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JEE Main 2022 · 27 Jul · Shift 2, Chemistry question 82

For a real gas at 25°C temperature and high pressure (99 bar) the value of compressibility factor is 2, so the value of Vander Waal's constant 'b' should be ______ ×10−2\times10^{-2} L mol−1^{-1} (Nearest integer)

(Given R = 0.083 L bar K−1^{-1}mol−1^{-1})

Answer and worked solution

Official answer: 25

High pressure. Z=1+PbRTZ=1+\frac{Pb}{RT}, so b=0.083×29899≈0.25b=\frac{0.083\times298}{99}\approx0.25 L mol−1^{-1}.

Answer: 25

Topics:Real gases and van der Waals equation
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