta U_3$