One mole of an ideal gas passes… - JEE Main 2019 Physics PYQ

JEE Main 2019 — Physics Thermodynamics

2019mcqhard

One mole of an ideal gas passes through a process where pressure and volume obey the relation $P={P}_{o}[1-\frac{1}{2}{(\frac{{V}_{o}}{V})}^{2}]$ . Here ${P}_{o}$ and ${V}_{o}$ are constants. Calculate the change in the temperature of the gas if its volume changes from ${V}_{o}$ to $2{V}_{o}$ .

Official previous-year question

Held on 10 Apr 2019 · Verified 6 Jul 2026.

Options

A
$\frac{1}{4}\frac{{P}_{0}{V}_{0}}{R}$
B
$\frac{5}{4}\frac{{P}_{0}{V}_{0}}{R}$
C
$\frac{1}{2}\frac{{P}_{0}{V}_{0}}{R}$
D
$\frac{3}{4}\frac{{P}_{0}{V}_{0}}{R}$

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Solution

For ideal gas, $PV=nRT$

Therefore, $\frac{nRT}{V}={P}_{0}[1-\frac{1}{2}{(\frac{{V}_{0}}{V})}^{2}]$

$\Rightarrow T=\frac{{P}_{0}V}{R}[1-\frac{1}{2}{(\frac{{V}_{0}}{V})}^{2}]$

Therefore, the change in temperature is:

$\Delta T={T}_{f}-{T}_{i}$

$=\frac{{P}_{0}({2V}_{0})}{R}(1-\frac{1}{2}{(\frac{{V}_{0}}{{2V}_{0}})}^{2})-\frac{{P}_{0}V}{R}(1-\frac{{V}_{0}^{2}}{2{V}_{0}^{2}})$

$=\frac{{5P}_{0}{V}_{0}}{4R}$

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Solution

For ideal gas, $PV=nRT$

Therefore, $\frac{nRT}{V}={P}_{0}[1-\frac{1}{2}{(\frac{{V}_{0}}{V})}^{2}]$

$\Rightarrow T=\frac{{P}_{0}V}{R}[1-\frac{1}{2}{(\frac{{V}_{0}}{V})}^{2}]$

Therefore, the change in temperature is:

$\Delta T={T}_{f}-{T}_{i}$

$=\frac{{P}_{0}({2V}_{0})}{R}(1-\frac{1}{2}{(\frac{{V}_{0}}{{2V}_{0}})}^{2})-\frac{{P}_{0}V}{R}(1-\frac{{V}_{0}^{2}}{2{V}_{0}^{2}})$

$=\frac{{5P}_{0}{V}_{0}}{4R}$

Options

Did you get this right?

Solution

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JEE Main Thermodynamics in other years

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Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics