# Trying to understand entropy as a novice in thermodynamics

I recently had my second lecture in Thermodynamics, a long lecture which involved the first law and a portion of the second law. At some point during the lecture we defined entropy as the change of heat energy per unit temperature. From this we derived a general expression for entropy (using laws derived for ideal gases) in which it was clear that it depended on the change of temperature and volume through the process as well as the number of moles.

I have also learned that entropy is a measure of disorder in a system which was nonsense to me especially that I don't understand how disorder (chaotic movement of particles) is related to the change in heat energy per unit temperature, it's more related to specific heat if you ask me, nonetheless in attempts to understand what's useful in knowing what's the amount of disorder in a system I learnt that it measures the state of reversibility of the process which still doesn't make sense when trying to relate it with the "change in heat energy per unit temperature".

TL;DR:

I need an answer these questions:

1. A process has an entropy of X what does this tell me?
2. Another process has higher entropy what does this tell me?
3. How can I relate the definition of entropy to "change in heat energy per unit time"?

Please don't explain using statistical thermodynamics.

• Unfortunately, everything made sense to me once I studied statistical mechanics. I remember being able to solve exercises without the feeling of "understanding" what the second law was about though. But everyone is different. – Alexandre C. Apr 21 '19 at 9:28
• This link was very helpful for me – gen-ℤ ready to perish Apr 21 '19 at 20:15
• See my latest update (Addendum 2) for you to reconsider my answer. – Bob D Apr 22 '19 at 22:11
• As someone with only a high-school background in physics, I'm a fan of Steve Mould's Youtube Explanation. However, I feel like his explanation is more of a description of entropy (and of the consequences of entropy) than a "proper" definition. If nothing else, relating heat differentials to sterling engines to the heat death of the universe requires far less mathematical work compared to statistically modeling energy states (though both approaches give the same result, for the same reasons). – Brian Apr 25 '19 at 22:17

This is a big topic with many aspects but let me start with the reason why entropy and the second law was needed.

You know the first law is conservation of energy. If a hot body is placed in contact with a cold body heat normally flows from the hot body to the cold . Energy lost by the hot body equals energy gained by the cold body. Energy is conserved and the first law obeyed.

But that law would also be satisfied if the same amount of heat flowed in the other direction. However one never sees that happen naturally (without doing work). What's more, after transferring heat from hot to cold you would not expect it to spontaneously reverse itself. The process is irreversible.

The Clausius form of the second law states that heat flows spontaneously from hot to cold. Clausius developed the property of entropy to create this as a general state function that could eventually be determined independently of trying to map just heat flow.

Found a little more time to bring this to the next level. This will tie in what I said above to the actual second law and the property of entropy.

So we needed a new law and property that would be violated if heat flowed naturally from a cold body to a hot body. The property is called entropy, $$S$$, which obeys the following inequality:

$$\Delta S_{tot}=\Delta S_{sys}+\Delta S_{surr}≥0$$

Where $$\Delta S_{tot}$$ is the total entropy change of the system plus the surrounding (entropy change of the universe) for any process where the system and surroundings interact. The equality applies if the process is reversible, and the inequality if it is irreversible. Since all real processes are irreversible (explained below), the law tells us that the total entropy of the universe increases as a result of a real process.

The property of entropy is defined as

$$dS=\frac {dQ_{rev}}{T}$$

where $$dQ$$ is a reversible differential transfer of heat and $$T$$ is the temperature at which it is transferred. Although it is defined for a reversible transfer of heat, it applies to any process between two states. If the process occurs at constant temperature, we can say

$$\Delta S=\frac{Q}{T}$$

where Q is the heat transferred to the system at constant temperature.

We apply this new law to our hot and cold bodies and call them bodies A and B. To make things simple, we stipulate that the bodies are massive enough (or the amount of heat Q transferred small enough) that their temperatures stay constant during the heat transfer Applying the second law to our bodies:

$$\Delta S_{tot}=\frac{-Q}{T_A}+\frac{+Q}{T_B}$$

The minus sign for body A simply means the entropy decrease for that body because heat is transferred out, and the positive sign for body B means its entropy has increased because heat is transferred in.

