Chemistry - Real Gas Question with Solution | TestHub
ChemistryReal GasCompressibility FactorMedium2 minQB
ChemistryMediumsingle choice
One mole of a gas occupies 0.5 l at 27°C. The compressibility factor of gas at this temperature is 0.8. If b = 0.04 l mol-1, what is the value of van der Waal's constant a for the gas?
Options:
Answer:
A
Solution:
Using the truncated virial form of the van der Waals equation, Z = 1 + (b - a/RT)/(Vm): 0.8 = 1 + (0.04 - a/24.63)/0.5 (with RT = 0.0821 × 300 = 24.63). So 0.04 - a/24.63 = -0.1, giving a/24.63 = 0.14 and a = 3.44 atm l2 mol-2. Ans: (a)
Stream:JEESubject:ChemistryTopic:Real GasSubtopic:Compressibility Factor
⏱ 2mℹ️ Source: QB
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