Chemistry - Real Gas Question with Solution | TestHub
ChemistryReal GasCompressibility FactorMedium2 minQB
ChemistryMediumsingle choice
At moderate pressure, the compressibility factor for a gas is given as: Z = 1 + 0.35P − (168/T)·P, where P is in bar and T is in Kelvin. What is the Boyle's temperature of the gas?
Options:
Answer:
B
Solution:
Z = 1 + (0.35 - 168/T)P. At the Boyle temperature the coefficient of P vanishes (initial slope of Z vs P is zero): 0.35 = 168/T, so T = 168/0.35 = 480 K. Ans: (b)
Stream:JEESubject:ChemistryTopic:Real GasSubtopic:Compressibility Factor
⏱ 2mℹ️ Source: QB
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