Adiabatic expansion of ideal gas | JEE Main

JEE MAIN 2024

06-04-24 S1

Question

An ideal gas, $\overline{\mathrm{C}}_{\mathrm{V}}=\frac{5}{2} \mathrm{R}$, 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 $\_\_\_\_$ $K$ (nearest integer).
[ $\overline{\mathrm{C}}_{\mathrm{V}}$ is the molar heat capacity at constant volume]

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Solution

$ \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 $\mathrm{T}_1=298$ and $P_2=\frac{5 \mathrm{~T}_2}{2 \times 298}$

Solve and we get $T_2=274.16 \mathrm{~K}$
$ \mathrm{T}_2 \approx 274 \mathrm{~K} $

Question Tags

JEE Main

Chemistry

Hard

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