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ChemistryThermodynamics - IFirst law of ThermodynamicsHard2 minPYQ_2023
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1mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of27°C. The work done is3 kJ mol-1. The final temperature of the gas is _____K(Nearest integer). GivenCv=20 J mol-1 K-1

Answer:
150
Solution:

From first law of thermodynamics:
U = q + W
We know for adiabatic process,
q=0

ΔU=wm × Cv × T = w

1×20×T2-300=-3000

T2-300=-150

T2=150 K

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics
2mℹ️ Source: PYQ_2023

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