Question

One mole of an ideal gas undergoes a process, whose process equation is given as $P=\frac{P_0}{1+{\left(\frac{V_0}{V}\right)}^2}$ Here $P_0$ and $V_0$ are constants. When the volume of the gas is changed from $V_0$ to $3V_0,$ find the change in the temperature of the gas.

• A. $\frac{7P_0{V}_0}{2R}$
• B. $\frac{11P_0{V}_0}{5R}$
• C. $\frac{7P_0{V}_0}{R}$
• D. $\frac{11P_0{V}_0}{R}$

Answer 1

The correct option is B $\frac{11P_0{V}_0}{5R}$

At initial volume, $V_1=V_0,$ we have pressure $P_1=\frac{P_0}{2}$
and when volume becomes $3V_0,$ we have
Pressure $P_2=\frac{P_0}{1+{\left(\frac{V_0}{3V_0}\right)}^2}=\frac{9P_0}{10}$

Change in temperature of the gas is,
$△T=\frac{P_2{V}_2}{nR}-\frac{P_1{V}_1}{nR}=\frac{1}{nR}[\frac{27P_0{V}_0}{10}-\frac{P_0{V}_0}{2}](\because n=1)$
$\Delta T=\frac{11P_0{V}_0}{5R}$