JEE Main 2026 — Thermodynamics Question with Solution

The given relation between pressure and volume is:

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

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

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

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

$T_A=\frac{P_A{V}_A}{nR}=\frac{\left(\frac{P_o}{2}\right)V_o}{2R}=\frac{P_o{V}_o}{4R}$

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

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

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

$T_B=\frac{P_B{V}_B}{nR}=\frac{\left(\frac{9P_o}{10}\right)(3V_o)}{2R}=\frac{27P_o{V}_o}{20R}$

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

$T_B-T_A=\frac{27P_o{V}_o}{20R}-\frac{P_o{V}_o}{4R}$

$T_B-T_A=\frac{27P_o{V}_o-5P_o{V}_o}{20R}=\frac{22P_o{V}_o}{20R}=\frac{11P_o{V}_o}{10R}$

Answer: $\frac{11P_o{V}_o}{10R}$