# Free Energies: What Does a Physical Chemist Mean when He/She Talks of Needing Work to Throw Excess Entropy Out of a Reaction?

The statement is often made, typically by a physical chemist, that the Gibbs (or other free) energy is the total energy let slip by a chemical reaction less the amount of work we need to do to “throw the excess entropy of the reactants relative to the products out of the system”. Now I must say I feel nervous around physical chemists talking about the second law. Not that I’m not confident of myself – without being cocky, a frame of mind which is always ends in tears when talking about something as simple yet fiendishly subtle as entropy – but I’m sure I’ve thought every bit as much about these ideas over the years as many of these chemists have, albeit in different words and different concepts. It’s simply that some of them – I’ve met quite a few – don’t seem to like physicists or mathematicians probing the grounding of thermodynamics in chemistry. Some of them seem especially offended by the idea that some upstart from Bell Labs whose inventions included a flame-throwing trumpet and a philosopher, physicist, theoretical statistician and group theorist who looked eerily (and unflatteringly) like Biff Tannen from “Back to The Future” could teach us all a thing or two about the second law. It’s an attitude that’s baffling to me: why wouldn’t one think that people like Claude Shannon or Edwin Jaynes would have at least something worthwhile to say on a topic they have thought deeply about?

Now, however batty such attitudes drive me, I can’t deny one thing. Physical chemists live and breathe reactions. Even if they insist on your subscribing to seemingly maddeningly handwavy ideas such as:

The free energy is the total energy let slip by a chemical reaction less the amount of work we need to do to “throw the excess entropy of the reactants relative to the products out of the system

before they’ll even speak to you, then, by dint of their sheer experience alone, you have to admit at some level these guys have got to know what they are talking about. Especially at a practical level. Take any falsifiable concept of entropy to a physical chemist and, even if they don’t understand a word of what you’re saying, they’ll come up with practical examples to put the Popper blowtorch test to your ideas and conceptions. If you can explain your conception of the second law to a physical chemist and it survives this test, you know you’re onto something, no matter how superior you may think your own theoretical prowess. So, as maddening as I have found the above statement about what the Free Energy means, it’s one I’ve heard many times from physical chemists, and so I have been determined in the last year to really probe what this statement means. And it actually makes a great deal of sense. Looking at it in the right way, I now agree, that this seemingly handwavy statement is actually a pretty sound concept of the second law of thermodynamics as applied to a chemical reaction.  Here’s how.

## Landauer’s Principle

A great deal of intuition for the statement can be built by thinking about the Landauer principle formulation of the second law of thermodynamics. In this post here, I explore the principle, which comes from asking the question how can the Szilard Engine / Maxwell Daemon be made to comply with the second law? Maxwell himself never meant the Daemon to comply with the second law; indeed he dreamt the Daemon up to show the law’s statistical nature; that it could, in principle, be overcome by “intelligence” or “choice”. I think it’s fairly safe to say that he never imagined we humans would actually build and test the Maxwell Daemon in the laboratory as we have done. Szilard thought the daemon would comply with the second law, but wrongly believed that it would do so because the measurements that the daemon makes its decisions by would need work. It turns out that, in principle, however, we can get a measurement for free, or at least as near to free as we want, and as accurately as we want by slowing it down and doing it with ever heavier molecules in the Szilard engine (it’s notable that the real Maxwell Daemon uses nanoscale polystyrene beads being buffetted in Brownian motion – huge “molecules” forming a simulated “gas”).

No, the reason the Daemon must comply with the second law is this: information is physical – in the physical World information must be written in some kind of real “ink”. We can’t simply think of it abstractly and disembodied from that ink as we often do in signal and information theory – even though in those fields it is often perfectly sound to do so. That ink is the physical miscrostates of a system. Moreover, the laws of physics are reversible at microscopic scales: the state of the World at any time is mapped one-to-one to the state of the World at any other. So, as the Daemon works, it gleans information about the state of the molecule in the Szilard engine – which half of the Szilard engine the molecule was in. When it re-intialises its memory to begin a new cycle, it can’t truly forget what that state was. That “forgotten” state somehow has to end up encoded in the physical states of the “stuff” that makes up the Daemon’s hardware, otherwise the mapping between states of the World at one time and any other times could not be one-to-one. To apply the second law properly to a “practical closed” system (i.e. one which swaps no particles with the outside World, but which can swap heat with the outside World – the so-called Canonical Ensemble ), you must return your system precisely to its same macrostate before tallying up entropy changes, as I shall show in even more detail when I get to talking about chemical reactions. The Daemon with this “forgotten” information encoded in its hardware is ever-so-subtly in a different macrostate. The state information encoded in the Daemon’s increasingly thermalised (by the encoding of molecule states “forgotten” by memory re-initialisation) cannot be infinite: see my post here. So, sooner or later, if the Daemon is to keep working, we must “throw this excess entropy out of the system”. That is, we must force somehow the “forgotten” molecule states to be encoded in the states of the Daemon’s environment, not in its hardware. They have to be encoded somewhere, by the fundamental reversibility of physics. Only then can we truly say that the Daemon comes back to its beginning macrostate and can we then validly tally up entropies and apply the second law. So, in “reinitialising” the stuff of the Daemon, we would find ourselves needing to do work on it to shift this microstate encoding into the environment.

