An ideal gas undergoes a process maintaining relation between...

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: $\dfrac{11P_o V_o}{10R}$