An ideal gas undergoes a process maintaining relation between pressure (P) and volume (V) as P = P_o (1 + ( V_o V )^2 )^-1, where P_o and V_o are…

The given relation between pressure and volume is:

$$P = P_o\left(1 + \left(\dfrac{V_o}{V}\right)^2\right)^{-1}$$P=Po​(1+(VVo​​)2)−1

For sample A, the initial volume is $V_A = V_o$VA​=Vo​. Substituting this into the given relation, the initial pressure $P_A$PA​ is:

$$P_A = P_o\left(1 + \left(\dfrac{V_o}{V_o}\right)^2\right)^{-1} = \dfrac{P_o}{2}$$PA​=Po​(1+(Vo​Vo​​)2)−1=2Po​​

Using the ideal gas equation $PV = nRT$PV=nRT for $n = 2$n=2 moles, the initial temperature $T_A$TA​ is:

$$T_A = \dfrac{P_A V_A}{nR} = \dfrac{\left(\dfrac{P_o}{2}\right) V_o}{2R} = \dfrac{P_o V_o}{4R}$$TA​=nRPA​VA​​=2R(2Po​​)Vo​​=4RPo​Vo​​

For sample B, the initial volume is $V_B = 3V_o$VB​=3Vo​. Substituting this into the relation, the initial pressure $P_B$PB​ is:

$$P_B = P_o\left(1 + \left(\dfrac{V_o}{3V_o}\right)^2\right)^{-1} = P_o\left(1 + \dfrac{1}{9}\right)^{-1} = \dfrac{9P_o}{10}$$PB​=Po​(1+(3Vo​Vo​​)2)−1=Po​(1+91​)−1=109Po​​

The initial temperature $T_B$TB​ for $n = 2$n=2 moles is:

$$T_B = \dfrac{P_B V_B}{nR} = \dfrac{\left(\dfrac{9P_o}{10}\right) (3V_o)}{2R} = \dfrac{27P_o V_o}{20R}$$TB​=nRPB​VB​​=2R(109Po​​)(3Vo​)​=20R27Po​Vo​​

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

$$T_B - T_A = \dfrac{27P_o V_o}{20R} - \dfrac{P_o V_o}{4R}$$TB​−TA​=20R27Po​Vo​​−4RPo​Vo​​

$$T_B - T_A = \dfrac{27P_o V_o - 5P_o V_o}{20R} = \dfrac{22P_o V_o}{20R} = \dfrac{11P_o V_o}{10R}$$TB​−TA​=20R27Po​Vo​−5Po​Vo​​=20R22Po​Vo​​=10R11Po​Vo​​

Answer: $\dfrac{11P_o V_o}{10R}$10R11Po​Vo​​