The need to apply the second law to a true cycle in macro-state space before applying the second law is the reason why reactant and product enthalpies, entropies and all the rest of it must always be specified at some standard conditions. We must compare all these quantites at the same conditions. Why can’t we turn all the heat let slip by an exothermic reaction in to work? Well, actually we can! If we’ve got a dirty great big cool reservoir at absolute zero temperature, then no problem, as we shall see! But in doing the conversion, we would change the macrostate of the reservoir. Our system would not be in its beginning macrostate. Practical reservoirs cannot absorb infinite heat, so we would degrade or “use up” a bit of our reservoir each time we ran the reaction. Sooner or later the reservoir would heat up to the temperature of the true “environment” and our shonky accounting would be Enronned. Such a process can’t go on indefinitely, which is the reason why Clausius and Carnot formulated the second law as the impossibility of building a perpetual motion machine of the second kind.

## The Free Energy of a Chemical Reaction

Now onwards towards these ideas applied to a chemical reaction. Already, applying Landauer’s principle, we have gotten some intuition for the chemist’s conception of the free energy: if the reaction products have lower entropy at the standard conditions than the products, what do we mean? From the Landauer principle standpoint we mean that the products can, at the standard conditions in question, encode fewer bits of information in their internal states than the reactants. Therefore, if the laws of physics are to stay truly reversible and state transition mappings truly one-to-one, the particular microstate encoded in the reactants has to show up somewhere. It can’t all be encoded in the “ink” of the products’ internal states, by definition of “lower entropy”. So somehow we have to force it to be encoded in the state of the environment. This is where we shall need to use some of the heat of reaction to do work to shift this “entropy”. But we can see these ideas in even more concrete words by understanding them in Clausius’s and Carnot’s conception of the second law of thermodynamics, to wit, that heat cannot flow spontaneously from a cold to a hotter body and Carnot’s theorem. For, following straight from this statement, is the fact that the minimum amount of work needed to raise heat from a colder to a hotter body is expended when one does the raising with a reversible heat pump (i.e. a reversible heat engine simply running backwards).

Before leaping into this discussion, it is well to take heed that one never talks about a truly isolated system when one speaks of free energies. There is always an “Outside World”, “Environment” or “Universe” implicitly there surrounding the system being talked about and a good deal of heat can and wontedly does flow across the boundary between the system and the outside world. Sometimes (although not in the discussion of free energies) even particles  can cross the boundary (the so called Grand Canonical Ensemble). In many discussions the outside world becomes highly implicit so that one must be careful to heed carefully the exact conditions any discussion happens under; my own experience that people can get pretty slack in specifying those. This non-isolated framework is in stark contrast almost the rest of physics, where one seeks to isolate the system of interest from the outside world to simplify discussions, so it takes a bit of getting wonted with if you’re new to thermodynamics. But the lack of isolated systems in most of thermodynamic discussions has a very compelling statistical grounding: isolation of a system forces statistical correlation between its particles because, for example, amongst other things, their energies must all sum to a constant, and so they cannot be statistically independent. If energy can flow freely in and out of surrounding reservoir, and the reservoir is very big, then this correlation is removed, thus affording the discussion a great simplification. Mostly, as in the discussion of free energy use, the “Outside World” is at thermochemical equilibrium, at a uniform temperature and is so big that no deal of heat flowing either into or out of it can change its temperature or equilibrium measurably. Sometimes, though, it comprises several big heat reservoirs, each at a different tempature; one then speaks of a “piecewise equilibrium outside World”.

As we have seen, the thermodynamic macrostate of the reactants and products must be carefully specified along with any discussion of free energy. It makes sense to behest that the reactants before the reaction on the products afterwards should be in thermodynamic equilibrium with the outside world. That means that free energies are always specified for chemical systems that begin and wind up at the same uniform temperature, to wit, that of the outside World. There are two common free energies in use: the Helmholtz free energy $A$ (standing for Arbeit, in Helmholtz’s mother tongue) and the Gibbs free energy $G$. Both talk about reactions that happen at a constant temperature and beginning and ending in equilibrium at that temperature with the outside world. The former  imagines things happening at a constant volume: if you will, inside a box that is infinitely stiff and strong so that it neither shrinks nor swells with pressure changes and through whose walls, naturally, heat can freely flow. The latter  imagines things happening at a constant pressure: if you will, inside a heat-permeable sack to keep everything together but which can freely (i.e. without work) shrink or swell according to whether the reaction implodes or explodes. The difference between them is, of course,  $P\,\Delta V$ where $P$ is the constant pressure assume for the Gibbs free energy and  the volume shift wrought by the reaction. All free energies are meant to quantify the maximum work gettable from a reaction, and we shall see soon why this is almost always less than the simple energy given from a reaction.

To keep things reasonably concrete, we talk about the simple reaction of the burning of hydrogen:

$$H_2+\frac{1}{2} O_2\to H_2O\tag{1}$$

This is a pretty energetic one (highly exothermic), and also the reaction of choice if you want to throw three astronauts, one hundred and twenty tons of kit and a fourth of the most advanced-economy-in-the-world’s-1960s GDP at the Moon. Since we are talking about the energy we can get out of this reaction, we line everything up in Figure 1 and think about things like an accountant does: there are now three energy “buckets” we have to think about:

1. The potential energy difference between the reactants and the products ;
2. The heat held in the reactants, say a mole of hydrogen and half a mole of oxygen;
3. The heat held in the products, i.e. one mole of water.

Figure 1: The energy chart of accounts for our reaction

I have drawn the heats like fluids in literal buckets, and the potential difference I show by putting the reactants at a height $\Delta U_0$ higher than the products: we can imagine tying the bucket to a rope on a pulley and letting it fall through this height to get at the available work. Indeed, if we did everything at absolute zero temperature, there would be no vibrational, rotational, translational or otherwise distractional states that our precious work would hide inside; we assume that an advanced enough technology could get it all this work at zero Kelvin. We might imagine nanobots or powerful X-ray laser tweezers doing work to pull the hydrogen and oxygen dimers asunder, and then hanging onto the individual atoms, having the atoms do work on the nanobots as the former combined into the lower potential energy state they find themselves in as water. Indeed, we don’t have to imagine a technology such a far away off in the to-come: fuel cell technology where catalysts loan work to split the dimers, separate charges, thus setting up electrical potential differences to pay back the loan and do further work withal, do a pretty good job of this.

Actually, as an aside, the buckets would look more like those in Figure 2, where there are phase change temperatures at which the heat capacity rises to very high values to represent the high values of the latent heats of melting and boiling.

Figure 2: More accurate profiles of the heat buckets

Now we talk about the all-important heat buckets. We imagine filling such a bucket with heat (i.e. all that energy that goes off and hides in the vibrational, rotational, translational and otherwise distractional states) to bring the stuff in question from zero Kelvin up to the temperature $T_0$ we wish to define the reaction’s free energy at. It so happens that the hydrogen and oxygen in their configurations form a rather bigger bucket than does the water, so we might expect that this energy difference would be available as work. However, as Carnot’s statement of the second law asserts, not all heat is equal and all of the heat that went into the hydrogen and oxygen to bring them from absolute zero to $T_0$ was added at temperatures lower than $T_0$ and so the leftover heat i.e. the heat in the hydrogen and oxygen “bucket” that we don’t use to fill the water “bucket” cannot spontaneously flow into the outside world at our defined temperature without breaking the second law. Indeed, we have to use up some of our precious work $\Delta U_0$ in actively throwing this heat out. This is the key to understanding free energy and why it is less than the simple energy difference. Let us now imagine making the reaction reversibly follow a readily analysed path to show these thoughts in more detail. If the path is reversible, and if it begins with the reactants and ends with the products all in the defined thermodynamic conditions, then the heat transfers and work done by this path must be equal to the free energy. Our thought experiment is made up of four stages:

1. Emptying of the reactant heat bucket;
2. Running of the reaction and thus extracting the potential difference at absolute zero;
3. Warming of the products back to the specified temperature ;
4. Restoring all thought experimental apparatus back to the thermodynamic state it was in at the beginning of the thought experiment.

In the first stage of this path, we reversibly “empty” the reactant heat bucket, as schematically shown below.

Figure 3: Stage 1: Reversibly cooling the reactants to absolute zero

To do this, we imagine a gigantic array of enormous heat pads (all individually like an equilibrium “outside world) representing all temperatures between absolute zero and  $T_0$ with a very fine temperature spacing $\Delta T$ between them. On my darker days I find myself imagining an experimental kit that looks eerily like a huge pallet on wheels of mortuary shelves, sliding in and out as needed by the experimenter! We bring the reactants into contact with the first heat pad, which is at a temperature $T_1 = T_0 – \Delta T$ a teeny-tiny bit cooler than $T_0$  thus reversibly drawing some heat $\Delta Q(T_1)=C_R(T_1)\,\Delta T$  off into the heat pad, where $C_R(T)$ is the specific heat of the reactants at temperature $T$. Next, we bring the reactants into contact with the second heat pad at temperature $T_2 = T_0 – 2\,\Delta T$, thus reversibly drawing heat $\Delta Q(T_2)=C_R(T_2)\,\Delta T$ off into that heat pad. We keep cooling then shifting to the next lowest heat pad until we have visited all the heat pads and thus sucked all the heat off into our heat pads, thus storing away the total heat in the bucket:

$$Q_R(T_0) = \int_0^{T_0} \,C_R(T)\,\mathrm{d}\,T\tag{2}$$

Now we look at Stage 2 of our thought experiment path. We fetch our nanobots or whatever and, working at absolute zero, we let the hydrogen and oxygen become water and store away our work $\Delta U_0$  in a suitable “work bucket”. Take heed again: at absolute zero there are no vibrational rotational, translational or otherwise distractional states for heats to gather in. Everything is at its ground state. Imagine a weight on a pulley storing work as gravitational potential energy, or maybe a superconducting coil, storing it as the energy of genesis of a magnetic field as good examples of work buckets.

Now, having put our work aside, we consider Stage 3; i.e. we must heat the water back up to the standard temperature $T_0$, as in Figure 4 below.

Figure 4: Stage 3: Heating the products

We bring the water into contact with the lowest temperature heat pad, which is at a temperature $\Delta T$ a tiny bit warmer absolute zero thus reversibly adding some heat $\Delta q(\Delta T)=C_P(\Delta T)\,\Delta T$ to the water, where $C_P(T)$ is the specific heat of the products (water) at temperature $T$. Note that, so far, we have added a nett quantity $\Delta Q(\Delta T) = \left(C_R(\Delta T) – C_P(\Delta T)\right) \Delta T$ of heat to the heat pad. Then we shift the water to the next highest temperature heat pad and wait for equilibrium, which is reached when a quantity $\Delta q(2\,\Delta T)=C_P(2\,\Delta T)\,\Delta T$ of heat has flowed from this heat pad into the water, and so forth. As we go forward in this way, we take heat $\Delta q(\Delta T_N)=C_P(\Delta T_N)\,\Delta T$ to the water from the heat pad ta temperature $T_N$, thus leaving a nett quantity $\Delta Q(\Delta T_N) = \left(C_R(\Delta T_N) – C_P(\Delta T_N)\right) \Delta T$ of heat added to that heat pad by the time we have reached the end of stage 3 and warmed the water back to the standard temperature $T_0$.

At the end of Stage 3, we have added heat:

$$Q_P(T_0) = \int_0^{T_0} \,C_P(T)\,\mathrm{d}\,T\tag{3}$$

to the water and there is heat:

$$Q_R(T_0) -Q_P(T_0)= \int_0^{T_0} \,\left(C_R(T)-C_P(T)\right)\,\mathrm{d}\,T\tag{4}$$

leftover and put aside in the heat pads.

So far, so good. We have gotten all the work $\Delta U_0$ into our work bucket without losing any. However, our experimental apparatus is still not at its beginning state, so we haven’t reckoned the free energy at the standard conditions, rather we have simply calculated the free energy $\Delta U_0$ available in the presence of our unrealistic heat sink array. To restore the system to its beginning state and calculate what work we could get if there were no heat sink array here, we must take away the nett heat flows we added to all the heat pads and send them into the outside World at temperature $T_0$. But the outside World at $T_0$ is warmer than any of the heat pads, so of course this heat transfer can’t happen spontaneously, simply by dent of Carnot’s statement of the second law! We must therefore bring in a reversible heat pump and use some of our work to pump this heat into the outside world to restore standard conditions. This excess of specific heat (Equation 4), or rather, as we shall see below a temperature weighted specific heat integral, is the reason why we cannot get at all the work $\Delta U_0$ and still leave the reaction products at the standard temperature $T_0$.

