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Chemistry · Thermodynamics & Thermochemistry

JEE Main 2024 — 6 April, Shift 1 — Question 84

An ideal gas, C‾V=52R\overline{\mathrm{C}}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}, is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm , respectively then the final temperature is \qquad K (nearest integer). [ C‾V\overline{\mathrm{C}}_{\mathrm{V}} is the molar heat capacity at constant volume]

Answer: 274

Numerical answer — enter this value.

Step-by-step solution

ΔU=q+w(q=0)\quad \Delta \mathrm{U}=\mathrm{q}+\mathrm{w}(\mathrm{q}=0)

nCVΔT=−Pext(V2−V1)\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}=-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)

V2=2 V1\mathrm{V}_{2}=2 \mathrm{~V}_{1} nRT2P2=2nRT1P1\frac{\mathrm{nRT}_{2}}{\mathrm{P}_{2}}=\frac{2 \mathrm{nRT}_{1}}{\mathrm{P}_{1}}

P1=5, T1=298\mathrm{P}_{1}=5, \mathrm{~T}_{1}=298 P2=5 T22×298\mathrm{P}_{2}=\frac{5 \mathrm{~T}_{2}}{2 \times 298} n52R(T2−T1)=−1(nRT2P1−nRT1P1)\mathrm{n} \frac{5}{2} \mathrm{R}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)=-1\left(\frac{\mathrm{nRT}_{2}}{\mathrm{P}_{1}}-\frac{\mathrm{nRT}_{1}}{\mathrm{P}_{1}}\right)

Put T1=298\mathrm{T}_{1}=298

and P2=5 T22×298P_{2}=\frac{5 \mathrm{~T}_{2}}{2 \times 298}

Solve and we get T2=274.16 K\mathrm{T}_{2}=274.16 \mathrm{~K}

T2≈274 K\mathrm{T}_{2} \approx 274 \mathrm{~K}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
6 April, Shift 1
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Internal Energy and the First Law