# How to reconcile the two definitions of work? (mechanical and thermodynamical)

When studying classical mechanics, work is defined as: $W_M=\int F_{tot} \hspace{2 mm} dx$.

However, for thermodynamics, work is defined as: $W_T=\int -F_{ext} \hspace{2 mm} dx$.

I'm having trouble reconciling both definitions. I have found the mathematical relationship between both of them (see below), but I wouldn't know how to interpret either this or the other thermodynamical functions in the light of this. For example, $\Delta U$ is usually related to the conservation of energy, but since the definition of $W$ is different, it isn't that direct a relationship now (for me at least).

$$W_M=\int F_{tot} \hspace{3pt} dx \hspace{3pt} =\int F_{int}-F_{ext} \hspace{3pt} dx$$ $$W_M=\int F_{int}\hspace{3pt} dx+W_T \hspace{3pt}$$

The books I've seen usually just throw the definition and start talking about the state functions, but do not spend any time reflecting on this definition or analyzing why is it different from the classical one. I found this source which tries to deal with this same issue: it is very interesting, but it gets a little confusing towards the end when he tries to relate his mechanical w and q with his thermodynamical W and Q (I don't understand how he combines equations 1 and 3 or 1 and 5 to get his results). Does anyone understand that last step?

Has anyone seen or thought about this issue before? Does anyone know of any other source or book where this issue is dealt with more detail?

EXAMPLE PROBLEM (to motivate the question): Consider a very big cylinder with a mobile piston in the middle separating two sections filled with gas (A and B). Both the cylinder walls and the piston are adiabatic, so there is no exchange of heat with the exterior or between sections. The pressure of the gas in A is higher than the pressure of the gas in B. The piston is originally being hold, but we let go of it for just a second and then hold it again (we assume the change in volume was small enough so as to not produce a change of pressures).

In this situation, I could consider the gas of section A as my system and claim it gave $p_B \Delta V$ of work to its exterior (the gas in section B, since it is the only exterior with which it interacts). But I could also consider the gas in section B as my system, and claim it received $p_A \Delta V$ of work from its exterior (gas in section A). But since $\Delta V$ is equal in both cases (only changes the sign, which is what determines is work is given or received) but the $P_i$ aren't, there is an energy difference I cannot account for.

My intuition tells me that the energy difference must come from the potential energy stored in the original pressure difference, but since thermodynamics (or the way I was taught thermodynamics) starts with a different definition of work and does not explain its relationship with the mechanical one, it is not that trivial to me how to incorporate this into a thorough thermodynamical analysis of this process.

-
Related: physics.stackexchange.com/q/37904/2451 and links therein. –  Qmechanic Dec 24 '13 at 2:56
No. My doubt is not about signs and conventions, but about a difference in the way work is calculated via termodynamics and via classic mechanics (and that is not just a sign, but a whole term) and how this translates into missing energy in a thermo analysis (did you see the example problem?). I didn't see any other links in that other thread except for a tex tutorial. I think my question is different and I still haven't found an answer to it. –  Nordico Dec 25 '13 at 5:48

Perhaps I can explain the difference. Consider a piston in a cylinder with the inside of the cylinder being the system. The surroundings are outside the cylinder.

Assume there is a gas inside the cylinder exerting a pressure $p$ on the piston from the inside. Let there be a vacuum on the outside, but several weights sit on the outside of the piston. These weights exert a gravitational force on the area of the piston. The result of this is a pressure $p_e$.

So $p$ is the internal pressure and $p_e$ is the external pressure. If the internal pressure is less than the external, the piston will descend and the external weights will do work on the system. On the other hand, if the internal pressure is greater, then the system will do work on the surroundings, that is, the weights.

Now we are set up. The thermodynamic definition of work is the work done ON the system by external weights (forces) or the work done by the system on the external weights. The first of these is traditionally counted as positive, the latter as negative.

Consider an expansion of the system. You can easily calculate the work done by the system on the weights if you know the acceleration of gravity (which we will assume you do), the mass $m$ of the weights, and the height $h$ to which they have been raised. That's ALL you need to know. Note that the pressure inside the cylinder does not come into it except to ensure that there is an expansion and that the pressure is great enough to raise the weights by $h$,

If there is a compression of the system, again all you need to know are the same quantities as for an expansion, except now $h$ is negative.

In one case we compute the work done to raise the weights; in the other we compute the work done when the weights are lowered.

How does this differ from the normal definition of work in physics? It doesn't. You only need to take all the forces into account if you want to compute the acceleration of the piston. Thermodynamics doesn't care about that, in fact the piston itself is usually assumed to be weightless.

-
Your setup is similar to the one described in the paper I quote as source. In that paper however, they try to apply the classic mechanical definition to arrive at the thermodynamical one. The reason why the internal pressure is not used in your reasoning, is because you are already using the thermodynamical definition which starts from that hypothesis ("To study the changes in energy of a system, you only need consider the work of external forces"). I would like to ARRIVE at that from a more simple base: only knowing the mechanical definition of work. –  Nordico Jun 5 '13 at 14:21
I do think you are making a mistake. The definition of infinitesimal work is unambiguously the dot product of the force vector and the infinitesimal distance vector. The mistake you are making is that the work is being done on (or done by) the WEIGHTS on the massless piston. I do not know what other definition of work one can use. We are not calculating the work done on or by the piston itself. –  Paul J. Gans Jun 6 '13 at 2:12
I don't think that is it. I understand that you can calculate the work done by a specific force instead of the resultant force; what I don't get is why are just the external forces considered when you want to analyze what happens to the system. There is nothing connecting all the concepts of energy that emerge from typical mechanics and the one that is used in thermodynamics (the energy of a system). –  Nordico Jun 6 '13 at 17:14

But they are the same. Work always represents a mechanical transfer of energy from one place to another. Everything else about it is just bookkeeping about where the "places" are and which one is ending up with more energy. I can't place a use for the distinction you make about $F_{tot} = F_{int} - F_{ext}$, so I suspect that it is special the a particular analysis. Could you say a little more about where you found it? –  dmckee Jun 3 '13 at 19:45
@dmckee I think I understand the OP since I have almost the same doubts. Let's look at a volume of gas at pressure $p$. If the gas expands because the external pressure $p_e<p$, then we will say that the work done by the system on the surrounding was $p_e \Delta V$. Now take $p_e>p$, the gas is compressed. In this case the work done by the gas is $p\Delta V$. In practice, whenever we have an irreversible transformation of the gas we use the less of the two pressures... why? –  pppqqq Oct 23 '13 at 17:34