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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2024 · 6 April · Morning Shift · Chemistry · Thermodynamics (C)

An ideal gas, ( C _ v = (5)/(2) R ), is expanded adiabatically...

An ideal gas, (\overline{\mathrm{C}}_{\mathrm{v}}=52R\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\mathrm{~K}) and (5  atm\mathrm{~atm}), respectively then the final temperature is _______ (K\mathrm{K}) (nearest integer). [(\overline{\mathrm{C}}_{\mathrm{v}}) is the molar heat capacity at constant volume]

Correct answer

0

Step-by-step solution

ΔU=q+w(q=0)nCVΔT=−Pext(V2−V1)V2=2 V1nRT2P2=2nRT1P1P1=5, T1=298P2=5 T22×298n52R(T2−T1)=−1(nRT2P1−nRT1P1)\begin{aligned} & \Delta \mathrm{U}=\mathrm{q}+\mathrm{w}(\mathrm{q}=0) \\ & \mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}=-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\ & \mathrm{V}_2=2 \mathrm{~V}_1 \\ & \frac{\mathrm{nRT}_2}{\mathrm{P}_2}=\frac{2 \mathrm{nRT}_1}{\mathrm{P}_1} \\ & \mathrm{P}_1=5, \mathrm{~T}_1=298 \\ & \mathrm{P}_2=\frac{5 \mathrm{~T}_2}{2 \times 298} \\ & \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) \end{aligned}
Put T1=298\mathrm{T}_1=298 and P2=5 T22×298\mathrm{P}_2=\frac{5 \mathrm{~T}_2}{2 \times 298} Solve and we get T2=274.16 KT2≈274 KT_2=274.16 \mathrm{~K} \mathrm{T}_2 \approx 274 \mathrm{~K}
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