Question

One mole of an ideal gas undergoes a process :
$P=\frac{P_0}{1+(V_0/V{)}^2}$
Here $P_0$ and $V_0$ are constants. Change in temparature of the gas when volume is changed from $V=V_0$ to $V=2V_0$ is

• A. $-\frac{2P_0{V}_0}{5R}$
• B. $\frac{11P_0{V}_0}{10R}$
• C. $-\frac{5P_0{V}_0}{4R}$
• D. $P_0{V}_0$

Answer 1

The correct option is B $\frac{11P_0{V}_0}{10R}$

At $V=V_0:P=\frac{P_0}{2}$
$\therefore T_1=\frac{PV}{nR}=\frac{\left(\frac{P_0}{2}\right)V_0}{R}=\frac{P_0{V}_0}{2R}(n=1)$
and at $V=2V_0:P=\frac{4P_0}{5}$

$\therefore T_f=\frac{PV}{nR}=\frac{\left(2V_0\right)\left(\frac{4P_0}{5}\right)}{R}=\frac{8P_0{V}_0}{5R}$

$\therefore \Delta T=T_f-T_i=(\frac{8}{5}-\frac{1}{2})\frac{P_0{V}_0}{R}=\frac{11P_0{V}_0}{10R}$
Hence, the correct answer is option (b).