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.

Answer

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.

ADDENDUM 1:

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.

ADDENDUM 2:

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.