Question bank/JEE Main/Physics

JEE MainPhysicsEasyThermodynamics

110 J of heat are added to a gaseous system where change in
internal energy is 40 J, then the amount of external work done is

Show answer and solution
ΔW=Δu+ΔW\Delta W = \Delta u + \Delta W
⇒110=40+ΔW\Rightarrow 110 = 40 + \Delta W
⇒ΔW=110−40=70J\Rightarrow \Delta W = 110 - 40 = 70J

Source: Grass-G / P-08 Physics paper 23 October.docx

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