Verified Solution • 2025–26 Session moderate

QJEEM22Ch9d85: JEE Main 2022 Chemistry - Class 11 JEEM22Ch9d85 Solution: The molar heat capacity for an ideal gas at constant pressure…

The molar heat capacity for an ideal gas at constant pressure is 20.785 J K−1 mol−120.785 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} . The change in internal energy is 5000 J5000 \mathrm{~J} upon heating it from 300 K300 \mathrm{~K} to 500 K500 \mathrm{~K} . The number of moles of the gas at constant volume is ____________. [Nearest integer] (Given: R=8.314 J K−1 mol−1\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} )

Step-by-Step Verified Solution

Cp=20.785 J K−1 mol−1\mathrm{C}_{\mathrm{p}}=20.785 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

amp; and ΔU=nCvΔTamp;∴nCv=5000200=25\begin{aligned}&\text { and } \Delta \mathrm{U}=\mathrm{nC} \mathrm{v} \Delta \mathrm{T} \\\\&\therefore \quad \mathrm{nC}_{\mathrm{v}}=\frac{5000}{200}=25\end{aligned}

and we know that

amp;Cp−Cv=Ramp;20.785−25n=8.314amp;n=25(20.785−8.314)=2\begin{aligned}&C_{p}-C_{v}=R \\\\&20.785-\frac{25}{n}=8.314 \\\\&n=\frac{25}{(20.785-8.314)}=2\end{aligned}

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