Question

The volume of one mole of the gas is changed from $V$ to $2V$ at constant pressure $P$. If $\gamma$ is the ratio of specific heats of the gas, change in internal energy of the gas is:

• A. $\frac{\gamma PV}{\gamma -1}$
• B. $\frac{R}{\gamma -1}$
• C. $PV$
• D. $\frac{PV}{\gamma -1}$

Answer 1

The correct option is D $\frac{PV}{\gamma -1}$

Given : $V_1=V$ $V_2=2V$ $P=constant$ $n=1$

Let initial temperature be $T$ i.e $T_1=T$

Using $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ $(\because P=constant)$

$\therefore$ $\frac{V}{T}=\frac{2V}{T_2}$ $⟹T_2=2T$

$\therefore$ Change in temperature $\Delta T=T_2-T_1=2T-T=T$ ..........(1)

From ideal gas equation : $PV=nRT$ where $n=1$

$⟹$ $RT=PV$ .........(2)

Also from equation $\gamma =\frac{f+2}{f}$ $⟹\frac{f}{2}=\frac{1}{\gamma -1}$

$\therefore$ Change in internal energy $\Delta U=\frac{f}{2}nR\Delta T$

$\therefore$ $\Delta U=\frac{1}{\gamma -1}\times 1\times RT$

$⟹$ $\Delta U=\frac{PV}{\gamma -1}$ (using (2))