# Thermodynamics, PV diagrams?

My teacher told me that the total amount of work done on or by a gas can be represented by the area enclosed in the process in a PV diagram. This is only valid for non isothermic processes, right?

-
why you think that? It's general. – ABC May 13 '13 at 14:02
Well in my book they have a PV diagram in the shape of an "right triangle, with the hypotenuse actually being an isothermal curve. They just find the area below the isobaric process to be 1600J. then they go on to say that since the last step of the total process is isothermal, there is no change in internal energy, so 1600J of heat flow out of the gas for the whole process. – Ovi May 13 '13 at 14:07
Isothermal processes mean no change in internal energy , all heat goes to work . I suggest you watch these lectures . youtube.com/watch?v=mb8LqNlHeLY. Your fundamentals are kind of messed up in this topic . – user23503 May 13 '13 at 14:12
Haha yea I know, I was absent for a few days and I'm trying to learn this by myself but the book isn't as clear as I'd like it to be. I'll watch those videos. – Ovi May 13 '13 at 14:19
Note that "enclosed" implies a cycle, some process that starts in one conditions passes through some others and ends up back where it started. For that case, and assuming that all the changes are quasi-static, then the relation is completely general. It is perhaps better to remember the rule that for any single path in the PV plane the work is the (possible negative) area under the curve---also general under the quasi-static assumption, but does not demand a cycle. – dmckee May 13 '13 at 14:48

There are several ways we can approach this, but I'll argue that the integral of the PV curve is a more general form of the force times distance concept of work:

$$W = F \Delta x$$

This applies for pretty much any action over a distance. If you compress a spring, lift a box, drive a car, the above equation applies to formalize the work done. To generalize this, let's consider a piston, which is a cylinder that has one moving wall. Force on the wall from the internal gas is pressure times area, which comes from the definition of pressure. The volume is the cross-sectional area times the distance between walls, and the change in volume is $\Delta V = \Delta x \times A$. Substitute these in our equation:

$$W = F \Delta x = \left( P A \right) \left( \frac{\Delta V }{A} \right) = P \Delta V$$

This has some major hand-waving, because the pressure changes as the position of the wall changes. Of course, there is a simple remedy to the situation, which is to write it as an integral.

$$W = \int_{1}^{2} P dV$$

I write it this way to indicate that there is a transition from state 1 to state 2. Pressure depends on the volume in a non-trivial way. To know exactly how pressure changes you need the thermodynamic specifiers mentioned in the comments - is it isothermal, isentropic, etc. Assuming you know the amount of gas in the piston and that it remains constant, the pressure is then a function of both volume and temperature, and this is why the additional specifier is needed.

-
You mean that It doesn't matter if during the process the temperature changes because $p=p(T,V)$ and you are only integrating over $V$? – Ana May 13 '13 at 15:30
@Anuar Here $T$ matters, but you don't integrate over $T$. You're somewhat confusing those two concepts. You could set $T$ equal to a constant for a particular type of process, or you could plug in some $T(V)$ function. In the latter case, the $V$ in the argument will affect the anti-derivative, so it has to be arithmetically included before the integration. – Alan Rominger May 13 '13 at 15:58