Why do we have to throw the heat out anyway? If we haven’t got the heat sink array in real life, why not leave this heat in the reaction products and get more work out? To do that, we’d be in even more strife! We’d have to heat the reaction products up to temperatures much higher than $T_0$, that is, we are proposing to shift heat from heat pads much cooler than $T_0$ to reaction products at temperatures much hotter than  $T_0$ spontaneously! We should have broken Carnot’s statement of the second law even worse than we did before! What all this analysis is saying is that the reaction simply can’t go forward unless it can give back to itself an amount of work needed to expel this leftover heat, otherwise the second law will be broken.

So as to keep everything in balance and calculate the free energy with everything beginning and ending at temperature $T_0$, there is no other way than to undertake Stage 4 shown in Figure 5.

Figure 5: Stage 4: Packing our playthings away

One after the other, we link each of our heat pads up to a reversible heat pump and pump the leftover heat in each of them into the outside world at temperature $T_0$ with some of our precious work $\Delta U_0$. By definition of the thermodynamic temperature (see my post here), the minimum (reversible) heat expelled into the outside world when the heat pump restores the heat pad at temperature to its beginning state is:

$$\left(C_R(T_N)-C_P(T_N)\right)\,\frac{T_0}{T_N}\,\Delta T\tag{5}$$

and amount of work needed to do this, by energy conservation, is simply this quantity less the heat $\left(C_R(T_N)-C_P(T_N)\right)\,\Delta T$ we are pumping from heat pad, namely:

$$\Delta W(T_N) = \left(C_R(T_N)-C_P(T_N)\right)\,\left(1-\frac{T_0}{T_N}\right)\,\Delta T\tag{6}$$

and so altogether we must lose an amount of work:

$$W_{Lost}= \int_0^{T_0} \,C_R(T)\,\mathrm{d}\,T – \int_0^{T_0} \,C_P(T)\,\mathrm{d}\,T – T_0\,\int_0^{T_0} \frac{C_R(T_N)-C_P(T_N}{T}\mathrm{d}\, T\tag{7}$$

after we have pumped all the excess heat out from the heat pads into the outside world so that the free energy, or the maximum work gettable, is:

$$\begin{array}{lcl}\Delta A(T_0) &=& \Delta U_0 – W_{Lost} \\&=& \Delta U_0 + Q_R(T_0) – Q_P(T_0) – T_0\,\left(S_R(T_0)-S_P(T_0)\right) \\&=& \Delta U_0 + Q_R(T_0) – Q_P(T_0) + T_0\,\Delta S(T_0)\end{array}\tag{8}$$

where, again, we meet the weird function of state $S$ we met whilst talking about Carnot’s theorem:

$$\begin{array}{lcl}S_R(T_0) &=& \int_0^{T_0}\,\frac{C_R(T)}{T}\,\mathrm{d}T\\ S_P(T_0) &=& \int_0^{T_0}\,\frac{C_P(T)}{T}\,\mathrm{d}T\\ \Delta S(T_0) &=& S_P(T_0)-S_R(T_0)\end{array}\tag{9}$$

We see therefore that the reaction cannot go forward spontaneously without breaking the second law unless it can “give itself back” at least the work needed to cast out the difference between the heat in the two heat buckets of Figure 1, that is, unless:

$$-\Delta A = \Delta U_0 – W_{Lost} = \Delta U_0 + Q_R(T_0) – Q_P(T_0) + T_0\,\Delta S(T_0)\geq 0\tag{10}$$

We use a negative sign in front of $\Delta A$ because the convention is that spontaneous reactions, i.e. those ones that give nett work after all four stages of our thought experiment, are said to have negative changes in free energy; the free energy of the reactants is said to have fallen in analogy with the gravitational potential of a falling body so that something in the outside world can use that free energy as useful work. Suppose now that $-\Delta A\geq 0$  and that we don’t wish to get any work out of our reaction. All we do is put our reactants into a bomb calorimeter and set them off, hoping that the calorimeter’s designers have done their job well: KABOOM! and then (if the bomb hasn’t become shrapnel and we’re still alive) wait for the heat to manifest itself as a small temperature rise in the calorimeter’s bath. Actually, in a perfect calorimeter, there would be no sound; all energy ends up as heat.

We now imagine doing our thought experiment again and ending up with our heat $\Delta U_0$ stored in the work bucket, then the products can conceptually be thought of as following the thought experiment’s path, working on themselves using some of the $\Delta U_0$ to expel the excess heat and then dumping the leftover $-\Delta A$  simply as heat at temperature $T_0$. Under these conditions, the amount of heat flowing into the outside world is what we would have calculated if we had thought like accountants and just did a simpleminded energy balance on Figure 1; the amount of heat gotten is $-\Delta U = \Delta U_0 + Q_R(T_0)-Q_P(T_0)$, where $\Delta U$ is the total change in internal energy (heat and potential) that happens as the reactants spontaneously become the products. This quantity, $-\Delta U\Delta U_0 +$Q_R(T_0)-Q_P(T_0)$is therefore seen to be the reaction’s total internal energy change; the energy released and all dumped as heat without drawing useful work from the reaction. Thus we arrive at Helmholtz’s “Arbeit”, the Helmholtz Free Energy: $$\Delta A(T_0) = \Delta U(T_0) – T_0\,\Delta S(T_0)\tag{11}$$ This is the figure we would get with an infinitely rigid bomb. If, however, our beginning and ending pressures with the same, the reaction would lose work thrusting outwards on the environment and so the minimum work could get from the reaction would be: $$\Delta G(T_0) = \Delta U(T_0) +P_0\,\Delta \,V- T_0\,\Delta S(T_0) = \Delta H(T_0) – T_0\,\Delta S(T_0)\tag{12}$$ where$\Delta H(T_0)$is the heat measured by a calorimeter with a non-rigid bomb allowing pressure equalisation after the whole setup has settled down, thus converting the explosive thrust to heat. Actually, in our hydrogen burning example, there is a considerable implosion. The quantity$H(T_0) + P\,V$for a substance called its enthalpy and$\Delta H(T_0)$is the enthalpy for the reaction. The excess of internal heats$Q_R(T_0) – Q_P(T_0)\$ thus, at first somewhat against intuition, hinders rather than helps to the extraction of useful work, for an excess must be gotten rid of – cast into the outside World – by our imaginary heat pump. So we can use the above discussion to strengthen insight into how an endothermic reaction happens. In this case, this is how our thought experiment would look:

1. The products bucket in Figure 1 is bigger than reactants bucket;
2. At stage 2 of our thought experiment we would have had to have a “work” bucket with some “initial capital” in it: we would have to borrow some work to raise the reactants’ potential energy to the products’ potential energy;
3. After Stage 3 of our thought experiment, there would have been a nett withdrawal of heat from each of the heat pads;
4. Well now it is easy to restore all of the heat pads to their beginning states: heat simply flows spontaneously from the outside world into the pads to do so. Indeed, we can use this heat to do work to pay back the work we borrowed from the work bucket, thus restoring the work bucket to its beginning state;
5. If the entropy change is highly positive so that outweighs a positive (i.e. a rise in internal heat and potential energy as reactants become products), then the endothermic reaction can still do work. In this case it is heat flowing from the outside world into the reactants/products that is doing the work. Consider towing a giant iceberg from Antarctica into a local harbour; one could get the iceberg to do useful work for us as it melted by running a heat engine drawing in heat at the ambient temperature from the outside world and then dumping it at the iceberg’s lower temperature.

There is one part of the discussion above that would not have washed with the classical thermodynamicists. Can you see what it is? We have tacitly assumed the Nernst heat theorem, or the third law of thermodynamics, that the entropy of anything at absolute zero Kelvin is nought. This was a very big deal when Nernst first proposed it, as for a long time people thought that only differences or relative entropies meant anything real. Nowadays however we are spoilt after many great men and women have worked out all the arguments that the classical thermodynamicists were grappling with and given us amazing machines that can bring substances so near to absolute zero that the talk of absolute zero and the behaviour of everything at this temperature can form part of the intuition we have used in our thought experiment. In the mid nineteenth century, atom-cooling and nano-Kelvins would have been pure science fiction. Actually there are some small modifications to the third law as systems with, e.g. lattice flaws, can get “frozen” in non-ground states at absolute zero and thus have some entropy. However, the quantities talked about are almost always utterly tiny compared with the quantities in Equation 9 at wonted temperatures and so Equation 9 defines, for all practical purposes, what is known as the entropy of formation of substances at a given temperature.