Chemistry - Real Gas Question with Solution | TestHub
ChemistryReal GasLiquification & Critical Constants of GasesMedium2 minQB
ChemistryMediumnumerical
The critical temperature of a van der Waals gas is 150 K. Find its Boyle temperature (in K).
Answer:
506.25
Solution:
Tc = 8a/(27Rb) and TB = a/(Rb).
So TB = 27/8 Tc.
TB = 27/8 × 150 = 506.25 K.
Answer: 506.25 K.
Stream:JEESubject:ChemistryTopic:Real GasSubtopic:Liquification & Critical Constants of Gases
⏱ 2mℹ️ Source: QB
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