Chemistry - Thermodynamics - I Question with Solution | TestHub
ChemistryThermodynamics - IWorkEasy2 minQB
ChemistryEasynumerical
One mole of a van der Waals gas expands reversibly and isothermally at 27°C from 1 L to 10 L. The magnitude of work done (in J, nearest integer) if a = 2.0×1012 dynes cm4/mole2 and b = 40 ml/mole.
Answer:
5655.00
Solution:
w = RTln(V2-b)/(V1-b) + a(1/(V2)-1/(V1)). a = 0.20 Pa m6 mol-2. RTln9.96/0.96 = 2494.2×2.3395 = 5835 J. a term: 0.20 (100 - 1000) = -180 J. w = 5835 - 180 = 5655 J. Ans = 5655.
Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Work
⏱ 2mℹ️ Source: QB
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