Chemistry - Real Gas Question with Solution | TestHub
ChemistryReal GasBoyle TemperatureEasy2 minQB
ChemistryEasysingle choice
The compressibility factor (Z) of a gas is given by
Z = 1 + 0.20 P - 80P/T ('P' in atm, 'T' in Kelvin)
What is the Boyle temperature of this gas?
Options:
Answer:
B
Solution:
Z = 1 + (b - a/RT)P, with b = 0.20 and a/R = 80.
At the Boyle temperature the slope of Z against P at P → 0 is zero: 0.20 - 80/T = 0, so T = 80/0.20 = 400 K.
Answer: (B).
Stream:JEESubject:ChemistryTopic:Real GasSubtopic:Boyle Temperature
⏱ 2mℹ️ Source: QB
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