## Saturday, October 9, 2010

### The Definition of Entropy

This post is part of a series, Nonsense and the Second Law of Thermodynamics. The previous post is entitled: The Carnot Cycle.  This post is heavily dependent on the previous post; so I recommend reading it first.

Let q represent the heat transferred in a process, and qrev represent the heat transferred in a reversible process. Let T be the absolute temperature (in Kelvin).

The sum of  qrev/T for all steps of the process over a full Carnot cycle is equal to zero.  In fact, it is true for any reversible cyclic process.

This fact provides the definition of entropy.  The change in entropy (ΔS) for any process is equal to the integral from the initial state to the final state of the quantity: dqrev/T.

Entropy is the reversible heat divided by the absolute temperature.  That is the definition of entropy.  There is also a statistical mechanical description of entropy that I describe in a subsequent post.

For a reversible process, dq/T = dqrev/T, but not for an irreversible process.   For an isolated system (no heat exchange with the surroundings), the entropy of any process must increase or remain the same.

For any process, the entropy of the surroundings plus the entropy of the system must increase.  The entropy of the system can decrease, but only at the expense of the surroundings.

Efficiency of a Carnot Engine

Remember in the Carnot cycle that T2 is larger than T1, and that therefore, if all the heat, q, transferred from the hot body is transferred to the cold body, q/Tis less than q/T1.  That means that the entropy decrease from the hot reservoir is less than the entropy increase of the cold reservoir if the same amount of heat were transferred.

The difference can be used to do useful work, i.e., instead of transferring the same amount of heat to the cold reservoir, only the amount of heat that is required by the second law is transferred and the rest is converted to useful work.

It would be nice to convert as much heat as possible into work.  However, the entropy of the colder body must be increased by at least as much as the entropy of the hotter body was decreased.  That is to say that the change in entropy, ΔS, must not be negative (and for an irreversible process, it must be positive).

If q2 is the heat removed from the hot reservoir, what is the minimum amount of heat, q1 that can be added to the cold reservoir to ensure that ΔS is not negative?

The condition is that q1/T1 is greater than or equal to q2/T2.  In other words, q1 must be greater  than or equal to (q2 x T1/T2).

So the efficiency (maximum work divided by heat transferred) is equal to  (q2-q1)/q2

Substitute (q2 x T1/T2) for q1 and the efficiency is:

1 - T1/T2

Although the proof is a bit beyond the scope of this blog post , it is possible to prove that we cannot build a more efficient engine operating between temperatures T1, and T2.  It is also possible to prove that all reversible Carnot cycles operating between the same two temperatures have the same efficiency, and that any reversible process can be divided into a series of Carnot cycles!

Units of Entropy and the Third Law

Heat has units of energy that is Joules (J).  The temperature must be in Kelvin (K), i.e., the absolute scale of temperature.  So entropy is in units of J/K.

The third law of thermodynamics states that it is impossible to reach absolute zero in a finite number of steps (you cannot leave the game).  This fact allows defining an absolute scale of temperature and entropy (let's keep score).  At T=0, S=0 (neglecting residual entropy).

The next post entitled, Perpetual Motion, explains one of the consequences of the second law.

Sources
• Atkins, P. W. Physical Chemistry, W. H. Freeman and Company, New York, 3rd edition, 1986
• McQuarrie, Donal d A., Statistical Thermodynamics,  University Science Books, Mill Valley, CA, 1973
• Bromberg, J. Philip, Physical Chemistry, Allan and Bacon, Inc., Boston, 2nd Edition, 1984
• Anderson, H.C., Stanford University, Lectures on Statistical Thermodynamics, ca. 1990.