# How does dividing by $T$ makes $\delta Q_{rev}$ an exact differential?

The change of entropy is defined as:

$$dS = \frac{dQ_{rev}}{T}$$

With $dQ_{rev}$ being the infinitesimal change of heat in a reversible process. However since $Q_{rev}$ is not a state function the integral,

$$\int_{Z1}^{Z2} dQ_{rev}$$

with $Z_{1,2}$ being states of our system, is path dependant, making $dQ_{rev}$ an inexact differential.

Entropy $S$ however is a state function meaning:

$$\int_{Z1}^{Z2} \frac{dQ_{rev}}{T}$$

is path independent, hence $\frac{dQ_{rev}}{T}$ became an exact differential.

• How does dividing $dQ_{rev}$ by $T$ makes it an exact differential?
• And how does this definition consider work done, since a change of state does not rely on a change of heat?
• Are you looking for a demonstration of the fact that $dQ_{rev}/T$ is an exact differential or for the physical intuition behind it? Jun 11, 2017 at 12:20
• I was looking for a more mathematical explanation, but now as I've seen WetSavanna's answer it seems to me as I underestimated the extent of this question somewhat. Jun 11, 2017 at 13:51

In general, you should not be surprised that path dependence of an integral is changed through multiplication of the integrand by simple functions. Only exact differential one forms have this behavior when integrated, by definition. Take almost any exact differential form, such as $\nabla\phi(x,\,y,\,z,\,\,\cdots)\cdot\mathrm{d}\mathbf{r}$ and multiply it by almost anything else to get $\psi(x,\,y,\,z,\,\,\cdots)\,\nabla\phi(x,\,y,\,z,\,\,\cdots)\cdot\mathrm{d}\mathbf{r}$. You can tell whether the resultant is an exact differential whose integral will be path independent by calculating $\nabla\times(\psi\,\nabla\phi)) = \nabla\psi\times \nabla\phi$ and checking whether it vanishes. You should easily see that, unless we have the very special case where $\nabla\psi$ is parallel to $\nabla\phi$, the path independence of the original integral of $\nabla\phi$ is ruined!

So this is part of the answer, but really we need physics - thermodynamics - insight. The reason comes from the definition of temperature and its relationship with the second law of thermodynamics. $T$ can be thought of as being defined to make $dQ_{rev}/T$ an exact differential.

As I discuss further in this answer here and here, temperature was defined by Carnot through the efficiencies of reversible heat engines. Carnot's theorem, that all reversible heat engines have the same efficiency, follows from the second law, because otherwise you could connect your more efficient engine to the least efficient one, configure the latter as a heat pump and thus spontaneously pump heat from a colder to a hotter reservoir. Given this theorem, we have a way to compare the "hotnesses" of reservoirs through the efficiency of a reversible heat engine running between them because the proportions of heat converted to work will be the same for any reversible heat engine working between the same two reservoirs. If we run a heat engine between a hot reservoir and a colder, but standard, reservoir, and if we define the latter to be "unity" temperature, then if the ratio of the proportion of heat drawn from the hot to that exhausted into the cold is $T$, then we define the temperature of the hot reservoir to be $T$. Note that this definition also gives us the work done as the difference between the intake and exhaust heats. The higher $T$, the bigger this ratio, and thus the greater the fraction of intake heat converted to work.

From this definition, we now look at any closed loop in state space. We can work out $\oint\frac{\mathrm{d}Q_{rev}}{T}$ by decomposing the loop into the traversal of little Carnot cycles and work out the limit as the areas of the little cycles is taken to nought. Therefore, the only way that $\oint\frac{\mathrm{d}Q_{rev}}{T}$ can differ from nought is if it differs from nought around at least one of the little Carnot cycles: what this means is that we'd have a reversible heat engine with a different efficiency from the Carnot cycle, which gainsays Carnot's theorem and thus the second law of thermodynamics. We must have $\oint\frac{\mathrm{d}Q_{rev}}{T}=0$, otherwise we violate the second law.

So by defining things in this way, we force the differential to be exact. The soundness of this definition is implied by the second law.

You should be able now to see the answer to your other question:

And how does this definition consider work done, since a change of state does not rely on a change of heat?

This comes from Carnot's theorem: the difference between the intake and exhaust heat equals the work output, and this must always the same for any reversible heat engine working between the same two reservoirs, otherwise Carnot's theorem is violated.

You are touching on some important though your formula is not exactly correct.

The correct one (refer to Wikipedia) is $$\triangle S = \int \frac{\delta Q}T$$ Here $\delta$ is used which means it is path dependant.

Note the second law shows that the above entropy is a state variable.