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QJEEAdv23Chc481: JEE Advanced 2023 Chemistry - Class 11 JEEAdv23Chc481 Solution: A gas has a compressibility factor of 0.5 and a molar volume of …

A gas has a compressibility factor of 0.5 and a molar volume of 0.4 dm3 mol−10.4 ~\mathrm{dm}^3 \mathrm{~mol}^{-1} at a temperature of 800 K800 \mathrm{~K} and pressure x\mathbf{x} atm. If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be y dm3 mol−1\mathbf{y} ~\mathrm{dm}^3 \mathrm{~mol}^{-1} . The value of x/y\mathbf{x} / \mathbf{y} is __________.

[Use: Gas constant, R=8×10−2 L\mathrm{R}=8 \times 10^{-2} \mathrm{~L} atm K−1 mol−1\mathrm{K}^{-1} \mathrm{~mol}^{-1} ]

Step-by-Step Verified Solution

For gas : Z=0.5, Vm=0.4 L/mol\mathrm{Z}=0.5, \mathrm{~V}_{\mathrm{m}}=0.4 \mathrm{~L} / \mathrm{mol}

T=800 K,P=X atm. ⇒Z=PVmRT⇒X(0.4)0.08×800=0.5⇒X=80

For ideal gas, PVm=RT\mathrm{PV}_{\mathrm{m}}=\mathrm{RT}

⇒Vm=RTP=0.08×80080=0.8 L mol−1=y

Then, xy=800.8=100\frac{x}{y}=\frac{80}{0.8}=100 .

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