An ideal gas undergoes a process maintaining relation between pressure $(P)$ and volume $(V)$ as $P = P_0\left(1 + \left(\dfrac{V_0}{V}\right)^2\right)^{-1}$, where $P_0$ and $V_0$ are constants. If two samples $A$ and $B$ (two moles each) with initial volumes $V_0$ and $3V_0$ respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, $T_B - T_A$ is ______.
($R =$ gas constant)
Answer: (B) $\dfrac{11P_0V_0}{10R}$
The official answer compares the two samples at their initial volumes, using $T = \dfrac{PV}{nR}$ with $n = 2$.
Sample A ($V = V_0$): $P = \dfrac{P_0}{2}$, so $T_A = \dfrac{P_0V_0}{4R}$.
Sample B ($V = 3V_0$): $P = \dfrac{P_0}{1 + 1/9} = \dfrac{9P_0}{10}$, so $T_B = \dfrac{9P_0}{10} \times \dfrac{3V_0}{2R} = \dfrac{27P_0V_0}{20R}$.
$T_B - T_A = \dfrac{27 - 5}{20}\cdot\dfrac{P_0V_0}{R} = \dfrac{11P_0V_0}{10R}$.
Solution by Sreeraj P, M.Sc Physics