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JEE MAIN 2023
13-4-23 S1
Question
A certain quantity of real gas occupies a volume of 0.15 dm3 at 100 atm and 500 K when its compressibility factor is 1.07. Its volume at 300 atm and 300K (When its compressibility factor is 1.4) is ____ × 10–4 dm3 (Nearest integer)
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Solution
$\begin{aligned} & \text { Sol. } \mathrm{Z}=\frac{\mathrm{PV}}{\mathrm{nRT}} ; \mathrm{n}=\frac{\mathrm{PV}}{\mathrm{ZRT}} \\ & \mathrm{Z}_1= 1.07, \mathrm{P}_1=100 \mathrm{~atm}, \mathrm{~V}_1=0.15 \mathrm{~L}, \mathrm{~T}_1=500 \mathrm{~K} \\ & \mathrm{Z}_2=1.4, \mathrm{P}_2=300 \mathrm{~atm}, \mathrm{~T}_2=300 \mathrm{~K}, \mathrm{~V}_2=2 \\ & \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{Z}_1 \mathrm{R} \mathrm{T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{Z}_2 \mathrm{RT} \mathrm{T}_2}=\mathrm{n} \\ & \mathrm{V}_2= \frac{1.4}{1.07} \times .03=392 \times 10^{-4} \mathrm{dm}^3\end{aligned}$
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