JEE Main 2026 · 4 Apr · Shift 1 · Physics Q36

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An ideal gas undergoes a process maintaining relation between pressure ( $P$ ) and volume ( $V$ ) as $P=P_0\left(1+\left(\dfrac{V_0}V\right)^2\right)^{-1}$ , where $P_0$ and $V_0$ are constants. If two samples $A$ and $B$ (two moles each) with initial volumes $V_0$ and $3V_0$ 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)

\dfrac{9P_0V_0}{8R}

\dfrac{11P_0V_0}{10R}

\dfrac{7P_0V_0}{6R}

\dfrac{13P_0V_0}{11R}

Answer and worked solution

Official answer: Dropped: full marks to everyone

This question was dropped in NTA's final key. Below is the standard working, which shows the ambiguity.

P increases with V along the given process, so two samples (both 2 mol) at the same pressure have the same volume and therefore the same temperature: read literally, $T_B-T_A=0$ , which is not an option.

The likely intended reading compares the starting states. With $T=\dfrac{PV}{2R}$ :

A at $V_0$ : $P=\frac{P_0}2$ , so $T_A=\dfrac{P_0V_0}{4R}$ .
B at $3V_0$ : $P=\frac{9P_0}{10}$ , so $T_B=\dfrac{27P_0V_0}{20R}$ .

That gives $T_B-T_A=\dfrac{11P_0V_0}{10R}$ (B).

Topics:Thermodynamic processes Ideal gas equation

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JEE Main 2026 · 4 Apr · Shift 1, Physics question 36