From the equation, we observe that for all $$T_{A}>T_{B}$$, $$\Delta S_{tot}>0$$. We further note that as the two temperatures get closer and closer to each other, $$\Delta S_{tot}$$ goes to 0. But if $$T_{A} meaning heat transfers from the cold body to the hot body, $$\Delta S$$ would be less than zero, violating the second law. Thus the second law precludes that natural transfer of heat from a cold body to a hot body.

Note that for $$\Delta S_{tot}=0$$ the temperatures would have to be equal. But we know that heat will not flow unless there is a temperature difference. So we see that for all real heat transfer processes, such processes are irreversible.

Irreversibility and entropy increase is not limited to heat transfer processes. Any process goes from a state of disequilibrium to equilibrium. Beside heat, you have processes involving pressure differentials (pressure disequilibrium). These process are also irreversible and generate entropy.

This will focus on the specific questions no. 1 and 2 in you post, that is

1. A process has an entropy of X what does this tell me?

2. Another process has higher entropy what does this tell me?

Before answering this, it has been said that when the change in entropy, $$\Delta S$$, is positive, “heat has entered the system”. It should be noted that heat entering the system is a sufficient condition for a positive entropy change, but it is not a necessary condition.

As I said above, irreversibility and entropy generation is not limited to heat transfer processes. For example, an irreversible adiabatic expansion results in an increase in entropy, although no heat transfer occurs.

An example is the free adiabatic expansion of an ideal gas, a.k.a. a Joule expansion. A rigid insulated chamber is partitioned into two equal volumes. On one side of the partition is an ideal gas. On the other side a vacuum. An opening is then created in the partition allowing the gas to freely expand into the evacuated half. The process is irreversible since the gas will not all return to its original half of the chamber without doing external work (compressing it).

Since there was no heat transfer between the gas and the surroundings, $$Q=0$$, and since the gas expanded into a vacuum without the chamber walls expanding, the gas does no work, $$W=0$$. From the first law, $$\Delta U=Q-W=0$$. For an ideal gas, any process, $$\Delta U=C_{v}\Delta T$$. Therefore there is no change in temperature. The end result is the volume of the gas doubles, the pressure halves, and the temperature remains the same.

We can determine the change in entropy for this process by devising a convenient reversible path to return the system to its original state, so that the overall change in entropy for the system is zero. The obvious choice is a reversible isothermal (constant temperature) compression process. The work done on the case in the isothermal compression equals the heat transferred out of the gas to the surroundings (increasing its entropy) and the change in internal energy is zero. Since this occurs at constant temperature we have, for the gas (system),

$$\Delta S=-\frac{Q}{T}$$

Since we have returned the system to its original state, the overall change in entropy of the system is zero. Therefore, the change in entropy due to the free expansion had to be

$$\Delta S_{exp}=+\frac{Q}{T}$$

We could also determine $$\Delta S$$ by combining the first law and the definition of entropy. This gives the second equation in Jeffery’s answer, which for the case of no temperature change ($$dT=0$$) gives us, for one mole of an ideal gas,

$$\Delta S=Rln\frac{V_{f}}{V_i}$$

or, in the case of our free expansion where the volume doubles,

$$\Delta S=Rln2$$

Therefore,

$$\Delta S=\frac{Q}{T}=Rln2$$

Now, to answer your questions, what does this tell us? And what does another process having higher entropy tell us?

Or, to put it another way, why should we care?

One thing it tells us is that, in the case of an ideal gas, an irreversible (free) adiabatic expansion of an ideal gas results in a lost opportunity to do work. In the free adiabatic expansion, no work was done. If, however, the process was a reversible adiabatic process against a variable external pressure (constant entropy process), such that $$Pv^k$$=constant ($$k=\frac{C_{p}}{C_{v}})$$ the gas would have performed work on the surroundings equal to

$$W=\frac{(P_{f}V_{f}-P_{i}V_{i})}{(1-k)}$$

Bottom line: One of the ramifications of an irreversible expansion process is that the work performed will be less than that for the same process carried out reversibly, due to the generation of entropy in the irreversible process. Irreversible processes lower the thermal efficiency of a system in performing work.

Hope this helps.

