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Physics · 2026

JEE Main · 4 April 2026, Shift 1 · Q36

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…

An ideal gas undergoes a process maintaining relation between pressure $\displaystyle (P)$ and volume $\displaystyle (V)$ as $\displaystyle P=P_{\mathrm{o}}\left(1+\left(\frac{V_{\mathrm{o}}}{V}\right)^2\right)^{-1}$, where $\displaystyle P_{\mathrm{o}}$ and $\displaystyle V_{\mathrm{o}}$ are constants. If two samples $\displaystyle A$ and $\displaystyle B$ (two moles each) with initial volumes $\displaystyle V_{\mathrm{o}}$ and $\displaystyle 3 V_{\mathrm{o}}$ respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, $\displaystyle T_B-T_A$ is $\displaystyle \_\_\_\_$. ( $\displaystyle R=$ gas constant)
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.