# Chemistry - Why do chemistry and physics have different sign convention in thermodynamics?

## Solution 1:

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.

## Solution 2:

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 $T$.

## Solution 3:

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.

## Solution 4:

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.