• @JeffreyJWeimer I don't think I used the term "developed to show ..". I read somewhere that Clausius, in coming up with his 1854 definition of this property was based on the lack of a law that would preclude a process the sole effect was heat transfer from a cold body to a hot one. And I think I did basically say the second law starts from heat flow. In any case, thanks for your suggestion. The OP apparently liked your answer better than mine as he unaccepted mine and accepted yours. – Bob D Apr 21 '19 at 18:04
• @Jeffrey J Weiner no problem with your edit. It’s in line with what I was trying to say. Thanks. – Bob D Apr 21 '19 at 21:31

Entropy is overloaded term. However, in thermodynamics, it has simple meaning.

Entropy of system is a quantity that depends only on the equilibrium state of that system. This is by definition; entropy is defined for a state. If the system is not in equilibrium state, it may or may not have an entropy. But if it is in equilibrium state, it does have entropy.

The value of entropy for some state that we study does not tell us much. This value is rarely of practical interest.

More intereseting is entropy change during a process which brings the systems from one state to a different state.

The reason we talk about entropy is often because it has interesting behaviour in such process: if reversible process happens inside a closed thermally isolated system, entropy of the closed system remains constant, while if non-reversible process happens, its value increases (by how much cannot be universally stated, it may be negligible or huge, but it definitely cannot decrease). This is another way to state 2nd law of thermodynamics.

If you want to understand how this entropy is connected to things like order or information on the molecular level, you have to study statistical physics of molecules. In thermodynamics proper, there is nothing that would connect entropy to such things.

• All systems have entropy, even those in "non-equilibrium" (steady-state) states. The value of the absolute entropy of a system is important in practice. It will tell us the extent that we obtain useful other work from the system since $\Delta G = -S\Delta T$ at constant $p$. – Jeffrey J Weimer Apr 21 '19 at 17:42
• @JeffreyJWeimer "all systems have entropy" is very ambitious, I would like to see a proof of that. If dQ/T is not defined for sequence of states that occur in time (such as when system has no single temperature), standard thermodynamic entropy at those states is not obviously definable. Regarding your formula, I presume you mean reversible process where entropy of a system does not change. But then its value and value of $\Delta G$ are practically useless, they are arbitrary like electric potential, with arb. zero. Only if entropy changes during process T=const, $\Delta S$ is related to heat. – Ján Lalinský Apr 21 '19 at 18:06
• Absolute entropy is also defined through stat mech using $\Omega$. Are you saying that states exist with no defined value of $\Omega$? I would wonder how? Certainly, the equation for the change in Gibbs energy is for the maximum other work for any process at constant $p$. Reversible give the maximum. Otherwise, while absolute $G$ depends on an arbitrary zero, $\Delta G$ NEVER does! – Jeffrey J Weimer Apr 21 '19 at 18:26
• Jeffery and Jan. The points you are both making are good but I'm not sure it they are illuminating for the OP who is just trying to make heads or tails out of something just learned. You are at a whole higher level. – Bob D Apr 21 '19 at 22:43
• @JeffreyJWeimer you are right that in statistical physics there is a different definition of entropy that applies even if system has no single temperature. However, the connection to thermodynamic entropy is then less clear. One can define many different "stat mech" entropies based on some measure of microstates compatible with macroscopic constraints, but those constraints are murky in violently non-equilibrium states, and the resulting entropies do not relate to 2nd law in an obvious way. – Ján Lalinský Apr 21 '19 at 23:03

It is helpful to think of entropy as not as related to any specific process, but as a physical property of a system at equilibrium, like internal energy and enthalpy. The internal energy tells you the total energy that the molecules of a system contain. The entropy is a measure of how this energy is distributed among the molecules of the system. The more non-uniformly this energy is distributed among the molecules, the higher the entropy.

So the entropy does not depend in any way on the specific process (used to arrive at an equilibrium state of a system). The actual process could be reversible or it could be irreversible. However, the only method we have for determining the change in entropy from one equilibrium state to another (irrespective of the actual process used) is to devise an alternative reversible path between the same two states, and to calculate the integral of dq/T for that alternate reversible path.

• > "The more non-uniformly this energy is distributed among the molecules, the higher the entropy." Did you mean "the more uniformly"? – Ján Lalinský Apr 21 '19 at 22:57
• I guess, to be more precise, I should have said that the total energy is distributed among the molecules in such a way that the number of quantum mechanical energy states is maximized. – Chet Miller Apr 22 '19 at 0:51

# Background

Entropy is expressed in three different ways.

