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

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

Correct answer: 70 J

Solution

Using the first law of thermodynamics, ΔQ = ΔU + W, where ΔQ is the heat added, ΔU is the change in internal energy, and W is the work done. Substituting ΔQ = 110 J and ΔU = 40 J, we get W = ΔQ - ΔU = 110 J - 40 J = 70 J.

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