Reference: Daniel V. Schroeder, *An Introduction to Thermal Physics*, (Addison-Wesley, 2000) – Problem 1.33.

We can plot the state of an ideal gas on a plot of pressure versus volume (a PV diagram). Using this diagram we can work out a few facts about how much heat heat and work flows into or out of the gas.

As an example, a monatomic ideal gas follows a triangular cycle on a PV diagram, starting at pressure and volume . On the first leg (side A) of the triangle, the pressure is held constant while the volume increases to (so the path is a horizontal line). Then (side B) the volume is held constant and the pressure is increased to , giving a vertical line on the PV diagram. Finally the pressure is reduced back to and volume back to along side C, which is a straight, diagonal line with slope .

In a compression (or expansion) problem, the work done on the gas is

For this problem, the work done on side A is

On side B (since is constant)

On side C, the work done is the negative of that done on side A, plus the area of the (right-angled) triangle, so

The total work done on the gas is

That is, the total work is just the area of the triangle.

From the equipartition theorem, the thermal energy of the gas is

so along side A (since is constant and increases)

along side B

(since is constant and increases) and along side C

(since both and decrease). The net change in after going round all three sides is zero, since the gas is back in its original state.

From conservation of energy, we can get the heat on each side. On side A

On side B

And on side C

The total heat added to the gas is

Since this is negative, a net amount of heat is emitted by the process. Thus the overall process converts the net work done on the gas to heat.

### Like this:

Like Loading...

*Related*

Pingback: PV diagrams: a diatomic ideal gas undergoes a rectangular cycle | Physics pages

Pingback: Isothermal and adiabatic compression of an ideal gas | Physics pages

ArthurI have a question about (9). It seems to me that (P2-P1)(V2-V1) also gives -P1V2 and -P2V1. Did you leave them out on purpose? If so, what is the reason?

gwrowePost authorThe change in along side C is the final energy minus the initial energy , which gives equation 9.

ArthurIs this method allowed when the work is not zero? It is not isochoric isn’t it?

ArthurSorry, got lost. Please ignore this last question.

VitalI wanted to ask what would happen if the triangle was mirrored (so Isochronic on the left side instead of the right). 🙂

VitalAlso, just before equation (10), shouldn’t it be Q = (delta)U + W?

gwrowePost authorWith the definitions of and used by Schroeder, see equation 1 here, so . Here is the heat added to a system and is the work done on the system, so their sum is the change in internal energy of the system.

VitalYes, I posted that comment a while before I really understood the meaning of things in Thermal Physics. Still working at it, seems quite interesting so far. I also wanted to ask what would happen if the triangle was mirrored (so Isochronic on the left side instead of the right)? 🙂