Initially a gas of diatomic molecules is… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Thermodynamics

2020integermedium

Initially a gas of diatomic molecules is contained in a cylinder of volume ${V}_{1}$ at a pressure ${P}_{1}$ and temperature $250K$ . Assuming that $25%$ of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature $2000K,$ when contained in a volume $2{V}_{1}$ is given by ${P}_{2}$ . The ratio ${P}_{2}/{P}_{1}$ is -

Official previous-year question

Held on 6 Sept 2020 · Verified 6 Jul 2026.

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Solution

${P}_{1}={P}_{1}$

${v}_{1}={v}_{1}$

${T}_{1}=250$

${n}_{1}={n}_{0}$

${n}_{2}=0.75{n}_{0}+2\times 0.25{n}_{0}=1.25\mathrm{no}$

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

${T}_{2}=2000$

${P}_{1}{V}_{1}={n}_{1}{\mathrm{RT}}_{1}$

${P}_{2}{V}_{2}={n}_{2}{\mathrm{RT}}_{2}$

${P}_{1}{V}_{1}={n}_{0}R\times 250$

${P}_{2}(2{V}_{1})=(.25{n}_{0}R(2000))$

equation $(1)/(2)$

$\frac{{P}_{1}}{2{P}_{2}}=\frac{250}{1.25\times 2000}$

$\frac{{P}_{1}}{{P}_{2}}=\frac{1}{5}$

$\frac{{P}_{2}}{{P}_{1}}=5$

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