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Question
Gas in a container absorbs 300 J of heat. After this, 100 J of work is done on it which is followed by 50 J of work done by the gas. The increase in the internal energy of the gas is:
Gas in a container absorbs 300 J of heat. After this, 100 J of work is done on it which is followed by 50 J of work done by the gas. The increase in the internal energy of the gas is:
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Solution
The correct option is B 350 J
Amount of heat absorbed Q=+300 JNet amount of work done on the gas W=−100+50=−50 JFrom 1st law of thermodynamics : Q=ΔU+W∴ 300=ΔU+(−50) ⟹ΔU=+350 J
Amount of heat absorbed Q=+300 J
Net amount of work done on the gas W=−100+50=−50 J
From 1st law of thermodynamics : Q=ΔU+W
∴ 300=ΔU+(−50) ⟹ΔU=+350 J
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