An ideal gas undergoes a process maintaining relation between pressure $$(P)$$ and volume $$(V)$$ as $$P = P_0\left(1 + \left(\frac{V_0}{V}\right)^2\right)^{-1}$$, 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 of the temperatures of these samples, $$T_B - T_A$$ is : 

(R = gas constant)

$$P = \frac{P_0}{1 + \frac{V_0^2}{V^2}} = \frac{P_0 V^2}{V^2 + V_0^2}$$

For sample A at $$V_A = V_0$$: $$P = \frac{P_0 V_0^2}{V_0^2 + V_0^2} = \frac{P_0}{2}$$

For sample B at $$V_B = 3V_0$$: $$P = \frac{P_0 (3V_0)^2}{(3V_0)^2 + V_0^2} = \frac{9P_0 V_0^2}{9V_0^2 + V_0^2} = \frac{9}{10}P_0$$

$$T_A = \frac{P_A V_A}{2R} = \frac{\left(\frac{P_0}{2}\right) V_0}{2R} = \frac{P_0 V_0}{4R}$$

$$T_B = \frac{P_B V_B}{2R} = \frac{\left(\frac{9}{10}P_0\right) (3V_0)}{2R} = \frac{27P_0 V_0}{20R}$$

$$T_B - T_A = \frac{27P_0 V_0 - 5P_0 V_0}{20R} = \frac{22P_0 V_0}{20R} = \frac{11P_0 V_0}{10R}$$