## Laws

In the laws of thermodynamics, entropy is defined as $$dS \equiv \frac{\delta q}{T}$$. Entropy is a state function. This means, differences $$\Delta S$$ are independent of the path. This also means, any expression for the derivative of entropy $$dS = f(...)$$ is exact. Since we always only need two parameters to define the state of a pure system, the result of a state function derivative for entropy is a set of Maxwell-equivalent expressions. For an ideal gas, the molar entropy change of a process in a close system without a phase change or chemical reaction can be determined using one of two general expressions.

$$\Delta \bar{S} = \int \bar{C}_p\ d \ln T - R \ln(p_f/p_i)$$ $$\Delta \bar{S} = \int \bar{C}_V\ d \ln T + R \ln(V_f/V_i)$$

For a real substance, the entropy change of the same process will include an excess term $$\Delta_{EX} \bar{S}$$. Expressions for the excess term can be derived from a mechanical equation of state of the real substance.

The Maxwell form of the second law expresses a spontaneity criteria using the total entropy change of the universe. This can be derived based on the differences between heat flow in reversible processes (where the system and surroundings have the same temperature at all times) and irreversible processes (where heat flow must follow the Clausius statement of the second law that says it is spontaneous from hot to cold temperatures).

The most common approach to teach entropy in thermodynamics at an undergraduate college level is through the laws. Applications continue in engineering to entropy balances with a term $$\Delta_{irr} S$$ to account for the irreversibility of processes in real systems.

As a reference for a novice, I have thought this book was interesting if not necessarily rigorous.

Engines, Energy, and Entropy: A Thermo Primer, J. B. Fenn

## Postulates

The postulate form of thermodynamics proposes the existence of entropy in its second postulate $$S = f(U, V, n)$$. The utility of this paradigm is the ability to obtain all four combined laws of thermodynamics using mathematics (Legendre transforms). Support is also given by this approach to the results from the mathematics applied to the various laws of thermodynamics.

The postulate form of thermodynamics is generally taught in graduate engineering courses, especially mechanical or chemical engineering.

## Statistical Mechanics

In statistical mechanics, we thank Ludwig Boltzmann for the definition $$S = k \ln \Omega$$. Unlike in the laws, which only deals with bulk behavior, statistical mechanics deals with matter as particles. Here we learn about entropy and its relationship to disorder. In truth, disorder is better stated as a measure of the number of ways that we can make a system with the equivalent energy using different arrangements of the particles inside of it.

Statistical mechanics is taught in undergraduate and graduate physical chemistry or chemical physics courses on thermodynamics.

1) The entropy CHANGE of a process is expressed as $$\Delta S$$. When this is positive, heat has entered the system (laws). When this is positive, the level of disorder in the system has increased (statistical mechanics). The reverse is true for negative entropy changes.

2) The entropy differences between two systems gives $$S_A > S_B$$. When all else is equivalent (same amounts, temperature, and pressure), this statement says that system A is more disordered than system B (statistical mechanics). We can also propose a process to exchange internal energy between the two systems at constant temperature and volume since $$\Delta U = T\Delta S - p\Delta V = T\Delta S$$ (laws and postulates). At constant temperature and pressure, we could propose a process to exchange enthalpy because $$\Delta H = T\Delta S + V\Delta p$$.

3) You can propose to use a change in entropy per time as a map of change in heat per time using its definition under the laws. By example, for an ideal gas, you would obtain the expression below for a process at constant pressure with moles $$n$$ and molar heat capacity $$\bar{C}_p$$.

$$\dot{S} = \frac{\delta \dot{q}}{T} = n \frac{\bar{C}_p}{T} \frac{dT}{dt}$$

Entropy is heat flow is temperature change. When you take your perspective as the surroundings and state that it is an infinite heat sink or source, you can directly determine that $$dT/dt = 0$$. This means that no entropy change occurs even though heat flows in or out of the (infinite sink/source) surroundings.

Extra) The most common use of entropy is to consider its change during a process $$\Delta S$$ rather than its absolute value $$S$$. The absolute value however has merit. At constant pressure, we determine the maximum amount of other work (non-mechanical work) that we can obtain from a system by the expression below.

$$\Delta G = w_{o} = -S \Delta T$$

A system that is more disordered (statistical mechanics) can give or will need a greater amount of other work for a process that must change its temperature than a system that is highly ordered. Only reversible processes will give/need that maximum. Irreversible will have a lower value of other work.