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Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IFirst law of ThermodynamicsEasy2 minPYQ_2004

ChemistryEasysingle choice

The work done during expansion of a gas from a volume of $4 dm^{3}$ to $6 dm^{3}$ against a constant external pressure of $3 atm$ is $(1 L atm = 101.32 J)$ :NEET 2004

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Answer:

Solution:

Work $= - P_{e x t} \times$ volume change $= 3 \times 101.32 \times (6 - 4) = 6 \times 101.32 = - 607.92 J = - 608 J$

Stream:NEETSubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics

⏱ 2mℹ️ Source: PYQ_2004

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