# PV diagrams: a monatomic ideal gas follows a triangular cycle

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 ${P_{1}}$ and volume ${V_{1}}$. On the first leg (side A) of the triangle, the pressure is held constant while the volume increases to ${V_{2}}$ (so the path is a horizontal line). Then (side B) the volume is held constant and the pressure is increased to ${P_{2}}$, giving a vertical line on the PV diagram. Finally the pressure is reduced back to ${P_{1}}$ and volume back to ${V_{1}}$ along side C, which is a straight, diagonal line with slope ${\left(P_{2}-P_{1}\right)/\left(V_{2}-V_{1}\right)}$.

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

$\displaystyle W=-\int_{V_{i}}^{V_{f}}P\left(V\right)dV \ \ \ \ \ (1)$

For this problem, the work done on side A is

$\displaystyle W_{A}=-P_{1}\left(V_{2}-V_{1}\right)<0 \ \ \ \ \ (2)$

On side B (since ${V}$ is constant)

$\displaystyle W_{B}=0 \ \ \ \ \ (3)$

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

$\displaystyle W_{C}=P_{1}\left(V_{2}-V_{1}\right)+\frac{1}{2}\left(V_{2}-V_{1}\right)\left(P_{2}-P_{1}\right)>0 \ \ \ \ \ (4)$

The total work done on the gas is

$\displaystyle W=W_{A}+W_{B}+W_{C}=\frac{1}{2}\left(V_{2}-V_{1}\right)\left(P_{2}-P_{1}\right)>0 \ \ \ \ \ (5)$

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

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

$\displaystyle U=\frac{3}{2}NkT=\frac{3}{2}PV \ \ \ \ \ (6)$

so along side A (since ${P}$ is constant and ${V}$ increases)

$\displaystyle \Delta U_{A}=\frac{3}{2}P_{1}\left(V_{2}-V_{1}\right)>0 \ \ \ \ \ (7)$

along side B

$\displaystyle \Delta U_{B}=\frac{3}{2}V_{2}\left(P_{2}-P_{1}\right)>0 \ \ \ \ \ (8)$

(since ${V}$ is constant and ${P}$ increases) and along side C

$\displaystyle \Delta U_{C}=-\frac{3}{2}\left(P_{2}V_{2}-P_{1}V_{1}\right)<0 \ \ \ \ \ (9)$

(since both ${P}$ and ${V}$ decrease). The net change in ${U}$ 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 ${Q=\Delta U-W}$ on each side. On side A

 $\displaystyle Q_{A}$ $\displaystyle =$ $\displaystyle \frac{3}{2}P_{1}\left(V_{2}-V_{1}\right)+P_{1}\left(V_{2}-V_{1}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{5}{2}P_{1}\left(V_{2}-V_{1}\right)>0 \ \ \ \ \ (11)$

On side B

$\displaystyle Q_{B}=\frac{3}{2}V_{2}\left(P_{2}-P_{1}\right)+0>0 \ \ \ \ \ (12)$

And on side C

$\displaystyle Q_{C}=-\frac{3}{2}\left(P_{2}V_{2}-P_{1}V_{1}\right)-P_{1}\left(V_{2}-V_{1}\right)-\frac{1}{2}\left(V_{2}-V_{1}\right)\left(P_{2}-P_{1}\right)<0 \ \ \ \ \ (13)$

The total heat added to the gas is

$\displaystyle Q=Q_{A}+Q_{B}+Q_{C}=-\frac{1}{2}\left(V_{2}-V_{1}\right)\left(P_{2}-P_{1}\right)=-W<0 \ \ \ \ \ (14)$

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.

## 10 thoughts on “PV diagrams: a monatomic ideal gas follows a triangular cycle”

1. Arthur

I 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?

1. gwrowe Post author

The change in ${U}$ along side C is the final energy ${U_{f}=\frac{3}{2}P_{1}V_{1}}$ minus the initial energy ${U_{i}=\frac{3}{2}P_{2}V_{2}}$, which gives equation 9.

2. Vital

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

1. gwrowe Post author

With the definitions of ${Q}$ and ${W}$ used by Schroeder, see equation 1 here, so ${Q=\Delta U-W}$. Here ${Q}$ is the heat added to a system and ${W}$ is the work done on the system, so their sum is the change in internal energy of the system.

1. Vital

Yes, 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)? 🙂