110 J of heat is added to a gaseous system, whose internal energy change is 40 J, then the amount of external work done is
Correct answer: B. 70 J
- A. 150 J
- B. 70 J
- C. 110 J
- D. 40 J
Explanation
Correct. Applying the first law of thermodynamics: ΔU = Q - W W = Q - ΔU Q = 110 J ΔU = 40 J W = 110 J - 40 J = 70 J
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