Chemistry - Real Gas Question with Solution | TestHub
ChemistryReal GasLiquification & Critical Constants of GasesMedium2 minQB
ChemistryMediumsingle choice
Passage / Comprehension
The equation of state for a gas is: P = RT/(V − b) − a/(T·V2), where V is the molar volume of the gas and a and b are constants different than van der Waal's constant.
The minimum pressure needed for liquefaction of gas at the temperature calculated above, is
Options:
Answer:
C
Solution:
Pc = RTc/(Vc - b) - a/(TcVc2) = RTc/2b - a/(9b2Tc) with Tc = √(8a/27Rb), Vc = 3b. First term: RTc/2b = √(2aR/27b3) = (√(6))/9√(aR/b3). Second term: a/(9b2Tc) = 1/(9b2)√((27aRb)/)8 = (√(6))/12√(aR/b3). So Pc = √(aR/b3)((√(6))/9 - (√(6))/12) = (√(6))/36√((aR)/)b3 = √((aR)/)216b3. Hence Ans (c).
Stream:JEESubject:ChemistryTopic:Real GasSubtopic:Liquification & Critical Constants of Gases
⏱ 2mℹ️ Source: QB
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