# Integrating factor $1/T$ in 2nd Law of Thermodynamics

How would you prove that $1/T$ is the most suitable integrating factor to transform $\delta Q$ to an exact differential in the second law of thermodynamics: $$dS = \frac{\delta Q}{T}$$

Where $dS$ is the change in entropy, $\delta Q$ is the change in heat energy, and $T$ is the equilibrium temperature.

• I am not sure I understand the question...From which premices would you want to prove it? – gatsu Apr 25 '13 at 10:39
• @gatsu From the basics foundations of thermodynamics. – Ana S. H. Apr 25 '13 at 12:05
• Can someone clarify what does the term "integrating factor" mean here? :\ I saw the same phrase used in another book, for exactly the same case. – seavoyage Oct 21 '16 at 3:29

There's a group I call "thermodynamic purists" who think that thermodynamics is a self-contained system based on semi-mathematical "axioms". I disagree! I think that thermodynamics is fundamentally a consequence of statistical mechanics, and that this is the best way to think about it and understand it. I acknowledge that reasonable people can differ on this, but anyway, here is my non-thermodynamic-purist way to address your question. (Someone else can give a thermodynamic-purist answer, which I bet is what you're looking for.)

What is heat $\delta Q$? It's energy flow (in the form of microscopic kinetic energy etc.). What is entropy change $dS$? It's the increase in disorder (more precisely, increase in $k_B$ * log(number of microstates)). What is temperature $T$? It's a parameter describing how unlikely it is that a high-energy microstate will occur (Boltzmann distribution etc.).

Then the appropriate question is: Why (according to these definitions) does $T dS = \delta Q$? It's a good question without a very obvious answer. This was discussed in a recent question, see here.

• Pure thermodynamic is underappreciated. I don't see it as mere consequence of statistical physics, but rather as the more detailed theory as some sort of a model of it. In any case, if you come from the hardcore statistical side, then temperature $(\partial S/\partial E)^{-1}$ is a definition and not a mystery. Here, to me, heat flow is just energy flow which is not due to volumetric change and alike. – Nikolaj-K Apr 26 '13 at 12:09
• You can of course take this path and define a parameter, suggestively named $T$, as the derivative of the logarithm of the number of microstates with respect to the energy. But then you still owe us a very general explanation why that thing is identical to what we in ordinary life call a temperature... – Markus Deserno May 13 '13 at 11:13
• Markus, yes the "very general explanation" you're looking for does exist. It can be found in statistical mechanics textbooks---for example, Kittel & Kroemer. – Steve Byrnes Nov 19 '18 at 15:08

I apologize "basics foundations of thermodynamics" still does not make a lot of sense to me.

Steve B already provided some answer associated to one way to interprete the word "foundation" that is from statistical mechanics. I will kinda play here devil's advocate and assume that you are refering to axiomatic thermodynamics.

As far as I am concerned, the foundations of axiomatic thermodynamics are simply experimental facts but there are many ways to choose the set of fundamental axioms to derive all the other results of thermodynamics and this may be related then to the question you ask.

The second principle of thermodynamics tells you that there is a quantity called entropy and labeled $S$ that is a function that depends on the state of the system only. Now, it also says that for an isolated system under a spontaneous evolution towards thermodynamic equilibrium, the entropy can only increase.

Let us consider for a minute the case of an isolated system that comprises of two identical sub-systems 1 and 2 but that have different internal energies $U_1^0$ and $U_2^0$ such that, say, $U_1^0 > U_2^0$.

Once these sub-systems are put in thermal contact, the whole system is out of equilibrium and there should be a heat flux from sub-system 1 to sub-system 2.

Let us describe it more formally:

The total entropy of the system at any moment (if we imagine the heat transfer slow enough) is given by:

$S_{tot}(t) = S_1(t)+S_2(t) = S(U_1^t,V,N)+S(U_2^t,V,N)$

Note that I used the same function $S$ for both systems since I have said previously that they were the same. However, they are not in the same state and therefore their entropies are different.

Let us assume that over a time interval $\delta t$, $U_1^t \rightarrow U_1^t + \delta U_1$ and $U_1^t \rightarrow U_2^t + \delta U_2$ then, the entropy of the total has to change by an amount:

$\delta S_{tot}(t) = S_{tot}(t+\delta t) - S_{tot}(t) = \left(\frac{\partial S}{\partial U_1}\right)_{V,N}\delta U_1 + \left(\frac{\partial S}{\partial U_2}\right)_{V,N}\delta U_2$

Note also that because of its extensive property $\delta S_{tot} = \delta S_1 + \delta S_2$

By identification (this step may not be very rigorous) we then have that:

$\delta S_i = \left(\frac{\partial S}{\partial U_i}\right)_{V,N}\delta U_i$

Since the sub-systems are only in thermal contact, there is not work exchanged between them and the first principle of thermodynamic says then that $\delta U_i = \delta Q_i$ we thus have:

$\delta S_i = \left(\frac{\partial S}{\partial U_i}\right)_{V,N}\delta Q_i$

Finally, to evaluate what is $\left(\frac{\partial S}{\partial U_i}\right)_{V,N}$, you can look at two cases:

• At time $t$ the system is at equilibrium and therefore $\delta S_{tot}=0$ implies that $\left(\frac{\partial S}{\partial U_1}\right)_{V,N}=\left(\frac{\partial S}{\partial U_2}\right)_{V,N}$, it thus corresponds to an intensive variable that has to "thermalize" between systems at thermal equilibrium...it has to be related to the temperature

• At time $t$ the system is still out of equilibrium and $\delta S_{tot} > 0$ implies that $\left(\frac{\partial S}{\partial U_1}\right)_{V,N} < \left(\frac{\partial S}{\partial U_2}\right)_{V,N}$

Since the subsystems are the same it implies that the higher the temperature of a system the smaller the intensive variable, the simplest quantity that does the job is $1/T$

This derivation does not exclude more complicated - always decreasing - functions of $T$ and I do not know if they should be completely excluded at this stage anyway...I guess that it can be made pretty sure that it should be $1/T$ then by using Schwartz theorem of equality of cross partial derivatives byt changing then the volume or the number of particles, but I am too lazy for that now, sorry.

Yes, it can be proved. But the mathematical rigour only applies to ideal gas, so you might feel disillusioned after all the calculations, considering how much this can be valid in reality.

If you insist you want to know, then you may read on. From the first law of thermodynamics,

$$\delta Q = \frac{3}2 nRdT + \frac{nRT}V dV,$$

where internal energy $U=\frac{3}2 nRT$ from kinetic theory of gas, pressure $P=nRT/V$ from ideal gas equation. The differential equation is exact if and only if $$\frac{\partial\left(\frac{3}2 nR\right)}{\partial V} = \frac{\partial\left(\frac{nRT}V\right)}{\partial T},$$

which is not the case here, that's why $\delta Q$ is inexact differential. If we now multiply a funtion $f(T)$ to $\delta Q$ to make it exact, then $f(T)$ is called integrating factor (in hindsight, we can only say that an integrating factor exists if $f$ is a function of $T$ alone). Simplifying $f(T)$ as $f$ for brevity, it must be true that

$$\frac{\partial\left(\frac{3}2 nRf\right)}{\partial V} = \frac{\partial\left(\frac{nRT}V f\right)}{\partial T}$$

The LHS of the equation is zero:

$$0 = \frac{n R}{V} f + \frac{n R}{V} T f'$$

where $f'=df/dT$. The constants can be eliminated, what remains is just a first order differential equation, and you can easily solve to $f=1/T$.

If you want the most general version of abstract proof you can look into this article, but I doubt you will gain you any further insight from it. Just my two cents.

• You could probably do the same thing, just positing a general equation of state and thermal equation of state, and find the restrictions. After all, the formula for entropy is general, the first law is general, and so, the result better not be equation of state-dependent. – Jerry Schirmer Apr 25 '13 at 17:18
• Perhaps it can be expressed in more general form, but I never heard that something dS=dQ/T can be proven in total without referring to first law. Yes, entropy may be alternatively defined in terms of microstates, but that would be off topic. – Neoh Apr 25 '13 at 17:33
• sure, you'll have to refer to the first law (which is very, very general), but you won't have to refer to the ideal gas law (which is not at all general). – Jerry Schirmer Apr 25 '13 at 17:52
• I'd love to see if there is a proof without referring to ideal gas too. – Neoh Apr 25 '13 at 17:55
• @Neoh Thank you for your answer and for the article, I'll check it. – Ana S. H. Apr 26 '13 at 0:10

Another approach is to $\mathbf{define}$ temperature to be what you've just asked it to be: that is almost the approach that pops out of Carnot's original arguments about the efficiency of heat engines and his formulation of the second law. One can rework the following idea of defining temperature in terms of efficiency to being a function of the scale factor one needs to multiply $\delta Q$ by to make it an exact differential.

The ideas in question, as with many things, are beautifully explained in the chapter "The Laws of Thermodynamics" (Ch 44) in the first volume of the Feynman lectures.

As soon as you realise that all reversible heat engines working between the same two "big"(see footnote) heat reservoirs at equilibrium have the same efficiency $i.e.$ that the proportions of heat drawn from the "hotter" reservoir, that dumped to the "colder" one and work derived are always the same if the engine is reversible (regardless of the engine's workings), you have proven that this universal, uniquely defined efficiency is a way to compare all heat reservoirs at equilibrium, thus: we take a particular reservoir as a standard and call its temperature unity. Then, if we run a reversible heat engine between a hotter reservoir and this one, and three units of heat is taken from hotter one for each unit of heat dumped to our standard reservoir (thus producing 2 units of work), then we shall call the temperature of the hotter one 3 units, by definition. Likewise, if we run a reversible heat engine between our standard reservoir and a colder one and we find that our reservoir delivers three units of heat to the engine for every unit dumped to the colder reservoir, then the colder one is by definition at a temperature of $\frac{1}{3}$ units.

From this definition, it then follows that $\frac{\delta Q}{T}$ is an exact differential because $\int_a^b \frac{d\,Q}{T}$ between positions $a$ and $b$ in phase space must be independent of path (otherwise one can violate the second law). What I've just said is not altogether obvious: you'll have to look at Feynman for details. The Wikipedia entry for "temperature" also does a fair job of explaining the same ideas, in the section "Second law of thermodynamics".

footnote: $i.e.$ big enough that finite heat flows to and from them do not change their macrostate appreciably).

One way to derive the existence and expression for entropy is to consider a system that undergoes arbitrary cyclic process $\gamma$ in which its temperature is changing and is the same as temperature of a heat reservoir it is connected to.

The cyclic process can be decomposed into a set of infinitesimal Carnot cycles, where the change of state $s$ during both isotherms and adiabats is infinitesimal (during isotherms, heat transferred is infinitesimal and during adiabats, the temperature changes by infinitesimal amount). Let the temperature $T$ be measured by ideal gas thermometer (obeying $PV=nRT$).

Since the infinitesimal cycles are Carnot cycles, they obey $$\frac{dQ_1}{T_1} + \frac{dQ_2}{T_2} = 0$$ as Carnot showed.

Let any isotherm of the infinitesimal Carnot cycle be called section.

When the left-hand side of the last equation is summed for all infinitesimal cycles, the contributions from all pairs of sections which coincide in the space of states cancel each other, because they have opposite sign of heat $dQ$ and the same $T$. Only the boundary sections remain and contribute to the sum. We thus obtain quantity which is a sum of $dQ/T$ over all states $s$ in the great cyclic process $\gamma$:

$$\oint_\gamma\frac{dQ/ds}{T(s)}ds = 0.$$ Since this equation is valid for all closed paths $\gamma$, the expression

$$\int_i^f \frac{dQ}{T}$$ is a function of the states $i$,$f$ only. We can thus define entropy of the state $f$ as

$$S(f) = S(i) + \int_i^f\frac{dQ}{T}.$$