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

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  3. 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

        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)? 🙂


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