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Work is defined as:

a force acting over a distance.

In everyday language, force is a push or pull. (Please understand, force has a very precise definition. I'm avoiding it until near the end of this file.) Suppose you have some object which you wish to move from point A to point B. So you begin to exert a force on the object, you begin pushing it and it moves to point B, whereupon you stop pushing.

By exerting this pushing action ("Luke, Luke use the fork" said Obi Wan ChemTeam!!) over a physical distance, you have performed a certain amount of work. Rendering this mathematically, we write:

F = the force and L = the length of displacement of the force's point of action. The calculus integral simply means that we get the work value by summing up the force times distance over the ENTIRE distance.

Fine, fine, fine. What does all this mean from a chemical point of view? And how is work involved in all of this? Fair enough.

One of the things a chemical system can do is expand, thereby increasing its volume. However, to do that, the system must push against the opposing pressure of the atmosphere. After all, the system IS NOT expanding into a vacuum, it is expanding and pushing the atmosphere out of the way. And that opens the door to a key definition - that of pressure:

pressure = force per unit area

or

P = F / A

The usual way to explain this is by using a piston:

Notice that the chemical system never actually touches the atmosphere, what the system is doing is pushing against a movable piston (which, under ideal conditions, offers no resistance to movement). In truth, the chemical system does push against the resisting atmosphere, but it is a very close approximation to use an ideal piston interposed between the system and the atmosphere.

So, since F = PA, we write this equation:

Then, since A is an area (with square units like cm^{2}) and dL has length units (like cm.), when you multiply them together, you get volume units (cm^{3}). So, we write a new equation:

This last equation has two more points about it: (1) we make an assumption that P remains constant (a fairly defensible one, I think). As a constant, P passes through the integral, since it will affect the ENTIRE volume change and (2) the negative sign comes from the fact that work is being performed BY the system and flowing out of it into the surroundings. The system has gone down in its work capacity, it has lost some of its ability to do work. This, by convention, is assigned a negative sign.

The last step is to think about dV. It represents the change in volume as it expands from V_{1} to V_{2}. Typically, it is written thus:

ΔV = V_{2}- V_{1}

By the way, there is a difference between the calculus symbol "d" and the delta, but I'm going to ignore it. Consequently, we arrive at the final equation, the one for "PV work:"

w = - PΔV

The unit on w is liter-atmospheres, a unit of energy.

Yes, liter-atmospheres is a unit of energy, just like Joules.