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Heat of `20` K cal is supplied to the system and `8400 J` of external work is done on the system so

Heat of `20` K cal is supplied to the system and `8400 J` of external work is done on the system so Heat of `20` K cal is supplied to the system and `8400 J` of external work is done on the system so that its volume decreases at constant pressure. The change in internal enregy is `(J=4200 J//kcal)`

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