Question

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

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

$1$ is isobaric
$V\propto T$
$T\uparrow \Rightarrow \Delta U>0$
$2$ is isothermal
$\Rightarrow \Delta U=0$
$3$ is adiabatic
$\Delta Q=0\Rightarrow \Delta U=-\Delta W\Delta W>0\Rightarrow \Delta U<0$
$\therefore {\left(\Delta U\right)}_1>{\left(\Delta U\right)}_2>{\left(\Delta U\right)}_3$