In physics class, we write the first law of thermodynamics as $\mathrm dU =\mathrm dQ - \mathrm dW$ and in the physical chemistry class, we write the same law as $\mathrm dU =\mathrm dQ + \mathrm dW$.

The reason being the sign convention is different in both the cases.

In physics we take work done by the system as positive and in chemistry work done on the system.

I realize that this does not cause any change in the actual law of nature but I just want to know why we have different sign conventions. Wouldn't just one convention make life easier?

Is there a historical reason? Or is this is to differentiate between subjects?

  • 8
    $\begingroup$ The standardized definition according to ISO 80000-5 Thermodynamics of internal energy $U$ for a closed thermodynamic system is $\Delta U=Q+W$, where $Q$ is amount of heat transferred to the system and $W$ is work done on the system, provided that no chemical reactions occur. $\endgroup$ – Loong Jan 12 '17 at 11:31
  • 2
    $\begingroup$ Personally I would describe the version with the "-" as the engineering convention. When you are building engines (as were the people who invented thermodynamics) the process you care about putting heat in and getting work out, and you define both those quantities as positive. Pure physics thermodynamics often (mostly?) uses the "+" version. $\endgroup$ – dmckee --- ex-moderator kitten Sep 28 '17 at 18:04
  • 1
    $\begingroup$ Even in engineering, there are two conventions... Mechanical engineers think work leaving the system is positive work, chemical engineers think work done on the system is positive... Other engineers just fake it. $\endgroup$ – Charlie Crown Feb 19 '19 at 17:41

This is not a simple physics versus chemistry distinction. I taught Physics for 25 years and saw many examples of either usage in multiple textbooks. In fact, at some point in my tenure, the AP Physics committee swapped conventions on the equation sheet for the AP Exam.

Just my take here: I've always attributed the work-done-by-the-system camp as being more prone to be used by engineering types who want to know "what the system can do for us" in practical applications. On the other hand, work-done-on-the-system seems to foster the view of an experimenter or theoretician operating on a system from without.

  • 4
    $\begingroup$ "Ask not what your system can do for you..." :-P Also, for what it's worth, only the second version makes sense to me. In the first version you're considering positive heat to be something that is supplied by the surroundings, and positive work to be something that is supplied by the system. That's obviously inconsistent, and by fixing the inconsistency you get the second equation. Just my (correct) opinion... ;) $\endgroup$ – user541686 Jan 12 '17 at 0:40
  • 1
    $\begingroup$ @Mehrdad Agreed. That's the same argument I gave my students. $\endgroup$ – bpedit Jan 12 '17 at 1:49
  • $\begingroup$ As I said above, only mechanical engineers are greedy and care about what the system can do for them :D $\endgroup$ – Charlie Crown Feb 19 '19 at 17:42

There is no difference and the equation will yield the same answer. The difference is by deciding the point of reference for the work done.

Where we have $W = \pm P\Delta{V}$, you just have to decide where do you look from. Am I the system and work is done on me, or am I outside and doing work on the system? anyhow, once you choose your point of reference you should stick to it and do all your calculations according to it. I guess, as in most of these cases it's historical.

Another example is different notation, chemists use $A$ to symbolize the free Helmholtz energy while in physics it's $T$.

  • 1
    $\begingroup$ Seems like using T for Helmholtz energy would be a nightmare, T is already an overloaded symbol even just sticking to chemists notation (temperature, kinetic energy, certain tensor operators). $\endgroup$ – Tyberius Feb 19 '19 at 16:03
  • $\begingroup$ I've never seen "T" being used for Helmholtz energy. I think you mean "F". $\endgroup$ – David Sousa Aug 20 '19 at 18:40

Physicists are interested in getting work done by the system. In physics we always look for what system is giving to us (surrounding). So, in physics we study work done by the system.

Whereas chemists are interested to know what work have the surroundings done on the system, in order to get the reaction completed. So, in chemistry we study work done on the system.


Chemistry made the switch over to the IUPAC convention back in the early 90's which dictates the first law as the sum of $Q$ and $W$ and in turn requires $W$ to be defined as the negative integral of $P(V)\mathrm{d}V$. See the recommendations of the International Union of Pure and Applied Chemistry.

As pointed out, physicists use $Q-W$ but define $W$ as the positive integral. Same final results though.

  • $\begingroup$ Perhaps it'd be great if you also link what section of that 250-page document was supposed to be read? $\endgroup$ – Gaurang Tandon May 15 '18 at 6:51

This is because.... In physics , we are concerned about the surroundings And in chemistry, we concerned about the system.

That's why , In physics , work done by the system is positive because change in energy (due to the work done) of the surroundings is positive ! (Energy is released from the system to the surroundings)

Whereas In chemistry , work done by the system is negative because the system loses some amount of energy to do work !

That's it!


Take it simple; it's nothing but a reference in different field for different uses that changes the formulae. If system is doing work on surrounding, U will decrease if W is taken positive – and vice versa.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.