Chemistry - Real Gas Question with Solution | TestHub
ChemistryReal GasBoyle TemperatureMedium2 minQB
ChemistryMediumsingle choice
The expression for compressibility factor for one mole of a van der Waal's gas at Boyle temperature is
Options:
Answer:
A
Solution:
Z = (PV)/(RT) = V/(V-b) - a/(RTV). At T = TB = a/(Rb), a/(RTB) = b, so Z = V/(V-b) - b/V = (V2 - b(V-b))/(V(V-b)) = (V2 - bV + b2)/(V(V-b)) = 1 + (b2)/(V(V-b)). Ans: (a)
Stream:JEESubject:ChemistryTopic:Real GasSubtopic:Boyle Temperature
⏱ 2mℹ️ Source: QB
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