Temperature Difference for Ideal Gas Process with Given Relation

To find the difference between the temperatures of the two gas samples, $T_B - T_A$, we use the equation of state for an ideal gas, $P V = n R T$, alongside the given process equation relating pressure $P$ and volume $V$:

$P(V) = P_0 \left(1 + \left(\frac{V_0}{V}\right)^2\right)^{-1} = \frac{P_0}{1 + \frac{V_0^2}{V^2}}$

Given:

• Number of moles for both samples: $n_A = n_B = 2$
• Volume of sample $A$: $V_A = V_0$
• Volume of sample $B$: $V_B = 3V_0$

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Step 1: Calculate the parameters for Sample $A$

1. Pressure of Sample $A$ ($P_A$): $P_A = \frac{P_0}{1 + \left(\frac{V_0}{V_0}\right)^2} = \frac{P_0}{1 + 1} = \frac{P_0}{2}$

2. Temperature of Sample $A$ ($T_A$): Using the ideal gas law $P_A V_A = n_A R T_A$: $T_A = \frac{P_A V_A}{n_A R} = \frac{\left(\frac{P_0}{2}\right) V_0}{2 R} = \frac{P_0 V_0}{4 R}$

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Step 2: Calculate the parameters for Sample $B$

1. Pressure of Sample $B$ ($P_B$): $P_B = \frac{P_0}{1 + \left(\frac{V_0}{3V_0}\right)^2} = \frac{P_0}{1 + \frac{1}{9}} = \frac{P_0}{\frac{10}{9}} = \frac{9 P_0}{10}$

2. Temperature of Sample $B$ ($T_B$): Using the ideal gas law $P_B V_B = n_B R T_B$: $T_B = \frac{P_B V_B}{n_B R} = \frac{\left(\frac{9 P_0}{10}\right) (3 V_0)}{2 R} = \frac{27 P_0 V_0}{20 R}$

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Step 3: Find the temperature difference $T_B - T_A$

$T_B - T_A = \frac{27 P_0 V_0}{20 R} - \frac{P_0 V_0}{4 R}$

Making a common denominator of $20$:

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

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Conclusion:

The correct option is B: $\frac{11P_0 V_0}{10R}$.