Problem — Carnot in p-V (adiabatics vs isotherms)

Problem

A $p$-$V$ graph shows a cycle made of two isotherms ($T_c$ high, $T_f$ low) and two adiabatics. How can the adiabatics be visually distinguished from the isotherms? In which direction is the cycle traversed for an engine that produces work?

Solution

Isotherms and adiabatics compared. The two families of curves follow different laws:

• isotherm: $pV = \text{const}$ (hyperbola, fixed $T$);
• adiabatic: $pV^\gamma = \text{const}$, with $\gamma > 1$.

Since the exponent $\gamma$ is greater than $1$, along an adiabatic the pressure drops more steeply as the volume grows. Graphically:

$\ev{\text{the adiabatics are steeper than the isotherms}}$

Starting from the same point, the adiabatic falls below the isotherm passing through that same point. In the drawing these are the curves $B\!\to\!C$ and $D\!\to\!A$ (in gold), steeper than the red and blue hyperbolas.

Direction of the cycle. An engine that produces work yields positive net work, so the enclosed area must be traversed clockwise in the $p$-$V$ plane: expansion (at the top, high pressure) followed by compression (at the bottom, low pressure), so that the expansion work exceeds the compression work.

$\ev{\text{work-producing engine} \Rightarrow \text{cycle traversed clockwise } (A\to B\to C\to D)}$

Traversing the cycle counterclockwise would instead give a refrigerator/heat pump: work absorbed, heat pumped from cold to hot.

Collegamenti

Argomenti: Macchine termiche Concetti: Ciclo di Carnot · Trasformazioni termodinamiche Competenze: Lettura dei grafici Metodi: Diagramma P-V e ln P - ln V Oggetti: Gas ideale