110 joules of heat is added to a gaseous system whose internal

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110 joules of heat is added to a gaseous system whose internal energy is 40 J. Then the amount of external work done is

1. 150 J
2. 70 J
3. 110 J
4. 40 J

Correct Option: B

∆Q = ∆U + ∆W
⇒ ∆W = ∆Q – ∆U = 110 – 40 = 70 J