Lecture 6: Entropy

Note

The energy of the world is constant. The entropy of the world tends towards a maximum. Rudolph Clausius

Warning

This lecture corresponds to Chapter 14 of the textbook.

Summary

Attention

When we studied the Carnot Cycle in Lecture 5: Second Law of Thermodynamics, we came up with an extraordinary result: the Clausius inequality. For a reversible process, it states that

$\oint \mathrm{đ} Q_{\mathrm{rev}} / T=0$

The fact the loop integral is zero translates into the fact that

$dS= \mathrm{đ} Q_{\mathrm{rev}} / T$

is an exact differential (see Complement 1: Exact Differential for important properties).

We also know that if a function corresponds to an exact differential, it is a function of state! This leads us to define $S$ as a new function of state, known as the entropy. Now, there is a difficulty that students often struggle with: while there is such a thing as an “entropy” as a function of state; the above equation only defines the entropy for a reversible process. It must be stressed that this does not mean the entropy is only defined for a reversible path! Instead, this means that to calculate the entropy change between two states, an equivalent reversible path has to be used if we want to calculate entropy using the equation above.

Since, there is, in practice, no truly reversible path, and using the following relationship (coming directly from Clausius inequality):

$\mathrm{d} S=\frac{\mathrm{d} Q_{\mathrm{rev}}}{T} \geq \frac{\mathrm{d} Q}{T},$

we realize that for a thermally isolated system, then

$\mathrm{d} S \geq 0.$

The strict equality will only be realized for a reversible process. All physical processes with see their entropy increase so long as the system is thermally isolated (this last sentence is another formulation of the second law of thermodynamics). We finally note that a system that is not thermally isolated can see its entropy decrease.

We study the Joule expansion as a key example to apply our newly discovered function of state $S$ (see definition below). Clearly, the Joule expansion is not a reversible process! It therefore corresponds to an increase in entropy. Finding the change in entropy is, at first sight, a difficult problem to solve. However, as mentioned above, the trick is to consider an equivalent reversible process between the two end points. This allows us to determine that the entropy changes as:

$\Delta S=R \ln 2.$

This allows us to highlight a possible paradox: in the thermally isolated system where no work is done, it would appear that no change in internal energy should take place (which is the case) and also no entropy change would occur! This is not correct: the entropy must be calculated using a reversible process rather than the physical one (warning: this is a point that requires some thinking before moving on!).

One other important result we obtain in this lecture is the link between the number of microstates (corresponding to a given macrostate – the one corresponding to a given internal energy, in other words, we work in a microcanonical ensemble). Remembering our definition of temperature provided in Lecture 1: General Introduction, we get the famous result:

$S_{\rm Boltzmann}=k_{\mathrm{B}} \ln \Omega$

This is a crucial result that can be used to connect the thermodynamic definition of entropy with the “popular” definition that entropy is “disorder”: indeed, if a system (like your dorm room) is in high disorder, and because there are many more ways to have a disordered room than an ordered one, the number of microstates is large and the entropy is large as well! This definition also allows us to understand the change in entropy of the Joule expansion, as described in the slides and screencast linked below.

Note

One source of confusion is how we can compute the entropy related to a process that is not reversible. The problem is that the thermodynamic definition of entropy relies on a reversible transfer of heat (the reason is that the Clausius inequality is only a strict equality for reversible processes):

$\mathrm{d} S=\frac{\mathrm{d} Q_{\mathrm{rev}}}{T}$

Let examine this for the case of a Joule expansion (See definition below).

A Joule expansion of an ideal gas is an irreversible process. There is no heat or work exchanged and thus, no change in internal energy. We now remember that for an ideal gas, no change in internal energy means that the temperature is constant. So, the process we are looking at is between:

Starting point: one container at temperature T with pressure p and volume V

End point: one container at temperature T with pressure p/2 and volume 2V. (the change in volume is dictated by the experiment and the change in pressure is a direct consequence of the ideal gas law).

Now, to calculate the entropy, we need to find a path that has the following properties:

it has to be reversible.

it must have the same starting and end points (here we insist that the starting and end points are defined using variables of state; so work or heat are not adequate).

In other words, to use the definition of entropy above, we must work with a reversible isotherm between (V,p) and (2V,p/2).

