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← JEE Advanced 2023 · Paper 1
ChemistryInteger+4Question 43 of 51

JEE Advanced 2023 · Paper 1, Chemistry question 9

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

[Use: Gas constant, R=8×10−2 L atm K−1 mol−1R = 8\times10^{-2}\,\mathrm{L\,atm\,K^{-1}\,mol^{-1}}]

Answer and worked solution

Official answer: 100

Real gas. Z=pVmRTZ=\dfrac{pV_m}{RT}:
0.5=x×0.40.08×800  ⟹  x=80 atm.0.5=\frac{x\times0.4}{0.08\times800}\implies x=80\ \text{atm}.

Ideal gas at the same TT and pp:
y=RTp=6480=0.8 dm3 mol−1.y=\frac{RT}{p}=\frac{64}{80}=0.8\ \text{dm}^3\,\text{mol}^{-1}.

xy=800.8=100.\frac xy=\frac{80}{0.8}=100.

Topics:Real gases and van der Waals equationIdeal gas equation
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