Chemistry - Thermodynamics - I Question with Solution | TestHub
ChemistryThermodynamics - IFirst law of ThermodynamicsHard2 minQB
ChemistryHardnumerical
One mole of an ideal diatomic gas (U = 2.5 nRT) is in a vertical cylinder of cross section 10 cm2 closed by a frictionless piston of mass 10 kg. Atmospheric pressure is 100 kPa and g = 10 m/s2. The gas is heated slowly at constant pressure by supplying 140 J of heat. Find the distance moved by the piston in mm.
Answer:
200.00
Solution:
Gas pressure = 105 + (10×10)/(10×10-4) = 2.0×105 Pa (the piston weight must be included). Constant P: q = nCPΔ T = 3.5RΔ T, so RΔ T = 40 J and w = PΔ V = RΔ T = 40 J. Force PA = 2.0×105×10-3 = 200 N. d = 40/200 = 0.20 m = 200 mm. Ans = 0200.
Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics
⏱ 2mℹ️ Source: QB
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