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One of the specification points in my A-Level is to be able to interpret negative/positive values in the $\Delta U = Q - W$ equation. I can't seem to find an intuitive explanation in my textbook so could anyone else help?

EDIT - If you are pushing a piston into a cylinder, what is happening to the variables in this equation?

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  • $\begingroup$ This equation or one with a positive sign is in many textbooks and many websites. The change in internal energy of a system is equal to the heat input into the system minus the work done by the system. $\endgroup$
    – Farcher
    Commented Mar 27, 2018 at 14:08
  • $\begingroup$ Positive values account for increase in internal energy , work done by system and heat supplied . The negative accounts for opposite ones .. $\endgroup$ Commented Mar 27, 2018 at 14:11
  • $\begingroup$ I think that may have answered it. So work done is done BY the system not done on the system? $\endgroup$
    – user190536
    Commented Mar 27, 2018 at 14:13
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    $\begingroup$ Related: physics.stackexchange.com/q/37904/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Mar 27, 2018 at 14:21
  • $\begingroup$ When on the system , $\Delta W$ is negative ... $\endgroup$ Commented Mar 27, 2018 at 14:21

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Always know how to answer this

The key question you have to ask yourself is

What do these symbols mean?

Heat

The $Q$ means heat, of course, but does it mean heat added to or removed from the system.

In almost all cases it takes the sense of "added to".

Work

Should $W$ be the work done on the system or the work done by the system?

If you are primarily interested in the system than it should be the work do on the system which has the added bonus of meaning that both symbols mean energy added to the system.

However, the main practical application of early thermodynamics was in the building of engines. And when you are designing an engine the things you worry about are how much fuel you put in and how much work you get out. The obvious convention for a engine designer is to have $W$ represent the work done by their machine, meaning that $W$ represents energy leaving the system.

First Law

So the engineers version of the First law is $$ \Delta U = Q - W \tag{engineering} \;,$$ while most chemists and physicists prefer $$ \Delta U = Q + W \tag{science} \;.$$

Side effects of this choice

Note that the effects of this choice reverberate through the math. Taking the case of a fluid system the mechanical work is $$ W = +P \,\mathrm{d}V \tag{engineering} \;, $$ or $$ W = -P \,\mathrm{d}V \tag{science} \;, $$ and that means that the way you write the various thermodynamic potentials changes as well.

Comparing texts that use different sign conventions for work is a painstaking undertaking.


Aside: Reif's textbook is the only one I am familiar with to use two symbols to represent both meanings.

  • $W$ is the work done by the system
  • $\mathcal{W}$ is the work done on the system

That way you can write the first law in the manner that makes the easiest sign handling for each problem you come to. I've been using this approach in all the writing I do for my students and even adopts some of Reif's other notational conventions that I don't care much for ($\bar{E}$ for the internal energy? Really?) just so that my students can compare the notes I give them directly to one of their texts.

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  • $\begingroup$ Thank you for the detailed answer. So if you are pushing a piston into a cylinder, you are doing work on the system to therefore dW is -ve so dU increases? $\endgroup$
    – user190536
    Commented Mar 27, 2018 at 14:39
  • $\begingroup$ Right. Your resource is using what I have called the "engineering" conventions. With $\mathrm{d}V$ negative, your work is negative, and the change in internal energy is positive. (You should get the same sign for the change in internal energy is both conventions, BTW.) $\endgroup$ Commented Mar 27, 2018 at 14:43

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