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Re: The sign of work



Other situations come to mind. A system consists of two
positive charges. I do 5 J of work in bringing them closer,
the work done "by the system" is -5 J. Or vice versa, if you
prefer. Is there a convention about this? The energy of the
system increases (like when a spring is deformed).

Daniel Schroeder wrote:

In a cylinder of gas if energy is added to the system either the internal
energy (U)( temperature ) increases or the volume expands or both.
Assume no change in volume. Then delta T will be positive as will added
energy (Q) ....right
Now assume T is unchanged, then P( delta V) must equal the Q and delta V
is positive as well as the volume expands doing work ON the surroundings.

Thus, I have the equation as : Q = delta U + W

I like W>0 for work done by the system.

I am also all ears.

First a minor quibble: In your first line, I think you mean
"heat" (as defined by most books but considered a four-letter word
by some on this list), not energy.

Now on to the main point. It's certainly true, in this situation,
that work is done by the system and therefore W>0 according to
the "engineering" convention but W<0 according to the chemistry
convention. But now consider another process: pumping up a
bicycle tire. You are doing work on the air as you compress it,
so if the system is the air, W<0 by your convention but W>0
by mine. I don't see how such examples can be used in deciding
which convention is preferable. But if we adopt my convention
(W>0 when work is done on the system), I think it will be easier
for everyone to remember that there are two ways for energy to
enter a system, one called Q and the other called W. This
convention emphasizes the parallel between Q and W in a way
that I find helpful.

Dan