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My Question:

Is it possible to show that the Carnot cycle is the most efficient cycle in an ideal gas without using any form of the Second Law of Thermodynamics? Also. If this is possible, how can we prove it?

Gases may be limited to ideal gases. In the general cyclic process, the work done by the engine during one cycle is supposed to correspond to the area enclosed by the loop. But how can we estimate the amount of heat that has entered the general cycle? It also does not seem clear what the low and high temperature heat sources are for the general cycle.

I understand that if I were to draw a loop on a PV diagram, even if it were mathematically more efficient than the Carnot cycle, it would of course not be possible if it violated the second law.

In other words, my question is whether such a "non-physical loop" can mathematically exist. Of course, no matter how mathematical my question is, I do not allow P and V to be negative; P and V shall be positive.

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    $\begingroup$ Would the statement even be true if you didn't use the second law? Couldn't an extremely unlikely occurrence allow for a more efficient cycle? $\endgroup$ Apr 27 '21 at 17:53
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But how can we estimate the amount of heat that has entered the general cycle? It also does not seem clear what the low and high temperature heat sources are for the general cycle.

These quantities are much easier to see on a $TS$ diagram (a temperature-entropy diagram) than on a $PV$ diagram. Hopefully it should be obvious that the any closed cycle which returns to its original state will also form a closed loop in the $TS$ plane.
enter image description here By Eric_DUMINIL at English Wikipedia

To interpret this, note that:

  • Since $dQ = T \, dS$ for a reversible process, heat is being added to the working fluid at all points for which $dS > 0$, and removed at all points for which $dS < 0$.
  • This implies that the area under the "upper" part of the curve connecting $S_A$ and $S_B$ above is the heat $Q_H$ added to the fluid from the hot reservoir, while the area under the "lower" part of the curve is the heat rejected to the cold reservoir $Q_C$.
  • By conservation of energy, the work done by the gas is $Q_H - Q_C$, or the area enclosed by the loop (just like in a $PV$ diagram.)
  • The hottest temperature reached in the cycle and the coldest temperature reached in the cycle are quite obvious: they're the $T$ values at uppermost and lowermost points on the loop, respectively.

Finally, this perspective allows us to see why the Carnot Cycle is the most efficient. Suppose that we have some cycle in which the operating fluid cycles between $S_A$ and $S_B$, and is in contact with some hot reservoir $T_H$ and cold reservoir $T_C$. What kind of loop gives us the most efficiency?

Well, the most efficient loop will be the one for which the least heat is wasted, i.e., the one for which $Q_C$ is smallest when expressed as a fraction of $Q_H$. This implies that we want a loop that makes $Q_H$ as big as possible while making $Q_C$ as small as possible. In other words, the top part of the loop should be as far up as possible, while the bottom part should be as low down as possible. This implies that the top part of the curve should be an isothermal process at $T_H$ from $S_A$ to $S_B$, while the bottom part should an isothermal process at $T_C$ from $S_B$ to $S_A$. These two processes are connected by isentropic processes at $S_A$ & $S_B$, forming a rectangle in the $TS$ plane. This exactly describes the Carnot Cycle.

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  • $\begingroup$ Thank you for your answer, I now understand that the T-S diagram makes the meaning of QH and QL clearer. However, your answer seemingly uses the second law of thermodynamics of the form "dS>0". It seems to me that the question remains as to whether it is possible to derive an expression for the entropy of an ideal gas and "dS>0" from PV=nRT. $\endgroup$ Apr 27 '21 at 19:05
  • $\begingroup$ @BlueVarious: I didn't use the second law of thermodynamics; I just said that the entropy of the working fluid can increase or decrease ($dS > 0$ or $dS < 0$) depending on what is done to it. (It's not a closed system, so $S$ can change in either direction.) I suppose that my answer presupposes that there exists a state variable $S$ called the entropy with the property that $dQ = T dS$, but that's about it. $\endgroup$ Apr 27 '21 at 19:42
  • $\begingroup$ I don't understand the form of the entropy of an ideal gas. But, is it correct that in an isothermal reversible process, T_A is constant, but the state transitions from (S_A,T_A) to (S_B,T_A), and in an adiabatic reversible process, since it is an isentropic process, it transitions from (S_B,T_A) to (S_B,T_B), and in an isothermal reversible process, it transitions from (S_A,T_B) to (S_A,T_B), and in an adiabatic reversible process, it returns from (S_A,T_A) to (S_A,T_A)? $\endgroup$ Apr 27 '21 at 22:07
  • $\begingroup$ Probably the Carnot cycle seems to be the most efficient, at least under the assumption that the loop must be oval in the T-S diagram. $\endgroup$ Apr 27 '21 at 22:24

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