Thermodynamic property relations
Every steam table and every refrigerant chart in this course was built by someone who never measured internal energy, enthalpy or entropy directly, because no instrument does. What they measured was pressure, volume and temperature, and what stands between those two lists is a short piece of calculus that this article is about.
This is the one topic in the course that is mostly mathematics, and it earns the detour because it explains where every number in every table this course has used actually came from. Nothing here is a new physical law. The first and second laws were stated once, several topics ago, and everything below is what falls out of them when the ordinary rules of calculus are applied without mercy.
Why tables are not enough
Pressure is a force on a diaphragm. Temperature is the reading of a thermometer brought to equilibrium with the system. Volume is a length, or three of them. All three are properties you could in principle measure this afternoon with equipment sitting on a bench, and the state postulate says that any two of them, chosen independently, fix everything else about a simple compressible system. That is a remarkable claim and it is also an unhelpful one on its own, because it does not say how to get from the two numbers you have to the one you actually want.
Internal energy, enthalpy and entropy are not on that bench. There is no internal-energy gauge and no entropy meter, and there never will be, because neither quantity is defined as the reading of any instrument. Enthalpy was defined as the combination \(h = u + Pv\) because it made the first law tidy for a flow process. Entropy was defined, in entropy, by an integral that happened to vanish around a reversible cycle. Both are properties in the fullest sense, fixed by the state, path independent, and both were invented rather than found, which leaves open the practical question this whole article answers: how do you attach a number to one of them without redoing that derivation from scratch every time.
The answer is that every one of these quantities, together with the three measurable ones, is tied to the others by the first and second laws combined into a single differential. For a simple compressible substance undergoing an internally reversible change,
which is nothing but the first law with \(\delta q = T\,ds\) and \(\delta w = P\,dv\) substituted in, and which, because both sides are exact differentials of properties, holds for any process between the same two states, reversible or not, exactly as the two \(T\,ds\) relations did when they first appeared in the entropy topic. Three more combinations like it exist, built from the same two laws, and between the four of them lies every relation this article derives.
What makes the four worth separating out is that each pairs naturally with a different choice of independent variables, and each therefore answers a different kind of question cheaply. That is the whole plan for what follows: define four combinations of \(u\), \(h\) and \(s\), notice that each one is an exact differential, and read off what exactness demands. Nothing is measured in this article that was not already measurable. What is gained is a route from three numbers on a bench to three that never could be.
Pressure, temperature and volume can each be read off an instrument directly.
The Maxwell relations
Alongside internal energy and enthalpy, define two more combinations, each one built to make a particular pair of variables fall out as the natural independent ones. The Helmholtz function \(a = u - Ts\) and the Gibbs function \(g = h - Ts\) are properties because they are built entirely from other properties, and differentiating each one and substituting the first relation above gives all four together:
Read each one as naming its own natural variables: \(u\) wants \(s\) and \(v\), \(h\) wants \(s\) and \(P\), \(a\) wants \(T\) and \(v\), \(g\) wants \(T\) and \(P\). That pairing is not decoration. It is what lets the next step work.
Every one of the four right-hand sides has the shape \(dz = M\,dx + N\,dy\), and for any function with continuous second derivatives (every property in this course qualifies), the order of differentiation cannot matter:
which is a statement about arithmetic, not physics. Raise \(y\) a little and then \(x\), or \(x\) then \(y\), and a property cannot know or care which route brought it to the same corner. Applying that single fact to each of the four differentials in turn produces the four Maxwell relations:
The last two are the ones worth memorising, because they are the ones that do real work. Each one takes a derivative of entropy, the unmeasurable property, and rewrites it entirely in terms of \(P\), \(v\) and \(T\), which is precisely the substitution the previous section promised. An entropy change that looked like it needed an entropy measurement to pin down turns out to need nothing more than an equation of state, or a table of \(P\)-\(v\)-\(T\) data, which is exactly the kind of data a bench can produce.
None of this required a new law. It required only that \(u\), \(h\), \(a\) and \(g\) be properties, which was never in doubt, and that a mixed second derivative not care about order, which is a fact about smooth functions and not about thermodynamics at all. The next three sections spend that fact three separate ways.
Four combinations of u, h, s, P, v and T are exact differentials — no surprise there.
The Clapeyron equation
Inside the saturation dome, from properties of pure substances, pressure and temperature stop being independent: one saturation pressure for every saturation temperature, a single curve rather than a surface. That collapse, which was a nuisance for the state postulate, is exactly what the third Maxwell relation needs to become an equation for a real, tabulated property.
Take \((\partial s/\partial v)_T = (\partial P/\partial T)_v\) and evaluate it along the phase change at fixed \(T\). Because pressure is fixed by temperature alone on the saturation curve, the partial derivative on the right is also the ordinary derivative \(dP/dT\) of that curve, whatever \(v\) happens to be doing at the same time. And crossing the dome at constant \(T\) and \(P\) takes entropy from \(s_f\) to \(s_g\) while volume goes from \(v_f\) to \(v_g\), so the left side is simply the ratio of the two changes:
One more substitution finishes it. Along the constant-pressure phase change, \(T\,ds = dh\) integrates directly to \(s_{fg} = h_{fg}/T\), so
the Clapeyron equation. It connects the slope of a curve you could trace with a pressure gauge and a thermometer to a latent heat you would otherwise need a calorimeter to measure, and it does so exactly, with no approximation anywhere in the derivation.
Run it as a check rather than a derivation and the agreement is convincing on its own. Near \(100\) °C the vapour pressure curve of water has a measured slope \(dP/dT \approx 3.62\) kPa/K, and \(v_{fg} \approx v_g \approx 1.673\) m³/kg (the liquid's volume is thousands of times smaller and drops out of the difference). Then
against a tabulated value of \(2257\) kJ/kg: three measurements of the easy kind, pressure, temperature and volume, reproducing a latent heat to within a fraction of a per cent.
