JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

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)

A
$\frac{9 P _{o} V _{o}}{8 R}$
B
$\frac{11 P _{o} V _{o}}{10 R}$
C
$\frac{7 P _{o} V _{o}}{6 R}$
D
$\frac{13 P _{o} V _{o}}{11 R}$

Solution

Evaluate $P$ at the two volumes.

At $V = V_{0}$ : $P_{A} = \frac{P _{0}}{1 + 1} = \frac{P _{0}}{2} .$ Using $P V = n R T$ with $n = 2$ :

$\frac{P _{0}}{2} \cdot V_{0} = 2 R T_{A} \Rightarrow T_{A} = \frac{P _{0} V _{0}}{4 R} .$

At $V = 3 V_{0}$ : $P_{B} = \frac{P _{0}}{1 + 1/9} = \frac{9 P _{0}}{10} .$

$\frac{9 P _{0}}{10} \cdot 3 V_{0} = 2 R T_{B} \Rightarrow T_{B} = \frac{27 P _{0} V _{0}}{20 R} .$

$T_{B} - T_{A} = \frac{27 P _{0} V _{0}}{20 R} - \frac{P _{0} V _{0}}{4 R} = \frac{27 P _{0} V _{0} - 5 P _{0} V _{0}}{20 R} = \frac{22 P _{0} V _{0}}{20 R} = \frac{11 P _{0} V _{0}}{10 R} .$

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →