12.5 Thermodynamic Relations: Maxwell, Clausius-Clapeyron & Joule-Thomson

The Problem Thermodynamic Relations Solve

Pressure, volume and temperature can be measured directly. Internal energy, enthalpy and entropy cannot. Thermodynamic relations are the algebra that expresses the unmeasurable properties in terms of the measurable ones, which is how steam tables and refrigerant property charts are constructed in the first place.

The mathematical engine is the equality of mixed second partial derivatives established in the Engineering Mathematics chapter: if $z = z(x,y)$ is a well-behaved function, then

$\frac{\partial^2 z}{\partial x\,\partial y} = \frac{\partial^2 z}{\partial y\,\partial x}$

That identity, applied to four thermodynamic potentials, generates the whole topic.

The Four Thermodynamic Potentials

The Gibbsian Equations

Combining the first and second laws for a reversible process in a simple compressible system:

$du = T\,ds - p\,dv$

$dh = T\,ds + v\,dp$

$df = -s\,dT - p\,dv$

$dg = -s\,dT + v\,dp$

Each is an exact differential of the form $dz = M,dx + N,dy$, so each satisfies

$\left(\frac{\partial M}{\partial y}\right)_x = \left(\frac{\partial N}{\partial x}\right)_y$

The Four Maxwell Relations

Applying the exactness condition to each Gibbsian equation in turn:

$\left(\frac{\partial T}{\partial v}\right)_s = -\left(\frac{\partial p}{\partial s}\right)_v \qquad \text{(from } du\text{)}$

$\left(\frac{\partial T}{\partial p}\right)_s = \left(\frac{\partial v}{\partial s}\right)_p \qquad \text{(from } dh\text{)}$

$\left(\frac{\partial s}{\partial v}\right)_T = \left(\frac{\partial p}{\partial T}\right)_v \qquad \text{(from } df\text{)}$

$\left(\frac{\partial s}{\partial p}\right)_T = -\left(\frac{\partial v}{\partial T}\right)_p \qquad \text{(from } dg\text{)}$

The last two are the useful ones in practice, because they express the change of entropy — unmeasurable — in terms of $p$, $v$ and $T$ derivatives that come straight from an equation of state or from tabulated data.

A memory aid

The relations pair $T$ with $s$ and $p$ with $v$, always across the equality. The minus sign appears in the two relations where $p$ and $s$ sit on the same side, or equivalently in the first and fourth as written above.

The Tds Equations

Writing entropy as a function of two variables and substituting the Maxwell relations gives:

$T\,ds = c_v\,dT + T\left(\frac{\partial p}{\partial T}\right)_v dv$

$T\,ds = c_p\,dT - T\left(\frac{\partial v}{\partial T}\right)_p dp$

These are the working equations for computing entropy change for any substance, not merely an ideal gas. For an ideal gas, where $pv = RT$, they reduce to the familiar forms

$\Delta s = c_v\ln\frac{T_2}{T_1} + R\ln\frac{v_2}{v_1} = c_p\ln\frac{T_2}{T_1} - R\ln\frac{p_2}{p_1}$

Specific Heat Relations

From the Tds equations, the general relation between the two specific heats is

$c_p - c_v = -T\left(\frac{\partial v}{\partial T}\right)_p^{2}\left(\frac{\partial p}{\partial v}\right)_T$

or, in terms of the volume expansivity $\beta$ and isothermal compressibility $\kappa$,

$c_p - c_v = \frac{T v \beta^2}{\kappa}$

Three consequences worth stating explicitly, because each generates an objective item:

1. Since $\kappa > 0$ and $\beta^2 \geq 0$, the right-hand side is never negative, so $c_p \geq c_v$ always, for every substance.
2. As $T \to 0$, the difference vanishes, so the two specific heats converge at absolute zero.
3. When $\beta = 0$ — as for liquid water near 4 degrees Celsius, where density is maximum — the two specific heats are equal.

For an ideal gas, $\beta = 1/T$ and $\kappa = 1/p$, and the relation collapses to $c_p - c_v = R$.

The Clausius-Clapeyron Equation

During a phase change, pressure and temperature are not independent — they are linked by the saturation curve. Applying the third Maxwell relation across the two-phase region, where pressure depends only on temperature, gives

$\boxed{\frac{dp}{dT} = \frac{h_{fg}}{T\,v_{fg}}}$

where $h_{fg}$ is the latent heat and $v_{fg} = v_g - v_f$ the change in specific volume. This is the slope of the phase boundary on a $p$-$T$ diagram.

Consequences

• For vaporisation, $v_{fg}$ is large and positive, so $dp/dT$ is positive but modest. Boiling point rises with pressure — the principle of the pressure cooker and of the high-pressure boiler.
• For fusion of most substances, $v_{fg}$ is small and positive, so the melting line is very steep.
• For water freezing, $v_{fg}$ is negative because ice is less dense than water. So $dp/dT$ is negative and the melting point falls as pressure rises. This anomaly, unique among common substances, explains why ice skates glide and why deep glacial ice can flow.

Simplified form

Far below the critical point, $v_g \gg v_f$ and the vapour is nearly ideal, so $v_{fg}\approx v_g = RT/p$. Substituting and integrating gives

$\ln\frac{p_2}{p_1} = \frac{h_{fg}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

which is how saturation pressure at one temperature is estimated from a known value at another.

The Joule-Thomson Coefficient

Throttling through a valve or porous plug is adiabatic and involves no work, so it is an isenthalpic process: $h_1 = h_2$. Whether the gas cools or warms is measured by the Joule-Thomson coefficient:

$\mu_{JT} = \left(\frac{\partial T}{\partial p}\right)_h$

Since pressure always falls on throttling ($dp < 0$):

Using the Maxwell relations, the coefficient can be written entirely in measurable terms:

$\mu_{JT} = \frac{1}{c_p}\left[T\left(\frac{\partial v}{\partial T}\right)_p - v\right]$

The ideal-gas result

For an ideal gas, $v = RT/p$ so $T(\partial v/\partial T)_p = v$, and therefore

$\mu_{JT} = 0$

An ideal gas undergoes no temperature change on throttling. Any observed Joule-Thomson effect is a consequence of real-gas behaviour, and this is one of the cleanest demonstrations that real gases are not ideal.

The inversion curve

The locus of states where $\mu_{JT} = 0$ is the inversion curve, and the highest temperature on it is the maximum inversion temperature. A gas cools on throttling only when it enters the valve below its inversion temperature.

The practical implication is direct and important. Air and nitrogen are below their inversion temperatures at room temperature, so throttling cools them — which is the basis of the Linde liquefaction process and of the expansion valve in every vapour-compression refrigerator. Hydrogen and helium, however, are above their inversion temperatures at ambient conditions and therefore warm on throttling. They must first be pre-cooled, with liquid nitrogen for hydrogen and liquid hydrogen for helium, before throttling can be used to liquefy them at all.