For such a process, we know that (because U does not change):

$\mathrm{d} S=\frac{\dbar Q_{\mathrm{rev}}}{T} = \frac{p\mathrm{d} V}{T} =\frac{R\mathrm{d} V}{V}$

where the last equality is obtained by using the ideal gas law. We thus obtain:

$\Delta S==\int_{V_{0}}^{2 V_{0}} \frac{R \mathrm{~d} V}{V}=R \ln 2$

Note: we see from these equations that the reversible process is NOT adiabatic (in contrast to the real process!). This is ok since heat is not a function of state.

Note

Maxwell demon: Maxwell’s demon is introduced as a way to make us think! It seems capable of causing entropy to decrease in a system with no consequent increase in entropy anywhere else. Does Maxwell demon disprove the second law of thermodynamics? It turns out it doesn’t: the process the demon uses decreases the amount of information on the system and, as we shall see in the next lecture, the less information one has, the more entropy we get!

Learning Material

Copy of Slides

The slides for Lecture 6 are available in pdf format here: pdf

Screencast

Key Definitions

Note

Internal energy:

For any process, we have: $\mathrm{d} U=T \mathrm{~d} S-p \mathrm{~d} V$ . However, $T \mathrm{~d} S$ can only be understood as heat for a reversible process. Likewise, $-pdV$ is only interpreted as work for a reversible process.

Joule expansion:

Joule expansion of an ideal gas is the expansion of an ideal gas “against” vacuum and in a thermally isolated system.

Natural variable:

The natural variables are a set of appropriate variables that allow to compute other state functions by partial differentiation of the thermodynamic potentials. For example, $U(S,V)$ , which means that the internal energy is a function of entropy and volume. This allows to write:

$\mathrm{d} U=\left(\frac{\partial U}{\partial S}\right)_{V} \mathrm{~d} S+\left(\frac{\partial U}{\partial V}\right)_{S} \mathrm{~d} V$

Boltzmann’s entropy:

Boltzmann entropy is given by

$S_{\rm Boltzmann}=k_{\mathrm{B}} \ln \Omega$

where $\Omega$ is the number of microstates corresponding to a given macrostate (microcanonical).

Gibbs’ entropy:

Gibbs’ entropy is given by

$S_{\rm Gibbs}=-k_{\mathrm{B}} \sum_{i} P_{i} \ln P_{i},$

where the sum runs over all possible microstates (each with probability $P_{i}$ . This expression is a generalization of Boltzmann entropy (where the probability of each microstate is the same). In this sense, Gibbs’ definition is a generalization that also applies in out-of-equilibrium situations.

A full list of terms, including the ones provided here, can be found in the Index.

Test your knowledge

Thermodynamically, the entropy is…
1. an exact differential that can be calculated for any path
2. an exact differential that can be calculated for reversible paths only
3. an exact differential that can be calculated for irreversible paths only
Imagine an isolated system that undergoes an irreversible adiabatic change. What can be said for sure?
1. The entropy of the system increases or remains the same.
2. The entropy of the system strictly increases during the process.
3. The entropy of the system remains the same.
Consider a strictly irreversible system. Among the statements below, which one is correct?
1. The heat transfer can be expressed as $\dbar{Q}=TdS$ and work involved as $\dbar{W}=-pdV$ .
2. The sum of the heat transfer and work involved can always be written $TdS-pdV$ .
3. There is not enough information to answer this question conclusively.
Here are four statements, which one is correct?
1. The entropy of an open system always increases, unless the process is reversible.
2. The entropy of a closed system always decreases, unless the process is reversible.
3. The entropy is a function a state provided all processes involved are reversible.
4. The entropy is only a function of state at $T=0$ .
5. None of the other answers is correct.
Maxwell’s demon
1. is capable of defeating the second law of thermodynamics by decreasing the entropy of the closed system. This remains a major issue with thermodynamics.
2. seems to be capable of defeating the second law of thermodynamics by decreasing the entropy of the closed system. This paradox is resolved by the fact entropy of information is changed during the process.
3. simply shows that some microstates, in spite of their very low but non-zero probability, can spontaneously appear during an experiment; there is no real paradox.
4. None of the other options is correct.
Imagine you are cooling down a metallic rod from $T_\textrm{initial}=200K$ to $T_\textrm{final}=100K$ . In this thought experiment you have access to a infinite number of heat baths continuously varying in temperature between $T_\textrm{initial}$ and $T_\textrm{final}$ . In your opinion, do you have enough information to devise a cooling process for which the entropy does not change?
1. Yes
2. No
3. It depends on the type of metallic rod we are working with.