An ideal gas undergoes a process maintaining relation between pressure \((P)\) and volume \((V)\) as \(P = P_o\left(1 + \left(\dfrac{V_o}{V}\right)^2\right)^{-1}\), where \(P_o\) and \(V_o\) are constants. If two samples \(A\) and \(B\) (two moles each) with initial volumes \(V_o\) and \(3V_o\) 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)

The given relation between pressure and volume is:

\(P = P_o\left(1 + \left(\dfrac{V_o}{V}\right)^2\right)^{-1}\)

For sample A, the initial volume is \(V_A = V_o\). Substituting this into the given relation, the initial pressure \(P_A\) is:

\(P_A = P_o\left(1 + \left(\dfrac{V_o}{V_o}\right)^2\right)^{-1} = \dfrac{P_o}{2}\)

Using the ideal gas equation \(PV = nRT\) for \(n = 2\) moles, the initial temperature \(T_A\) is:

\(T_A = \dfrac{P_A V_A}{nR} = \dfrac{\left(\dfrac{P_o}{2}\right) V_o}{2R} = \dfrac{P_o V_o}{4R}\)

For sample B, the initial volume is \(V_B = 3V_o\). Substituting this into the relation, the initial pressure \(P_B\) is:

\(P_B = P_o\left(1 + \left(\dfrac{V_o}{3V_o}\right)^2\right)^{-1} = P_o\left(1 + \dfrac{1}{9}\right)^{-1} = \dfrac{9P_o}{10}\)

The initial temperature \(T_B\) for \(n = 2\) moles is:

\(T_B = \dfrac{P_B V_B}{nR} = \dfrac{\left(\dfrac{9P_o}{10}\right) (3V_o)}{2R} = \dfrac{27P_o V_o}{20R}\)

The difference in the initial temperatures of the two samples is:

\(T_B - T_A = \dfrac{27P_o V_o}{20R} - \dfrac{P_o V_o}{4R}\)

\(T_B - T_A = \dfrac{27P_o V_o - 5P_o V_o}{20R} = \dfrac{22P_o V_o}{20R} = \dfrac{11P_o V_o}{10R}\)

Answer: \(\dfrac{11P_o V_o}{10R}\)