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) $\dfrac{11P_0V_0}{10R}$
Concept. $T = \dfrac{PV}{nR}$ with $n = 2$; the official answer compares the samples at their initial volumes.
Step 1. Sample A ($V_0$). $P = \dfrac{P_0}{2}$: $T_A = \dfrac{P_0V_0}{4R}$.
Step 2. Sample B ($3V_0$). $P = \dfrac{P_0}{1 + 1/9} = \dfrac{9P_0}{10}$: $T_B = \dfrac{27P_0V_0}{20R}$.
Step 3. Difference.
$$T_B - T_A = \frac{27 - 5}{20}\frac{P_0V_0}{R} = \frac{11P_0V_0}{10R}$$
Answer. $\dfrac{11P_0V_0}{10R}$ (Option B).
Solution by Sreeraj P, M.Sc Physics
