The change in the magnitude of the volume… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Thermodynamics

2020integermedium

The change in the magnitude of the volume of an ideal gas when a small additional pressure $\Delta P$ is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $\Delta T$ at constant pressure. The initial temperature and pressure of the gas were $300K$ and $2\mathrm{atm}$ respectively. If $|\Delta T|=C|\Delta P|$ then value of $C$ in $(K/\mathrm{atm})$ is __________

Official previous-year question

Held on 4 Sept 2020 · Verified 6 Jul 2026.

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Solution

$\mathrm{PV}=\mathrm{nRT}$

$P\Delta V+V\Delta P=0$

$\Delta V=-\frac{\Delta P}{P}V....(i)$

In second case

$P\Delta V=-nR\Delta T$

$\Delta V=-\frac{nR\Delta T}{P}...(\mathrm{ii})$

equating $(i)$ and $(\mathrm{ii})$

$\frac{nR\Delta T}{P}=-\frac{\Delta P}{P}V$

$\Delta T=\Delta P\frac{V}{\mathrm{nR}}$

$c=\frac{v}{\mathrm{nR}}$

Putting the value of $V,n$ and $R,C=150$

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