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A gas absorbs $200$ J of heat and does $80$ J of work on its surroundings. The change in internal energy is:
A$280$ J, summing heat and work without sign convention
B$-120$ J, with both quantities working against the gas
C$120$ J, since $\Delta U = Q - W = 200 - 80 = 120$
D$80$ J, equal to the work done alone on the gas
Answer & Solution
Correct answer: C. $120$ J, since $\Delta U = Q - W = 200 - 80 = 120$
First law: $\Delta U = Q - W = 200 - 80 = 120$ J.
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