4 Classical Thermodynamics

From now on, we will be more gentle. We will add or subtract energy to the system very slowly, so that at every stage of the process the system is effectively in equilibrium and can be described by the thermodynamic variables $p$ and $V$. Such a process is called quasi-static.

For quasi-static processes, it is useful to write (4.156) in infinitesimal form. Unfortunately, this leads to a minor notational headache. The problem is that we want to retain the distinction between $E(p,V)$, which is a function of state, and $Q$ and $W$, which are not. This means that an infinitesimal change in the energy is a total derivative,

$dE={\displaystyle \frac{\partial E}{\partial p}}dp+{\displaystyle \frac{\partial E}{\partial V}}dV$

while an infinitesimal amount of work or heat has no such interpretation: it is merely something small. To emphasise this, it is common to make up some new notation⁸⁸ 8 In a more sophisticated language, $dE$, $-dW$ and $-dQ$ are all one-forms on the state space of the system. $dE$ is exact; $-dW$ and $-dQ$ are not.. A small amount of heat is written $-dQ$ and a small amount of work is written $-dW$. The first law of thermodynamics in infinitesimal form is then

$dE=-dQ+-dW$

(4.157)

Although we introduced the first law as applying to all types of work, from now on the discussion is simplest if we just restrict to a single method to applying work to a system: squeezing. We already saw in Section 1 that the infinitesimal work done on a system is

$\mathrm{\hspace{0.17em}}-dW=-pdV$

which is the same thing as “force $\times$ distance”. Notice the sign convention. When $-dW>0$, we are doing work on the system by squeezing it so that . However, when the system expands, $dV>0$ so and the system is performing work.

Suppose now that we vary the state of a system through two different quasi-static paths as shown in the figure. The change in energy is independent of the path taken: it is $\int 𝑑E=E(p_2,V_2)-E(p_1,V_1)$. In contrast, the work done $\int -dW=-\int p𝑑V$ depends on the path taken. This simple observation will prove important for our next discussion.

A reversible process must lie in equilibrium at each point along the path. This is the quasi-static condition. But now there is the further requirement that there is no friction involved.

Processes which move in a cycle like this, returning to their original starting point, are interesting. Run the right way, they convert heat into work. But that’s very very useful. The work can be thought of as a piston which can be used to power a steam train. Or a playstation. Or an LHC.

Second Law à la Kelvin: No process is possible whose sole effect is to extract heat from a hot reservoir and convert this entirely into work

Second Law à la Clausius: No process is possible whose sole effect is the transfer of heat from a colder to hotter body

It’s worth elaborating on the meaning of these statements. Firstly, we all have objects in our kitchens which transfer heat from a cold environment to a hot environment: this is the purpose of a fridge. Heat is extracted from inside the fridge where it’s cold and deposited outside where it’s warm. Why doesn’t this violate Clausius’ statement? The reason lies in the words “sole effect”. The fridge has an extra effect which is to make your electricity meter run up. In thermodynamic language, the fridge operates because we’re supplying it with “work”. To get the meaning of Clausius’ statement, think instead of a hot object placed in contact with a cold object. Energy always flows from hot to cold; never the other way round.

The statements by Kelvin and Clausius are equivalent. Suppose, for example, that we build a machine that violates Kelvin’s statement by extracting heat from a hot reservoir and converting it entirely into work. We can then use this work to power a fridge, extracting heat from a cold source and depositing it back into a hot source. The combination of the two machines then violates Clausius’s statement. It is not difficult to construct a similar argument to show that “not Clausius” $\Rightarrow$ “not Kelvin”.

Our goal in this Section is to show how these statements of the second law allow us to define a quantity called “entropy”.

The key to understanding this is to appreciate that a reversible cycle does more than just extract heat from a hot reservoir. It also, by necessity, deposits some heat elsewhere. The energy available for work is the difference between the heat extracted and the heat lost. To illustrate this, it’s very useful to consider a particular kind of reversible cycle called a Carnot engine. This is series of reversible processes, running in a cycle, operating between two temperatures $T_H$ and $T_C$. It takes place in four stages shown in cartoon in Figures 27 and 28.

• •

  Isothermal expansion $AB$ at a constant hot temperature $T_H$. The gas pushes against the side of its container and is allowed to slowly expand. In doing so, it can be used to power your favourite electrical appliance. To keep the temperature constant, the system will need to absorb an amount of heat $Q_H$ from the surroundings

• •

  Adiabatic expansion $BC$. The system is now isolated, so no heat is absorbed. But the gas is allowed to continue to expand. As it does so, both the pressure and temperature will decrease.

• •

  Isothermal contraction $CD$ at constant temperature $T_C$. We now start to restore the system to its original state. We do work on the system by compressing the gas. If we were to squeeze an isolated system, the temperature would rise. But we keep the system at fixed temperature, so it dumps heat $Q_C$ into its surroundings.

• •

  Adiabatic contraction $DA$. We isolate the gas from its surroundings and continue to squeeze. Now the pressure and temperature both increase. We finally reach our starting point when the gas is again at temperature $T_H$.

At the end of the four steps, the system has returned to its original state. The net heat absorbed is $Q_H-Q_C$ which must be equal to the work performed by the system $W$. We define the efficiency $\eta$ of an engine to be the ratio of the work done to the heat absorbed from the hot reservoir,

$\eta ={\displaystyle \frac{W}{Q_H}}={\displaystyle \frac{Q_H-Q_C}{Q_H}}=1-{\displaystyle \frac{Q_C}{Q_H}}$

Ideally, we would like to take all the heat $Q_H$ and convert it to work. Such an engine would have efficiency $\eta =1$ but would be in violation of Kelvin’s statement of the second law. We can see the problem in the Carnot cycle: we have to deposit some amount of heat $Q_C$ back to the cold reservoir as we return to the original state. And the following result says that the Carnot cycle is the best we can do:

Carnot’s Theorem: Carnot is the best. Or, more precisely: Of all engines operating between two heat reservoirs, a reversible engine is the most efficient. As a simple corollary, all reversible engines have the same efficiency which depends only on the temperatures of the reservoirs $\eta (T_H,T_C)$.

The work $W$ performed by Ivor now goes into driving Carnot. The net effect of the two engines is to extract $Q_H^{\prime }-Q_H$ from the hot reservoir and, by conservation of energy, to deposit the same amount $Q_C^{\prime }-Q_C=Q_H^{\prime }-Q_H$ into the cold. But Clausius’s statement tells us that we must have $Q_H^{\prime }\ge Q_H$; if this were not true, energy would be moved from the colder to hotter body. Performing a little bit of algebra then gives

$Q_C^{\prime }-Q_H^{\prime }=Q_C-Q_H\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }{\eta }_{\mathrm{Ivor}}=1-{\displaystyle \frac{Q_C^{\prime }}{Q_H^{\prime }}}={\displaystyle \frac{Q_H-Q_C}{Q_H^{\prime }}}\le {\displaystyle \frac{Q_H-Q_C}{Q_H}}={\eta }_{\mathrm{Carnot}}$

The upshot of this argument is the result that we wanted, namely

${\eta }_{\mathrm{Carnot}}\ge {\eta }_{\mathrm{Ivor}}$

The corollary is now simple to prove. Suppose that Ivor was reversible after all. Then we could use the same argument above to prove that ${\eta }_{\mathrm{Ivor}}\ge {\eta }_{\mathrm{Carnot}}$, so it must be true that ${\eta }_{\mathrm{Ivor}}={\eta }_{\mathrm{Carnot}}$ if Ivor is reversible. This means that for all reversible engines operating between $T_H$ and $T_C$ have the same efficiency. Or, said another way, the ratio $Q_H/Q_C$ is the same for all reversible engines. Moreover, this efficiency must be a function only of the temperatures, ${\eta }_{\mathrm{Carnot}}=\eta (T_H,T_C)$, simply because they are the only variables in the game. $\mathrm{\square }$.

Since the efficiency of a Carnot engine depends only on the temperatures $T_H$ and $T_C$, we can use this to define a temperature scale that is independent of any specific material. (Although, as we shall see, the resulting temperature scale turns out to be equivalent to the ideal gas temperature). Let’s now briefly explain how we can define a temperature through the Carnot cycle.

The key idea is to consider two Carnot engines. The first operates between two temperature reservoirs $T_1>T_2$; the second engine operates between two reservoirs $T_2>T_3$. If the first engine extracts heat $Q_1$ then it must dump heat $Q_2$ given by

$Q_2=Q_1\left(1-\eta (T_1,T_2)\right)$

where the arguments above tell us that $\eta ={\eta }_{\mathrm{Carnot}}$ is a function only of $T_1$ and $T_2$. If the second engine now takes this same heat $Q_2$, it must dump heat $Q_3$ into the reservoir at temperature $T_3$, given by

$Q_3=Q_2\left(1-\eta (T_2,T_3)\right)=Q_1\left(1-\eta (T_1,T_2)\right)\left(1-\eta (T_2,T_3)\right)$

But we can also consider both engines working together as a Carnot engine in its own right, operating between reservoirs $T_1$ and $T_3$. Such an engine extracts heat $Q_1$, dumps heat $Q_3$ and has efficiency $\eta (T_1,T_3)$, so that

$Q_3=Q_1\left(1-\eta (T_1,T_3)\right)$

Combining these two results tells us that the efficiency must be a function which obeys the equation

$1-\eta (T_1,T_3)=\left(1-\eta (T_1,T_2)\right)\left(1-\eta (T_2,T_3)\right)$

The fact that $T_2$ has to cancel out on the right-hand side is enough to tell us that

$1-\eta (T_1,T_2)={\displaystyle \frac{f(T_2)}{f(T_1)}}$

for some function $f(T)$. At this point, we can use the ambiguity in the definition of temperature to simply pick a nice function, namely $f(T)=T$. Hence, we define the thermodynamic temperature to be such that the efficiency is given by

$\eta =1-{\displaystyle \frac{T_2}{T_1}}$

(4.158)

We deal first with the isothermal changes of the ideal gas. We know that the energy in the gas depends only on the temperature⁹⁹ 9 A confession: strictly speaking, I’m using some illegal information in the above argument. The result $E=\frac{3}{2}N{k}_{B}T$ came from statistical mechanics and if we’re really pretending to be Victorian scientists we should discuss the efficiency of the Carnot cycle without this knowledge. Of course, we could just measure the heat capacity $C_V={\partial E/\partial T|}_V$ to determine $E(T)$ experimentally and proceed. Alternatively, and more mathematically, we could note that it’s not necessary to use this exact form of the energy to carry through the argument: we need only use the fact that the energy is a function of temperature only: $E=E(T)$. The isothermal parts of the Carnot cycle are trivially the same and we reproduce (4.160) and (4.161). The adiabatic parts cannot be solved exactly without knowledge of $E(T)$ but you can still show that $V_A/V_B=V_D/V_C$ which is all we need to derive the efficiency (4.162).,

$E={\displaystyle \frac{3}{2}}N{k}_{B}T$

(4.159)

So $dT=0$ means that $dE=0$. The first law then tells us that $-dQ=--dW$. For the motion along the line $AB$ in the Carnot cycle, we have

$Q_H={\displaystyle {\int }_A^B}-dQ=-{\displaystyle {\int }_A^B}-dW={\displaystyle {\int }_A^B}p𝑑V={\displaystyle {\int }_A^B}{\displaystyle \frac{N{k}_B{T}_H}{V}}𝑑V=N{k}_B{T}_H\log \left({\displaystyle \frac{V_B}{V_A}}\right)$

(4.160)

Similarly, the heat given up along the line $CD$ in the Carnot cycle is

$Q_C=-N{k}_B{T}_C\log \left({\displaystyle \frac{V_D}{V_C}}\right)$

(4.161)

Next we turn to the adiabatic change in the Carnot cycle. Since the system is isolated, $-dQ=0$ and all work goes into the energy, $dE=-pdV$. Meanwhile, from (4.159), we can write the change of energy as $dE=C_{V}dT$ where $C_V=\frac{3}{2}N{k}_B$, so

$C_{V}dT=-{\displaystyle \frac{N{k}_{B}T}{V}}dV\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em}}{\displaystyle \frac{dT}{T}}=-\left({\displaystyle \frac{N{k}_B}{C_V}}\right){\displaystyle \frac{dV}{V}}$

After integrating, we have

$T{V}^{2/3}=\mathrm{constant}$

Applied to the line $BC$ and $DA$ in the Carnot cycle, this gives

$T_H{V}_B^{2/3}=T_C{V}_C^{2/3}\mathit{\hspace{1em}\hspace{1em} },T_C{V}_D^{2/3}=T_H{V}_A^{2/3}$

which tells us that $V_A/V_B=V_D/V_C$. But this means that the factors of $\log (V/V)$ cancel when we take the ratio of heats. The efficiency of a Carnot engine for an ideal gas — and hence for any system — is given by

${\eta }_{\mathrm{carnot}}=1-{\displaystyle \frac{Q_C}{Q_H}}=1-{\displaystyle \frac{T_C}{T_H}}$

(4.162)

We see that the efficiency using the ideal gas temperature coincides with our thermodynamic temperature (4.158) as advertised.

We have made no mention of microstates in defining the entropy. Yet it is clearly the same quantity that we met in Section 1. From (4.163), we can write $dS=-dQ/T$, so that the first law of thermodynamics (4.157) is written in the form of (1.16)

$dE=TdS-pdV$

(4.164)

The above result holds for any engine operating between two temperatures. But by the same method of cutting corners off a Carnot cycle that we used above, we can easily generalise the statement to any path, reversible or irreversible. Putting the minus signs back in so that heat dumped has the opposite sign to heat absorbed, we arrive at a result is known as the Clausius inequality,

${\displaystyle \oint \frac{-dQ}{T}}\le 0$

The second law, as expressed in (4.166), is responsible for the observed arrow of time in the macroscopic world. Isolated systems can only evolve to states of equal or higher entropy. This coincides with the statement of the second law that we saw back in Section 1.2.1 using Boltzmann’s definition of the entropy.

In fact, for the simplest systems such as the ideal gas which require only two variables $p$ and $V$ to specify the state, we do not need the second law to infer to the existence of an adiabatic surface. In that case, the adiabatic surface is really an adiabatic line in the two-dimensional space of states. The existence of this line follows immediately from the first law. To see this, we write the change of energy for an adiabatic process using (4.157) with $-dQ=0$,

$dE+pdV=0$

(4.167)

Let’s view the energy itself as a function of $p$ and $V$ so that we can write

$dE={\displaystyle \frac{\partial E}{\partial p}}dP+{\displaystyle \frac{\partial E}{\partial V}}dV$

Then the condition for an adiabatic process (4.167) becomes

${\displaystyle \frac{\partial E}{\partial p}}dp+\left({\displaystyle \frac{\partial E}{\partial V}}+p\right)dV=0$

Which tells us the slope of the adiabatic line is given by

${\displaystyle \frac{dp}{dV}}=-\left({\displaystyle \frac{\partial E}{\partial V}}+p\right){\left({\displaystyle \frac{\partial E}{\partial p}}\right)}^{-1}$

(4.168)

The upshot of this calculation is that if we sit in a state specified by $(p,V)$ and transfer work but no heat to the system then we necessarily move along a line in the space of states determined by (4.168). If we want to move off this line, then we have to add heat to the system.

However, the analysis above does not carry over to more complicated systems where more than two variables are needed to specify the state. Suppose that the state of the system is specified by three variables. The first law of thermodynamics is now gains an extra term, reflecting the fact that there are more ways to add energy to the system,

$dE=-dQ-pdV-ydX$

We’ve already seen examples of this with $y=-\mu$, the chemical potential, and $X=N$, the particle number. Another very common example is $y=-M$, the magnetization, and $X=H$, the applied magnetic field. For our purposes, it won’t matter what variables $y$ and $X$ are: just that they exist. We need to choose three variables to specify the state. Any three will do, but we will choose $p$, $V$ and $X$ and view the energy as a function of these: $E=E(p,V,X)$. An adiabatic process now requires

$dE+pdV+ydX=0\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }{\displaystyle \frac{\partial E}{\partial p}}dp+\left({\displaystyle \frac{\partial E}{\partial V}}+p\right)dV+\left({\displaystyle \frac{\partial E}{\partial X}}+y\right)dX=0$

(4.169)

But this equation does not necessarily specify a surface in ${𝐑}^3$. To see that this is not sufficient, we can look at some simple examples. Consider ${𝐑}^3$, parameterised by $z_1$, $z_2$ and $z_3$. If we move in an infinitesimal direction satisfying

$z_{1}d{z}_1+z_{2}d{z}_2+z_{3}d{z}_3=0$

then it is simple to see that we can integrate this equation to learn that we are moving on the surface of a sphere,

$z_1^2+z_2^2+z_3^2=\mathrm{constant}$

In contrast, if we move in an infinitesimal direction satisfying the condition

$z_{2}d{z}_1+d{z}_2+d{z}_3=0$

(4.170)

Then there is no associated surface on which we’re moving. Indeed, you can convince yourself that if you move in a such a way as to always obey (4.170) then you can reach any point in ${𝐑}^3$ from any other point.

In general, an infinitesimal motion in the direction

$Z_{1}d{z}_1+Z_{2}d{z}_2+Z_{3}d{z}_3=0$

has the interpretation of motion on a surface only if the functions $Z_i$ obey the condition

$Z_1\left({\displaystyle \frac{\partial Z_2}{\partial z_3}}-{\displaystyle \frac{\partial Z_3}{\partial z_2}}\right)+Z_2\left({\displaystyle \frac{\partial Z_3}{\partial z_1}}-{\displaystyle \frac{\partial Z_1}{\partial z_3}}\right)+Z_3\left({\displaystyle \frac{\partial Z_1}{\partial z_2}}-{\displaystyle \frac{\partial Z_2}{\partial z_1}}\right)=0$

(4.171)

So for systems with three or more variables, the existence of an adiabatic surface is not guaranteed by the first law alone. We need the second law. This ensures the existence of a function of state $S$ such that adiabatic processes move along surfaces of constant $S$. In other words, the second law tells us that (4.169) satisfies (4.171).

In fact, there is a more direct way to infer the existence of adiabatic surfaces which uses the second law but doesn’t need the whole rigmarole of Carnot cycles. We will again work with a system that is specified by three variables, although the argument will hold for any number. But we choose our three variables to be $V$, $X$ and the internal energy $E$. We start in state $A$ shown in the figure. We will show that Kelvin’s statement of the second law implies that it is not possible to reach both states $B$ and $C$ through reversible adiabatic processes. The key feature of these states is that they have the same values of $V$ and $X$ and differ only in their energy $E$.

The upshot of this argument is that the space of states can be foliated by adiabatic surfaces such that each vertical line at constant $V$ and $X$ intersects the surface only once. We can describe these surfaces by some function $S(E,V,X)=\mathrm{constant}$. This function is the entropy.

The above argument shows that Kelvin’s statement of the second law implies the existence of adiabatic surfaces. One may wonder if we can run the argument the other way around and use the existence of adiabatic surfaces as the basis of the second law, dispensing with the Kelvin and Clausius postulates all together. In fact, we can almost do this. From the discussion above it should already be clear that the existence of adiabatic surfaces implies that the addition of heat is proportional to the change in entropy $-dQ\sim dS$. However, it remains to show that the integrating factor, relating the two, is temperature so $-dQ=TdS$. This can be done by returning to the zeroth law of thermodynamics. A fairly simple description of the argument can be found at the end of Chapter 4 of Pippard’s book. This motivates a mathematically concise statement of the second law due to Carathéodory.

Second Law à la Carathéodory: Adiabatic surfaces exist. Or, more poetically: if you want to be able to return, there are places you cannot go through work alone. Sometimes you need a little heat.

What this statement is lacking is perhaps the most important aspect of the second law: an arrow of time. But this is easily remedied by providing one additional piece of information telling us which side of a surface can be reached by irreversible processes. To one side of the surface lies the future, to the other the past.

Although ideas of “heat” date back to pre-history, a good modern starting point is the 1787 caloric theory of Lavoisier. This postulates that heat is a conserved fluid which has a tendency to repel itself, thereby flowing from hot bodies to cold bodies. It was, for its time, an excellent theory, explaining many of the observed features of heat. Of course, it was also wrong.

Lavoisier’s theory was still prominent 30 years later when the French engineer Sadi Carnot undertook the analysis of steam engines that we saw above. Carnot understood all of his processes in terms of caloric. He was inspired by mechanics of waterwheels and saw the flow of caloric from hot to cold bodies as analogous to the fall of water from high to low. This work was subsequently extended and formalised in a more mathematical framework by another French physicist, Émile Clapeyron. By the 1840s, the properties of heat were viewed by nearly everyone through the eyes of Carnot-Clapeyron caloric theory.

Yet the first cracks in caloric theory has already appeared before the turn of 19th century due to the work of Benjamin Thompson. Born in the English colony of Massachusetts, Thompson’s CV includes turns as mercenary, scientist and humanitarian. He is the inventor of thermal underwear and pioneer of the soup kitchen for the poor. By the late 1700s, Thompson was living in Munich under the glorious name “Count Rumford of the Holy Roman Empire” where he was charged with overseeing artillery for the Prussian Army. But his mind was on loftier matters. When boring cannons, Rumford was taken aback by the amount of heat produced by friction. According to Lavoisier’s theory, this heat should be thought of as caloric fluid squeezed from the brass cannon. Yet is seemed inexhaustible: when a cannon was bored a second time, there was no loss in its ability to produce heat. Thompson/Rumford suggested that the cause of heat could not be a conserved caloric. Instead he attributed heat correctly, but rather cryptically, to “motion”.

Having put a big dent in Lavoisier’s theory, Rumford rubbed salt in the wound by marrying his widow. Although, in fairness, Lavoisier was beyond caring by this point. Rumford was later knighted by Britain, reverting to Sir Benjamin Thompson, where he founded the Royal Institution.

The journey from Thompson’s observation to an understanding of the first law of thermodynamics was a long one. Two people in particular take the credit.

In Manchester, England, James Joule undertook a series of extraordinarily precise experiments. He showed how different kinds of work — whether mechanical or electrical – could be used to heat water. Importantly, the amount by which the temperature was raised depended only on the amount of work, not the manner in which it was applied. His 1843 paper “The Mechanical Equivalent of Heat” provided compelling quantitative evidence that work could be readily converted into heat.

But Joule was apparently not the first. A year earlier, in 1842, the German physician Julius von Mayer came to the same conclusion through a very different avenue of investigation: blood letting. Working on a ship in the Dutch East Indies, Mayer noticed that the blood in sailors veins was redder in Germany. He surmised that this was because the body needed to burn less fuel to keep warm. Not only did he essentially figure out how the process of oxidation is responsible for supplying the body’s energy but, remarkably, he was able to push this to an understanding of how work and heat are related. Despite limited physics training, he used his intuition, together with known experimental values of the heat capacities $C_p$ and $C_V$ of gases, to determine essentially the same result as Joule had found through more direct means.

The results of Thompson, Mayer and Joule were synthesised in an 1847 paper by Hermann von Helmholtz, who is generally credited as the first to give a precise formulation of the first law of thermodynamics. (Although a guy from Swansea called William Grove has a fairly good, albeit somewhat muddled, claim from a few years earlier). It’s worth stressing the historical importance of the first law: this was the first time that the conservation of energy was elevated to a key idea in physics. Although it had been known for centuries that quantities such as “$\frac{1}{2}m{v}^2+V$” were conserved in certain mechanical problems, this was often viewed as a mathematical curiosity rather than a deep principle of Nature. The reason, of course, is that friction is important in most processes and energy does not appear to be conserved. The realisation that there is a close connection between energy, work and heat changed this. However, it would still take more than half a century before Emmy Noether explained the true reason behind the conservation of energy.

With Helmholtz, the first law was essentially nailed. The second remained. This took another two decades, with the pieces put together by a number of people, notably William Thomson and Rudolph Clausius.

William Thomson was born in Belfast but moved to Glasgow at the age of 10. He came to Cambridge to study, but soon returned to Glasgow and stayed there for the rest of his life. After his work as a scientist, he gained fame as an engineer, heavily involved in laying the first trans-atlantic cables. For this he was made Lord Kelvin, the name chosen for the River Kelvin which flows nearby Glasgow University. He was the first to understand the importance of absolute zero and to define the thermodynamic temperature scale which now bears his favourite river’s name. We presented Kelvin’s statement of the second law of thermodynamics earlier in this Section.

In Germany, Rudolph Clausius was developing the same ideas as Kelvin. But he managed to go further and, in 1865, presented the subtle thermodynamic argument for the existence of entropy that we saw in Section 4.3.3. Modestly, Clausius introduced the unit “Clausius” (symbol Cl) for entropy. It didn’t catch on.

In our discussion we have ignored the particle number $N$. Yet both $F$ and $G$ implicitly depend on $N$ (as you may check by re-examining the many examples of $F$ that we computed earlier in the course). If we also consider changes $dN$ then each variations gets the additional term $\mu dN$, so

$dF=-SdT-pdV+\mu dN\mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }dG=-SdT+Vdp+\mu dN$

(4.174)

While $F$ can have an arbitrarily complicated dependence on $N$, the Gibbs free energy $G$ has a very simple dependence. To see this, we simply need to look at the extensive properties of the different variables and make the same kind of argument that we’ve already seen in Section 1.4.1. From its definition (4.173), we see that the Gibbs free energy $G$ is extensive. It a function of $p$, $T$ and $N$, of which only $N$ is extensive. Therefore,

$G(p,T,N)=\mu (p,T)N$

(4.175)

where the fact that the proportionality coefficient is $\mu$ follows from variation (4.174) which tells us that $\partial G/\partial N=\mu$.

The Gibbs free energy is frequently used by chemists, for whom reactions usually take place at constant pressure rather than constant volume. (When a chemist talks about “free energy”, they usually mean $G$. For a physicist, “free energy” usually means $F$). We’ll make use of the result (4.175) in the next section when we discuss first order phase transitions.

The other Maxwell relations are derived by playing the same game with $F$, $G$ and $H$. From the properties (4.172), we see that taking mixed partial derivatives of the free energy gives us,

${{\displaystyle \frac{\partial S}{\partial V}}|}_T={{\displaystyle \frac{\partial p}{\partial T}}|}_V$

(4.177)

The Gibbs free energy gives us

${{\displaystyle \frac{\partial S}{\partial p}}|}_T=-{{\displaystyle \frac{\partial V}{\partial T}}|}_p$

(4.178)

While the enthalpy gives

${{\displaystyle \frac{\partial T}{\partial p}}|}_S={{\displaystyle \frac{\partial V}{\partial S}}|}_p$

(4.179)

The four Maxwell relations (4.176), (4.177), (4.178) and (4.179) are remarkable in that they are mathematical identities that hold for any system. They are particularly useful because they relate quantities which are directly measurable with those which are less easy to determine experimentally, such as entropy.

It is not too difficult to remember the Maxwell relations. Cross-multiplication always yields terms in pairs: $TS$ and $pV$, which follows essentially on dimensional grounds. The four relations are simply the four ways to construct such equations. The only tricky part is to figure out the minus signs.