Problem — True or false on reversibility and cycles

Problem

True or false, with justification:

1. In a reversible adiabatic process the gas exchanges heat with the surroundings.
2. A cycle traversed clockwise in the $P$-$V$ plane produces positive net work.
3. The Carnot efficiency also depends on the type of gas used.
4. No real engine can exceed the Carnot efficiency between the same two temperatures.

Solution

1. FALSE. Adiabatic means, by definition, $Q = 0$: no heat crosses the walls. What changes is the internal energy, due to work: $\Delta E_\text{int} = -L_\text{gas} = nC_v(T_B - T_A)$. In the reversible version we also have $PV^\gamma = \text{const}$.

2. TRUE. The net work of a cycle is the area enclosed by the curve in the $P$-$V$ plane. Traversing the cycle clockwise makes this area positive ($L_\text{net} > 0$): the engine acts as a heat engine. Counterclockwise, $L_\text{net} < 0$ (refrigerator/heat pump).

3. FALSE. The Carnot efficiency depends only on the two temperatures:

$\eta_\text{Carnot} = 1 - \frac{T_C}{T_H}$

In the derivation, the terms containing $n$, $\gamma$, and the volumes cancel out: the working fluid disappears. This is the core of the second law discovered by Carnot.

4. TRUE. The Carnot efficiency is a universal ceiling imposed by the second law: any real engine, always irreversible, operates with $\eta < \eta_\text{Carnot}$. No engine, however refined, can exceed it between the same two temperatures.

$\ev{\text{F},\ \text{T},\ \text{F},\ \text{T}}$

Collegamenti

Argomenti: Termodinamica · Macchine termiche Concetti: Trasformazioni termodinamiche · Ciclo di Carnot · Rendimento · Secondo principio della termodinamica Competenze: Ragionamento di ranking