Chemistry - Real Gas Question with Solution | TestHub
ChemistryReal GasCompressibility FactorEasy2 minQB
ChemistryEasysingle choice
The expression for compressibility factor for one mole of a van der Waal's gas at low pressure, neglecting higher order terms, is
Options:
Answer:
A
Solution:
From P = (RT)/(V-b) - a/(V²), Z = V/(V-b) - a/(RTV). At low pressure (V large), V/(V-b) ≈ 1 + b/V, so Z ≈ 1 + b/V - a/(RTV) = 1 + (b - a/RT)/V. Ans: (a)
Stream:JEESubject:ChemistryTopic:Real GasSubtopic:Compressibility Factor
⏱ 2mℹ️ Source: QB
Doubts & Discussion
Loading discussions...
