For isobaric heating of 800 mol of gas per 500 Kelvin, a heat quantity of 9.4 mJ was reported.

For isobaric heating of 800 mol of gas per 500 Kelvin, a heat quantity of 9.4 mJ was reported. Determine the work of the gas and the increment of its internal energy.

Given:
isobaric p = const n = 800 mol ∆Т = 500K Q = 9.4 MJ = 9400000J
Find: A ‘=? ∆U =?
Decision:
isobaric => Q = ∆U + A ‘∆U = 3/2 * n * R * ∆T ∆U = 3/2 * 800 * 8.31 * 500 = 3324000 (J) A’ = Q – ∆UA ‘ = 9400000 – 3324000 = 6076000 (J)
Answer: A ‘= 6076000 J, ∆U = 3324000J