## Work of a System: On vs. By

avo
Posts: 36
Joined: Wed Jun 12, 2019 6:01 pm

### Work of a System: On vs. By

Hello! No matter how much I try to read the difference between work on a system vs. work by a system, I'm still finding myself not understanding how they vary or when to tell the work work would be positive or negative! I'm looking for some clarification or ways to practice this so I wont' get so tripped up in the future!
NS_Tutor_Mathias
Posts: 616
Joined: Sat Mar 30, 2019 8:39 pm

### Re: Work of a System: On vs. By

Let's clear a few things up: Energy is the ability to do work. And by unit analysis, you can without overthinking it, remember that W = -PdV. Short of some silly questions physics grad students get asked, that is what you need to know first and foremost.

So, imagine we have a spherical or cylindrical container filled with some gas, at a temperature above zero. The molecules within this container collide with the walls, and the walls collide with the molecules, both exerting equal and opposite forces on each other. Pressure is just the perpendicular component of force of these collisions divided by the surface area they are happening on. If the collisions are perfectly elastic, the walls of the container don't move and no energy is lost to heat, then the energy of this system (the energy of the gas) remains constant.

But what happens if the container can expand? Now these forces are in a very literal sense acting while moving something along a distance, meaning the distance that the walls of our tank are travelling. A force acting on a distance, conveniently parallel to the direction of expansion, is exactly what the standard definition of work describes: F x d = F * d * cos(90) = F*d

But how does the gas do work? Kinetic molecular theory of gases does can give us a hint: The kinetic energy of the gas is used for this expansion, to push against the walls of the container and whatever forces are limiting the size of the container (whether strength of the material or atmospheric pressure on the outside). By doing work, our system (the gas) is losing energy!

Now let's check in real quick with what sign convention would make sense to relate W, P and dV: The change in volume is positive, but the change in energy was negative. So we arrive at a non-systematic, somewhat intuitive way to re-build our initial equation W = -PdV, because expansion (work done BY the system) makes the system lose energy, so the sign of W should definitely be negative (Since I skipped over it a bit Einitial + W = Efinal /// therefore, if the sign of W is negative, Efinal is less than Einitial for the system).

Let's check out the converse real quick: If the outside does work on the system, say by atmospheric pressure being greater than that in our malleable container, then the container contracts. That means the sign of dV is negative, and W = -PdV will give us a positive sign of work. This is the case of work being done ON the system, and the system gaining energy.

You can further use this intuition to reason about adiabatic heating and cooling: Adiabatic processes are those in which neither heat nor mass is exchanged. If work is done on the system, it must have gained energy. The first law of thermodynamics reminds us that perfect energy transfer is impossible, and whenever work is done some proportional amount of heat is also released (and vice versa, heat can do work). So a system that gains energy must also gain heat. A system that loses energy must also lose heat. Contracting gases are bound to heat up, as in a bicycle pump, and expanding gases are bound to cool down, as in orographic lift causing cooling of air.

Let me know if this helps for starters!

Edit: Chemistry apparently sometimes uses the opposite sign convention. This is the one more common on the MCAT, and definitely the standard for physics. If you understand this one, you will have no trouble changing tracks and flipping conventions if a passage demands it.