Chemistry - Real Gas Question with Solution | TestHub
ChemistryReal GasCompressibility FactorMedium2 minQB
ChemistryMediummultiple choice
At very high pressure, the compressibility factor of 1 mole of a van der Waal's gas can be given as
Options:(select one or more)
Answer:
A, B
Solution:
At very high pressure the molar volume becomes comparable to b, so the attraction term a/V2 is negligible: (P)(V − b) = RT. Then Z = PV/RT = V/(V − b), which is option (b). Expanding, V/(V − b) = 1/(1 − b/V) ≈ 1 + b/V ≈ 1 + bP/RT, which is option (a). (c) has the wrong sign and (d) is the low-pressure (attraction-dominated) form. Hence (a) and (b).
Stream:JEESubject:ChemistryTopic:Real GasSubtopic:Compressibility Factor
⏱ 2mℹ️ Source: QB
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