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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.

  • 5
    $\begingroup$ They didn't want to give away the sign, so they specified "on or by". You have to know the sign convention in use in order to choose the correct answer. But I admit the wording is poor. $\endgroup$
    – garyp
    Oct 2 '17 at 19:03
  • $\begingroup$ Related: physics.stackexchange.com/q/37904/2451 , physics.stackexchange.com/q/39568/2451 and links therein. $\endgroup$
    – Qmechanic
    Oct 2 '17 at 21:39
  • $\begingroup$ Only if they mention on or they write.by the the sign has a meaning. Its like negative acceleration.is deceleration. If you say accelerates or decellerates then a minus sign has absolutely no meaning. Also the info that the gas is ideal is missing, for.a general.gas answer E is correct. $\endgroup$
    – lalala
    Jun 17 '18 at 8:16
  • $\begingroup$ I were never a fan of questions like these, because they basically only test how well you have read the chosen textbook (and how well you remember its choice of convention), and not how well you understand any of the physics. $\endgroup$
    – Steeven
    Sep 17 '18 at 5:31
  • $\begingroup$ This is an astoundingly bad question. They intentionally worded it to say, "if you didn't memorize our particular sign convention, too bad!" $\endgroup$
    – knzhou
    Sep 17 '18 at 9:37

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.


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