entropystatistical mechanicsthermodynamics
Can we talk about the entropy for just one electron, such as the electron in hydrogen atom by the famous Boltzmann's formula?
$$S=k_B\ln\Omega$$
I guess the answer is OK.
But in thermodynamics, we also define the relation closely related to entropy
$$ dS = \dfrac{dQ}{T} $$
But it will be nonsense to talk about the temperature. So What's the paradox in both definitions?
The Second Law of Thermodynamics states that the entropy of the universe always increases or stays constant. This means that we can reduce the entropy of the gas in the box (gas compressed from the whole box to half the box at constant temperature), only if we increase the entropy somewhere else. For example, we can compress the gas, doing work on it, and thus heating it. Then we have to allow the gas to lose this heat to the environment so that it ends at the original temperature, in half the box. The result is that by heating the environment I have increased the entropy of the environment by at least as much as I have decreased the entropy of the gas in the box.
Since most systems are not isolated (they can exchange energy with the environment), I always have to include both the system (box of gas etc) and the environment.
2) -> 1): This is more probable, and energy is set free.
If we allow the gas to expand suddenly into the whole box by removing the partition in the middle, the entropy increases, but there is no change in the energy. This is an irreversible process - entropy increases.
ΔQ=TΔS
This is not generally true. It is only correct when the change is reversible - in other words the change occurs slowly through a series of equilibrium states. You also have to interpret it correctly. ΔQ is the heat added to the system from the outside. ΔS is the increase in entropy of the system. If the process of taking this heat from the outside world is also reversible, the outside world loses ΔQ of heat, and loses ΔS in entropy, so the entropy of the universe is unchanged.
The case above (sudden expansion of the gas) is an example where this equation does not hold. There is no transfer of heat, but the entropy of the gas increases.
To summarise. The temperature of a gas is a fairly easy concept - it is proportional to the average random kinetic energy of the molecules. The entropy is more complicated. Entropy is increased by allowing the gas to occupy more volume, because the molecules can be arranged in space in a larger number of ways (more "states" as you describe it). Entropy is also increased by raising the temperature of the gas, because this gives more ways for the kinetic energy to be distributed.
When does a change in a system occur spontaneously (for example a chemical reaction, or the expansion of a gas)? It occurs either because the change increases the entropy of the system, or because it decreases the total energy of the system. The spare energy leaves the system, heating and thus increasing the entropy of the environment.
Of course, if the course you are taking is not in physics or physical chemistry, it is likely to include a great deal of nonsense when it comes to entropy. The subject can only be understood by disentangling the concepts as I have tried to do.
Always ask,
1. Energy or entropy? Energy is conserved, entropy is not.
2. Energy, heat, or temperature - all different concepts.
2. Is the change reversible or irreversible?
3. When the system changes, does the outside world also change?
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 $T_{A}<T_{B}$ 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,
$$\Delta S=R\ln\frac{V_{f}}{V_i}$$
or, in the case of our free expansion where the volume doubles,
$$\Delta S=R\ln2$$
Therefore,
$$\Delta S=\frac{Q}{T}=R\ln2$$
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.
Best Answer
The bare Boltzmann formula is a bit basic to make sense of for a lone electron. You need to talk about a generalization: one thinks in terms of the Shannon entropy of the state.
We also need to be careful in that the entropy becomes conditional on knowledge that we already have.
An electron in a known pure quantum state has zero entropy: we know its state perfectly and need no information to define it.
If however, if it is in a mixed state, then there is indeed a nonzero entropy. Let's say the spin up / spin down is unknown to us; that is, it is in a classical probabilistic mixture of pure spin states. This situation might have arisen because we have sampled the electron from an ensemble, or it might have arisen through something like the Wigner's Friend thought experiment, where it is known that the is in a spin eigenstate but not which one by Wigner. Wigner's friend has all the measurement results, so right after the measurement, the entropy conditioned on being Wigner's friend is nought. If there are probabilities $p$ and $1-p$ of the electron's being spin up / down respectively, then the entropy conditioned on being Wigner is $-p\,\log p -(1-p)\,\log p$ (multiply by the Boltzmann constant if you want to give it the same dimensions as $Q/T$. More generally, if a quantum system is in a classical mixture of pure states described by Density Matrix $\rho$, then the above formula generalizes to the von Neumann entropy:
$$S=-\mathrm{tr}(\rho\,\log\rho)$$
Now to the temperature of an electron. Temperature is a parameter of a statistical distribution, namely it defines the Boltzmann distribution of an equilibrium ensemble of particles. As such, the notion is not directly applicable to a lone electron: we simply don't have a system of particles in thermodynamic equilibrium! However, we might have sampled the lone electron from an ensemble of electrons in thermodynamic equilibrium at temperature $T$. We can therefore think of its energy state as a classical mixture of energy eigenstates, and the von Neumann entropy of this mixture is (modulo multiplication by the Boltzmann constant) precisely the Gibbs entropy of the ensemble, calculated as a per-particle average entropy. One can then say that the electron is from a population at temperature $T$. If, for example, we add a small amount of heat to the electron population just before the sampling, then the population's entropy changes by $\mathrm{d}Q/T$, and this further entropy, divided per particle, is the change in the entropy attributable to the heat addition of the electron as a mixture of energy eigenstates.