The same equation runs the other way in practice more often than not: given a latent heat and one point on the saturation curve, predict pressures nearby without a full table. Approximating the vapour as an ideal gas and neglecting \(v_f\) against \(v_g\), so that \(v_{fg} \approx RT/P\), turns the Clapeyron equation into the Clausius–Clapeyron equation,
assuming \(h_{fg}\) is roughly constant over the interval, which is the standard way of estimating how a boiling point shifts with altitude, or how a refrigerant's saturation pressure responds to a change in operating temperature, without reopening a full property table.
The saturation curve has one pressure for every temperature — a single line, not a surface.
Specific heat relations
Ideal and real gases showed \(c_p - c_v = R\) for an ideal gas by substituting \(h = u + RT\) into the two definitions and differentiating: three lines of algebra, and correct as far as it goes. What it could not show is why the relation takes exactly that shape for a substance that is not an ideal gas, because the shortcut it used, that internal energy depends on temperature alone, is precisely the assumption an ideal gas gets for free and a real one does not.
The general result again comes from an exact differential, this time by equating the two forms \(T\,ds\) takes when entropy is written first as a function of \(T\) and \(v\), then as a function of \(T\) and \(P\), and substituting the two specific-heat definitions \(c_v = T(\partial s/\partial T)_v\) and \(c_p = T(\partial s/\partial T)_P\) together with the Maxwell relations from the previous section. The algebra is longer than it is illuminating; the result is what matters:
Both factors on the right are built from an equation of state alone: no entropy, no internal energy, nothing that was not already on the bench. And the sign works out in your favour automatically: \((\partial v/\partial T)_P^2\) cannot be negative, and \((\partial P/\partial v)_T\) is negative for every stable substance (squeezing something at constant temperature raises its pressure, never lowers it), so the product of a positive square and a negative slope, times the leading minus sign, comes out positive. \(c_p \ge c_v\) is not an empirical observation in this account. It falls out of the formula whether or not you already believed it.
Put the ideal gas back in and watch the general result collapse to the familiar one. With \(Pv = RT\), \((\partial v/\partial T)_P = R/P\) and \((\partial P/\partial v)_T = -P/v\), so
using \(Pv = RT\) a second time to cancel the temperature. The three-line derivation was not wrong. It was a special case of this one, valid because the ideal-gas equation of state is simple enough that the general formula degenerates to a constant.
The same reasoning explains a fact used without proof earlier in the course: that an ideal gas's specific heats depend on temperature alone, never on pressure or volume. A short argument along the same lines gives \((\partial c_v/\partial v)_T = T(\partial^2 P/\partial T^2)_v\), and for \(Pv = RT\), which is linear in \(T\) at fixed \(v\), the second derivative on the right is identically zero. That single vanishing term is the entire reason ideal-gas tables list \(c_p(T)\) and \(c_v(T)\) as functions of one variable rather than two: not a simplification adopted for convenience, but a consequence forced by the shape of the equation of state.
Two isotherms, an instant apart in T, sit at different v for the same pressure.
The Joule–Thomson coefficient
A throttle (a valve, a porous plug, a length of capillary tube) does no work and, run fast enough to be adiabatic, passes no heat either, so the steady-flow energy balance collapses to \(h_2 = h_1\), exactly as it did for the refrigerant crossing the expansion valve in refrigeration and heat pumps. What is not fixed is the temperature. Whether it falls, rises, or does not move at all as the pressure drops is answered by a single derivative, the Joule–Thomson coefficient:
which looks unpromising to evaluate directly, since it asks how \(T\) moves along a path of constant \(h\), and nobody has an \(h\)-meter to hold fixed while nudging pressure. The standard trick for exactly this situation, a derivative at constant \(h\), wanted in terms of derivatives at constant \(T\) or \(P\), is the cyclic relation among partial derivatives, applied to \(h(T, P)\), together with \(dh = c_p\,dT + \left[v - T(\partial v/\partial T)_P\right]dP\) (itself a short derivation from \(T\,ds = dh - v\,dP\) and a Maxwell relation). The result is
built once again from nothing but an equation of state and a specific heat, both measurable. For an ideal gas the bracket vanishes identically (\(v = RT/P\) gives \(T(\partial v/\partial T)_P = TR/P = v\) exactly), so \(\mu_{JT} \equiv 0\) and an ideal gas cannot change temperature on throttling at all, which is the same fact already used, without derivation, when internal energy was declared a function of temperature alone.
Real gases are not ideal gases, and \(\mu_{JT}\) is exactly the number that says by how much. Where \(\mu_{JT} > 0\), dropping the pressure drops the temperature, throttling cools the gas, and where it is negative, throttling warms it. The boundary between the two regions, plotted on a \(T\)-\(P\) diagram, is the inversion curve, and a gas only cools on throttling if its state sits inside it.
This is not a footnote. It is how gases are liquefied. The Linde process compresses a gas, cools it by ordinary heat exchange, and throttles it repeatedly, relying on \(\mu_{JT}\) being positive at the operating point for each pass to buy a little more cooling than the one before, until liquid drops out. The process only works if the gas starts inside the inversion curve to begin with, and this is where hydrogen and helium cause trouble: their maximum inversion temperatures are around \(200\) K and \(40\) K respectively, both far below ordinary room temperature. Throttle either one starting from a laboratory bench and it warms up rather than cools, which is the opposite of what a naive reading of “throttling drops the pressure, so it should help” would predict. Both gases have to be pre-cooled by other means (heat exchange against an already-liquefied bath, in practice) until their state finally lies inside the inversion curve, at which point the Joule–Thomson effect can take over and finish the job.