A certain amount of gas of… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Thermodynamics

2022mcqeasy

A certain amount of gas of volume $V$ at $27^{\circ}C$ temperature and pressure $2\times {10}^{7}N{m}^{-2}$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use $\gamma =1.5$ )

Official previous-year question

Held on 25 Jul 2022 · Verified 6 Jul 2026.

Options

A
$3.536\times {10}^{5}\mathrm{Pa}$
B
$3.536\times {10}^{6}\mathrm{Pa}$
C
$1.25\times {10}^{6}\mathrm{Pa}$
D
$1.25\times {10}^{5}\mathrm{Pa}$

Did you get this right?

Solution

Given here, ${P}_{1}=2\times {10}^{7}N{m}^{-2}=2\times {10}^{7}\mathrm{Pa}$

Using Boyle's law,

${P}_{1}{V}_{1}={P}_{2}{V}_{2}$

Since ${V}_{2}=2{V}_{1}$ .

Hence, ${P}_{2}=\frac{{P}_{1}}{2}$ (isothermal expansion)

${P}_{2}=1\times {10}^{7}\mathrm{Pa}$

Now, during adiabatic expansion,

${P}_{2}{({V}_{2})}^{\gamma }={P}_{3}{(2{V}_{2})}^{\gamma }$

${P}_{3}=\frac{1\times {10}^{7}}{{2}^{1.5}}=3.536\times {10}^{6}\mathrm{Pa}$

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Solution

Given here, ${P}_{1}=2\times {10}^{7}N{m}^{-2}=2\times {10}^{7}\mathrm{Pa}$

Using Boyle's law,

${P}_{1}{V}_{1}={P}_{2}{V}_{2}$

Since ${V}_{2}=2{V}_{1}$ .

Hence, ${P}_{2}=\frac{{P}_{1}}{2}$ (isothermal expansion)

${P}_{2}=1\times {10}^{7}\mathrm{Pa}$

Now, during adiabatic expansion,

${P}_{2}{({V}_{2})}^{\gamma }={P}_{3}{(2{V}_{2})}^{\gamma }$

${P}_{3}=\frac{1\times {10}^{7}}{{2}^{1.5}}=3.536\times {10}^{6}\mathrm{Pa}$

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Solution

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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