Chemistry - Thermodynamics - I Question with Solution | TestHub
ChemistryThermodynamics - IWorkHard2 minQB
ChemistryHardnumerical
One mole of a van der Waals gas expands reversibly and isothermally at 27°C from 2 L to 20 L. The magnitude of work done (in J) if a = 1.42 × 1012 dynes cm4/mole2 and b = 30 ml/mole.
Answer:
5713.00
Solution:
P = (RT)/(V-b) - a/(V2), so w = RTln(V2-b)/(V1-b) + a(1/(V2)-1/(V1)). a = 1.42×1012 dyn cm4 = 0.142 Pa m6 mol-2. First term: 8.314×300×ln19.97/1.97 = 2494.2×2.3162 = 5777 J. Second term: 0.142 (50 - 500) = -63.9 J (V in m3: 0.02 and 0.002). w = 5777 - 64 = 5713 J. Ans = 5713.
Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Work
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