# Why is the sum of two inexact differentials exact?

From the first law of thermodynamics, $$\mathrm{d}U = đQ + đW$$, where $$\mathrm{d}$$ represents an exact differential and $$đ$$ an inexact differential. Exact differentials correspond to state functions and are path-independent.

How is it possible, and what does it imply, when we say that the sum of two inexact differentials is an exact differential? Is the sum of two path-dependent functions always a path-independent function? Also, what is the difference between a partial derivative and an inexact derivative / differential, apart from mathematical implications?

• I saw essentially the question (somehow the answer is still unclear to me) posted in Physics SE. Perhaps you'll be able find it. It might be a case of notation abuse as per Wikipedia. I also think that an elegant answer might be given here too. When I say the answer is not clear to me, I mean it is my problem... I always remain with the same taste even if at first I think to understand... – Alchimista Feb 8 at 12:31
• It obviously is not valid generally, but particularly for the Q and W case. BTW, it is exact/inexact differential, not derivative. – Poutnik Feb 8 at 12:33
• @Alchimista Sir,could you provide me that quesion link? – PV. Feb 8 at 14:41
• @Poutnik Sir, how do you differentiate between a derivative and a differential? – PV. Feb 8 at 14:42
• I have a pie. Amy and Bill eat pieces of pie. From the fact that the pie is gone there is no way to show who ate how much pie... – Jon Custer Feb 8 at 22:16

## 3 Answers

There's not much to do, as you nearly spotted the difference by yourself: $$U$$ is a state function, thus is path-independent. In other terms, one can estimate it at any moment only with an initial and a final value (and express it in terms of its variables only). But with $$Q$$ and $$W$$, it's different: these are path functions (or process functions) consequently depending on the path. Hence, their differential is not something one might express depending only of its variables. (these are not "real" functions for mathematicians...)

Is the sum of two path-dependent functions a path-independent function?

The sum of inexact differentials is an exact differential in that particular case. In other cases, it depends on the units involved: for instance, with entropy, we have $$\mathrm{d}S = \delta Q_\mathrm{rev} / T$$. So, no typical case here!

what is the difference between a partial derivative and an inexact derivative?

These are absolutely not the same! A partial derivative is a differentiation of a multi-variable function regarding one of its variables only, while what we have here is an inexact differential, thus not a derivative as 1) no variable is specified here, and 2) an inexact differential can be of a function depending on only 1 variable.

How is it possible, and what does it imply, when we say that the sum of two inexact differentials is an exact differential?

In the example at hand it means that we divide the ways we can change the energy of the system into exactly two bins. The first we call heat, and all the other ones we call work.

Heat is an energy transfer that occurs when two entities of different temperature are in contact. The difference in average kinetic energy of the particles causes an energy transfer during collisions which overall increases the temperature of the cooler one and decreases the temperature of the hotter one.

Work is everything else, from compression of a gas to stirring to powering a heating coil inside the system with an electrical power source outside of the system.

We can have different amount of heat and work depending how we run a process. If the starting and final state match for two processes, the change of the system's energy will be the same, no matter what path we choose (i.e. what the individual contributions are).

Is the sum of two path-dependent functions always a path-independent function?

We could have divided up the work into pV-work and non-pV-work. Both would be inexact differentials. If we add them up, we would still have an inexact differential.

So the answer to your question is no.

Also, what is the difference between a partial derivative and an inexact derivative / differential, apart from mathematical implications?

A partial derivative is when you form the derivative of a multi-variable function with respect to a single variable. So if my function T(r) describes the temperature distribution in 3D space, and I form the derivative with respect to the height $$z$$, that is a partial derivative. This is different from the total derivative, which would tell me the gradient of the temperature at a given point (i.e. in which direction it changes most). So a partial derivative and an inexact differential are quite distinct.

Imagine that you propose to climb Mount Everest. From your desk you can compute exactly how much energy you will need to reach the top of the mountain starting from base camp. That would simply be the gravitational energy difference between the two locations. With that information at hand you load up on supplies - dividing the gravitational potential energy difference by the energy density (calories/weight) of your favorite granola brand - and get underway. Not surprisingly however you find yourself out of granola before getting very far and decide to turn back. You then have some free time to puzzle over what went wrong.

After some thought you realize that not all of that granola was turned into work, and that different members of your climbing team exhibited different efficiencies converting granola into work. One of them seemed to convert it mostly into heat and did not get very far at all.

So every step required an exact amount of energy at minimum to raise your body up the gravitational potential (performing the desired work), but the amount of energy actually spent was inexact and varied between people, some requiring more energy just to stay warm in the Himalayan summer. As you dwell deeper into the subject you realize that climbing Everest was more complicated than envisioned. In fact much energy from the granola was spent changing your latitude and longitude, rather than your elevation, as well as in keeping your body warm and functioning.

After some days a couple of your climbing companions surprise you by announcing that they managed to reach summit and get back alive. More importantly they also monitored their metabolism, position and granola consumption on the way up. From this data you manage to estimate how much granola was used up to change their latitude and longitude and how much they used to keep warm and their bodies functioning. The energy accounted for by the change in height did not change. But now you have a better estimate of how much granola to pack next time you attempt the climb.