- For a reversible path between two states (1 and 2), entropy change of a system is NOT zero. It is
$$\Delta S = \int_a^b \frac{dQ}{T}$$
For reversible path between two states, entropy of the universe (Or any isolated system) is zero.
$$\Delta S + \Delta S_\text{surroundings} = 0$$
So You cannot just take any system and say that entropy change between two states for this system will be zero because it is zero for a reversible process. It is not. So when you say
Surely the total change of entropy is zero.
for reversible process of closed system, it is not true.
Answer to This question might help you here.
- As for the first part of your question, I don't understand what the question is. Could you edit it to be more specific?
Also, You said the following, which is false.
The entropy changes of the system are same for both cases, reversible
and irreversible processes because the first and final states are
unchanged. In this situation I think the surrounding also have the
same first and final states for both reversible and irreversible
processes.
We don't know whether surrounding has same first and final states or not. We only know about the system's first and final states. Think about it this way: In a reversible process, system is going from state A to B, and so is surrounding. Since it is reversible, $ \Delta S_{System} = - \Delta S_{Surrounding} $. So ultimately, $ \Delta S_{Universe} = 0$.
Now for an irreversible process, we know that through this irreversible path, the System goes from A to B. We don't know about surroundings. Now, since system's states are same, $ \Delta S_{System} $ will be same as above case. For the surrounding, you say that states are same as the reversible case. But then, here also $ \Delta S_{Surrounding} $ would be same as before and again $ \Delta S_{Universe} = 0$. But we know that that is not true for irreversible process. Hence, Surrounding's states must be different. So, in irreversible process, while the system goes A to B same as before, the surrounding must go from A to some C. There is no reason to believe that it would go from A to B again.
.
For reversible processes, we may derive that
$$ d S = \frac{\delta Q}{T} $$
(note that it's $dS$, not $\delta S$, on the left hand side because $S$ itself is actually a well-defined property of the state, namely the amount of disorder in the system, and not the "rate of chaos", as you misleadingly wrote; on the other hand, there is no unique $Q$ "overall accumulated heat" that would define a current state of a system because we don't know "from when" this heat would be counted and only the changes of $Q$ are well-defined which is why we use $\delta$ and not $d$) up to some changes of conventions or redefinitions of variables.
Why is it possible to choose conventions in which the equation above holds? Well, we may always transfer heat into a system or from the system at (nearly) the same temperature which is reversible. When we reverse such a heating by a cooling, there must be a function of the state $S$ that returns to the original value; if the processes were irreversible, it should increase to a higher level.
Now, the question is how such an $S$ and its changes have to be defined so that it has the properties demanded in the previous paragraph. Clearly, it has to be an extensive quantity. So the more heat we answer, the greater is $dS$. Because the heat may always go through an intermediate system of the same temperature, it doesn't matter where the heat is flowing from and into.
So it must be that $dS=\delta Q / B(T)$ where $B(T)$ is some universal function of temperature (and nothing else). How this function $B(T)$ may be determined using any system, for example the perfect gas which is really simple.
The gas obeys $pV=nRT$, the differential for the heat absorbed in the reversible expansion is
$$\delta Q = C_v dT + p\, dV.
$$
This differential is not exact which means that it cannot be written as $dM$ for some function of state (effectively, a function of $p,V,T$ etc.) $M$.
We want
$$dS = \frac{\delta Q}{B(T)} = \frac{C_v dT}{B(T)} + \frac{p\,dV}{B(T)}$$
to be an exact differential of something. The first term is $C_v$ times the differential of some function of $T$, whatever $B(T)$ is. However, the second term $p\,dV/B(T)$ is only an exact differential if the non-constant factor $p$ in the numerator, in front of $dV$, cancels. So $B(T)$ must be equal to $p$ times a function of $V$. But because it's a function of $pV/nR=T$ at the same moment, it must be equal to $pV/nR=T$, up to an overall scaling.
So up to the freedom to rescale the temperature or entropy by a factor (a choice of units), we may choose $B(T)=T$ i.e. equal to the absolute temperature of the ideal gas. No nonlinear function such as $B(T)=T^n$ for $n\neq 1$ would be acceptable. For this example, we would have $B(T)=p^n$ times a power of the volume (times a constant) so in the denominator, we would get $p^n$ which wouldn't cancel $p$ in the numerator and the ratio would still depend on $p$, leaving us with $f(p)dV$ which is not an exact differential.
The ideal gas calculation above was simpler than others but with a working knowledge of the microscopic model of any physical system, we could use any physical system to derive the same thing. In fact, in statistical physics, which is the explanation of thermal phenomena in terms of the statistical properties of many atoms and their jiggling, it is possible to derive $\delta Q = T\,dS$ for all reversible processes in general. But I am afraid that statistical physics is OK material for the college level audience only – but I surely don't want to underestimate you.
Best 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 $T_{A}<T_{B}$ 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.