Question

A gas expands from volume $V_0$ to $2V_0$ under three different processes. Process 1 is isobaric, 2 is isothermal and 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

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

Process 3:
This is an adiabatic expansion since volume increases from $V_0$ to $2V_0$.
In adiabatic expansion. internal energy decreases.
$\therefore \Delta U_3$ < 0
Process 2:
In isothermal expansion, since temperature remains constant, internal energy does not change. $\Delta U=0$

Process 1:
$\Delta U_1=n\frac{R}{\gamma -1}\Delta T$
Since $\gamma >1$
$\Delta U_1>0$
$\therefore \Delta U_1>\Delta U_2>\Delta U_3$