Question

An ideal gas is expanded from volume $V_0$ to $2V_0$ under three different processes. 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 respectively, 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

The correct option is A $\Delta U_1>\Delta U_2>\Delta U_3$

Internal energy for an ideal gas, $\Delta U=n{C}_V\Delta T$

Process $1$ is an isobaric ($P=$ constant) expansion.
Using ideal gas equation, $PV=nRT$,
$\Rightarrow V\propto T$
Since the volume of the gas increases, therefore temperature of the gas will increase,
$\Delta U_1$ = positive

Process $2$ is an isothermal process, thus $\Delta T=0$
$\therefore \Delta U_2=0$

Process $3$ is an adiabatic expansion, therefore gas will do work at the expense of internal energy.
$\Delta U=W$ $(Q=0)$
Hence, internal energy of the gas decreases,
$\Delta U_3$ = negative

Therefore, $\Delta U_1>\Delta U_2>\Delta U_3$ is the correct option.