Change in entropy for a reversible process

Question

The infinitesimal change in entropy of a reversible process is given by $\text{d}S=\frac{\delta Q}{T}$. How is this proven?
For a measurable change, $\Delta S = \int_{1}^{2} \frac{\delta Q}{T}$.
I've always read, for the latter, that the integration bounds 1 and 2 represent the initial and final states. However, I'm not quite sure exactly what this means. Are they the initial and final temperatures? What are the bounds exactly?

Answer 1

The historical proof of this result goes along the following lines (I will not write all the equations):

1. Start with the Kelvin-Planck postulate that “it is impossible to construct a device which operates on a cycle and produces no other effect than the transfer of heat from a single body to produce work.”

2. Show that no cycle can have higher efficiency than the Carnot cycle, because that would violate the Kelvin-Planck postulate.

3. Use an ideal gas to run a Carnot between two reservoirs ar $T_1$ and $T_2<T_1$ to show that the Carnot efficiency, which shouldn't depend on the working fluid, is $$-\frac{W}{Q_1} = 1-\frac{T_2}{T_1}$$

4. Use this result and the energy balance ($Q_1+Q_2+W=0$) to show that the Carnot satisfies $$ \frac{Q_1}{T_1}+\frac{Q_2}{T_2} = 0 $$ This proves that $\oint dQ/T=0$ along the special path of the Carnot cycle.

5. Create an arbitrary closed path and operate a series of Carnot cycles to show that $$ \oint\frac{dQ}{T} = 0 $$ is valid along any closed reversible path.

6. Conclude from #5 that a state function $S$ exists such that $$dS = \frac{dQ_\text{rev}}{T}\tag{1}$$ and call this function entropy

OR

avoid all this circuitous logic which drives students crazy (and weary of entropy) and use Eq. (1) as the definition of entropy, as @GiorgioP-DoomsdayClockIsAt-90 very correctly said. In this day and age, this is the most pedagogical way to talk about entropy (the founders can be excused for stumbling onto entropy in such indirect way).

Answer 2

The equality $$ dS = \frac{\delta Q}{T} \tag{1} $$ in the case of a reversible process, cannot be proven because it is the definition of $dS$. One starts with the fact that for every reversible cycle $$ \oint \frac{\delta Q}{T}=0, $$ to introduce a function of state $S$ by integrating equation $(1)$ between two states.

The difference in entropy between two states, A and B, is then given by $$ S(B)-S(A)=\int_A^B \frac{\delta Q}{T}, $$ where $A$ and $B$ are two arbitrary equilibrium states described by any set of thermodynamic variables relevant to the system of interest. For example, for a closed fluid system, one could use volume and energy as well as volume and temperature or pressure and temperature. Each choice will have a different expression of the differential form $\frac{\delta Q}{T}$ under integral.