Problem — Carnot cycle in the ln P – ln V plane

Problem

In the $\ln P$–$\ln V$ plane the adiabats and isotherms become straight lines. Draw a Carnot cycle on a single $\ln P$–$\ln V$ graph. What are the slopes of the four segments? How can one tell that in the $\ln P$–$\ln V$ plane the area no longer represents work?

Solution

Why they become straight lines. Taking the logarithm of the process equations:

• Isotherm: $PV = nRT = \text{const} \Rightarrow \ln P + \ln V = \text{const}$, a line of slope $-1$;
• Adiabat: $PV^\gamma = \text{const} \Rightarrow \ln P + \gamma\ln V = \text{const}$, a line of slope $-\gamma$ (steeper, since $\gamma > 1$).

The cycle. The two isotherms (slope $-1$) are parallel to each other, and so are the two adiabats (slope $-\gamma$): the Carnot cycle becomes a parallelogram with four straight sides.

Carnot cycle in the $\ln P$–$\ln V$ plane: a parallelogram. Isotherms of slope $-1$, adiabats of slope $-\gamma$ (steeper).

Slopes of the four segments.

$\ev{\text{isotherms: slope } -1; \qquad \text{adiabats: slope } -\gamma \text{ (steeper)}}$

Why the area is no longer the work. In the $P$-$V$ plane the area has units $[\,P\,]\cdot[\,V\,] = \text{Pa}\cdot\text{m}^3 = \text{J}$, i.e. an energy: it is the work. In the $\ln P$–$\ln V$ plane the axes are pure numbers (dimensionless logarithms), so the enclosed area is also dimensionless: it does not have the units of an energy. The logarithmic change of coordinates distorts the areas and erases their physical meaning. The $\ln P$–$\ln V$ diagram is useful for visualising the curves as straight lines with distinctive slopes, but to compute the work one must return to the $P$-$V$ plane or to the formulas.

Links

Topics: Macchine termiche Concepts: Ciclo di Carnot · Trasformazioni termodinamiche Skills: Lettura dei grafici Methods: Diagramma P-V e ln P - ln V Objects: Gas ideale