JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Heat & Thermodynamics

An ideal gas undergoes a process maintaining relation between pressure (P) and volume (V) as

P = P_{o} (1 + (\frac{V _{o}}{V})^{2})^{- 1}

, where

P_{o}

and

V_{o}

are constants. If two samples A and B (two moles each) with initial volume

V_{o}

and

3 V_{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)

Answer: (B)

Solution

V = V_{o}

the pressure is

P = \frac{P _{o}}{1 + 1} = \frac{P _{o}}{2}

, so for the two moles of A,

\frac{P _{o}}{2} V_{o} = 2 R T_{A}

gives

T_{A} = \frac{P _{o} V _{o}}{4 R}

. At

V = 3 V_{o}

the pressure is

P = \frac{P _{o}}{1 + ( 1/3 ) ^{2}} = \frac{9 P _{o}}{10}

, so

\frac{9 P _{o}}{10} (3 V_{o}) = 2 R T_{B}

gives

T_{B} = \frac{27 P _{o} V _{o}}{20 R}

. The difference is

T_{B} - T_{A} = \frac{27 P _{o} V _{o}}{20 R} - \frac{5 P _{o} V _{o}}{20 R} = \frac{11 P _{o} V _{o}}{10 R}

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option