I am a high school student. So, while studying about thermodynamics, I got a little curious about entropy. As I read, entropy is the rate of change of chaos. So, if the entropy change of a system is $\delta$S and the primary and final entropy is $S_1$ and $S_2$,
then, $\Delta$$S$ = $S_2$ – $S_1$
Now, come to the mathematical expression of $\Delta$$S$
I am really disgusted by the very brief answer of the mathematical expression in my book. It says that, in a reversible process, the amount of heat or $\Delta$$Qrev$ a system is emitting or producing, divided by the temperature at which the heat is emitted or produced of $T$, or $\Delta$$Qrev$$/$$T$, is equal to $\Delta$$S$.
So, $\Delta$$S$ = $\Delta$$Qrev$$/$$T$
Ok, so my book doesn't tell how does this equation arise, and that is the irritating thing. Also, there is another question regarding this topic, when we are talking about the rate of change of chaos: what is the relation of temperature and heat in here?
Although I know that this is a complex part of physics, I am asking for your help so that I can be able to have a clear idea about the matter. Thank you.
Best Answer
For reversible processes, we may derive that $$ 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.