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The question asks work for done by or on the gas. How does that make sense? If I do work $mgh$ on a block by raising it, the work done by the block on me is $-mgh$. They have the same magnitude and opposite sign.

In physics we use the sign convention of ∆W=P∆V for work done BY gas. So if it expansion it is Positive and vice versa. However in chemistry we use ∆W=-P∆V for work done by gas and expansion is negative. I understand that overall the sign convention give same answer for ∆U and ∆Q as one has Q=U+W whereas other one is U=Q+W. But doesn't this mean work done according to your sign convention will change?

The correct answer is supposed to be D.

knzhou
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1 Answers1

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If the answer is D, then the form of the first law being used is $$\Delta U=Q+W$$where $W$ is the work done BY the surroundings ON the gas. You know that $\Delta U$ is equal to zero because the process is isothermal. You know that Q is negative because heat is removed.

Chet Miller
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