Where Does the Joule–Thomson Effect Come From?

Pass a high-pressure gas through a porous plug in a pipe with adiabatic walls. No heat enters from outside. Nothing has been set up to extract work, either. Even so, the temperature of the gas that has passed through the plug changes — and, what is more, it does not necessarily go down. Here we start from the total differential of the enthalpy, define the Joule–Thomson coefficient μ = (∂T/∂p)_H, and follow what determines its sign. At the end, we will see that this phenomenon is, just as it stands, the expansion valve of an air conditioner.

1. The Total Differential of Enthalpy

First we prepare our tools. Regarding the enthalpy H as a function of p and T, its total differential is

d H = {(\frac{\partial H}{\partial p})}_{T} d p + {(\frac{\partial H}{\partial T})}_{p} d T

(1)

The coefficient on the right is easy to identify. Heat added at constant pressure goes directly into the increase of enthalpy, so

{(\frac{\partial H}{\partial T})}_{p} = C_{p}

The problem is the coefficient on the left, (∂H/∂p)_T. It is hard to measure this directly, so we replace it with a quantity that is easier to measure. This is where the Joule–Thomson coefficient comes in.

Definition μ ≡ (∂T/∂p)_H
How much does the temperature move when the pressure is changed while the enthalpy is held constant? Throttling expansion is precisely the operation of “lowering p at constant H” (next section), so this quantity can be measured.

The three variables H, p and T are linked by a chain relation (cyclic relation) that always holds mathematically.

{(\frac{\partial x}{\partial y})}_{z} {(\frac{\partial y}{\partial z})}_{x} {(\frac{\partial z}{\partial x})}_{y} = - 1

(2)

Apply this to (H, p, T).

{(\frac{\partial H}{\partial p})}_{T} {(\frac{\partial p}{\partial T})}_{H} {(\frac{\partial T}{\partial H})}_{p} = - 1

We already know the last two.

{(\frac{\partial T}{\partial H})}_{p} = \frac{1}{C_{p}} {(\frac{\partial p}{\partial T})}_{H} = \frac{1}{μ}

Therefore

{(\frac{\partial H}{\partial p})}_{T} = - \frac{1}{{(\frac{\partial p}{\partial T})}_{H} {(\frac{\partial T}{\partial H})}_{p}} = - \frac{1}{\frac{1}{μ} \cdot \frac{1}{C_{p}}} = - μ C_{p}

Putting this back into (1) completes the total differential of the enthalpy.

d H = - μ C_{p} d p + C_{p} d T

(3)

What do we gain? The change in H can now be written with only two measurable quantities (C_p and μ). C_p can be measured with a calorimeter, and μ by the throttling expansion of the next section. Even for a real gas whose equation of state is unknown, measuring these two lets us build a table of H — the property tables of refrigerants are made in exactly this way.

2. Why Is Throttling Expansion “Constant Enthalpy”?

Place a porous plug in the middle of an adiabatic pipe, push in gas at pressure p₁ from the left, and let it emerge on the right at pressure p₂ (< p₁). Suppose that a volume V₁ of gas is pushed in on the left side and a volume V₂ is pushed out on the right side.

The work W done on the gas is the work done by the left piston minus the work done on the right piston:

W = p_{1} V_{1} - p_{2} V_{2}

Because the process is adiabatic, Q = 0. By the first law, ΔU = W, that is,

\begin{array}{rcl} U_{2} - U_{1} & = & p_{1} V_{1} - p_{2} V_{2} \\ U_{1} + p_{1} V_{1} & = & U_{2} + p_{2} V_{2} \end{array}

∴ H_{1} = H_{2} (isenthalpic change)

(4)

Note: this is not a reversible process The flow through the plug is a violently turbulent, irreversible process. The intermediate states are not even in equilibrium, so the gas does not “travel along an isenthalpic line.” All that is certain is that H is equal at the inlet and the outlet. The isenthalpic lines drawn on a T–p diagram are “lines connecting equilibrium states that have the same H,” not the path the gas takes. The entropy always increases in this process.

3. The Isothermal Joule–Thomson Coefficient

Looking at (3), we can define one more natural quantity: how much the enthalpy changes when the pressure is lowered while the temperature is held constant —

μ_{T} \equiv {(\frac{\partial H}{\partial p})}_{T} = - C_{p} μ

(5)

This is called the isothermal Joule–Thomson coefficient. In Section 1 we found (∂H/∂p)_T = −μC_p, so this is merely a redefinition. Since C_p > 0, μ_T and μ always have opposite signs.

4. How Is μ Measured?

Trying to measure μ directly would mean “measuring a temperature difference precisely while keeping the system adiabatic,” which is troublesome. In practice, (5) is used the other way around.

Place a heater downstream of the throttle and heat the gas until the outlet temperature is the same as the inlet temperature.
Since there is no shaft work and the flow is a steady adiabatic flow apart from the heater (the same balance as in Section 2), the heat supplied then is exactly the difference between the inlet and outlet enthalpies, ΔH. Combined with the pressure difference Δp, this gives μ_T ≈ ΔH/Δp.
Using a separately measured C_p, obtain μ = −μ_T/C_p.

Because the temperature can be held constant, we are spared from worrying about the quality of the adiabatic insulation. This is the standard method.

Following the signs (i) A gas that cools when throttled: to bring the temperature back, heat must be added, so ΔH > 0. Since this is throttling, Δp < 0. Hence μ_T = ΔH/Δp < 0. From μ = −μ_T/C_p and C_p > 0 we get μ > 0.
(ii) A gas that warms when throttled: heat must be removed, so ΔH < 0 and Δp < 0, giving μ_T > 0, and therefore μ < 0.
In short, μ > 0 means cooling and μ < 0 means warming.

5. For an Ideal Gas, μ = 0

So far we have not used the equation of state at all. From here on we do. Using a thermodynamic relation (a general formula that follows from the Maxwell relations), μ can be written with only the equation of state and C_p. Moreover, only the contents of the square brackets determine the sign, so whether the gas cools or warms is decided by the equation of state alone.

μ = \frac{1}{C_{p}} [T {(\frac{\partial V}{\partial T})}_{p} - V]

(6)

Let us insert the ideal gas, V = nRT/p. Since (∂V/∂T)_p = nR/p = V/T, the contents of the square brackets are T·(V/T) − V = 0.

This phenomenon does not occur in an ideal gas μ_ideal = 0. Throttling an ideal gas does not change its temperature by even 1 K.
The two things that were thrown away in constructing the ideal gas — the size of molecules and the attraction between molecules — are exactly what the Joule–Thomson effect really is. That is why this is the one lesson in which nothing can get started without the equation of state of a real gas.

6. Calculating with a van der Waals Gas

We use the most straightforward model of a real gas.

p = \frac{R T}{v - b} - \frac{a}{v^{2}} (v is the molar volume)

a is the attraction and b is the size of the molecules (the excluded volume into which molecules cannot enter one another). Substituting into (6) and simplifying (we write w = v − b), we get

μ C_{p} = \frac{\frac{2 a}{v^{2}} - \frac{R T b}{w^{2}}}{\frac{R T}{w^{2}} - \frac{2 a}{v^{3}}}

(7)

For a dilute gas (v ≫ b), this reduces to an easy-to-read form.

μ \approx \frac{1}{C_{p}} (\frac{2 a}{R T} - b)

(8)

Eq. (8) says everything about this phenomenon 2a/RT is the attraction term, which acts in the cooling direction (positive). b is the size term, which acts in the warming direction (negative).
Only the attraction term acts as 1/T. So as the temperature is raised the attraction term withers away, and eventually loses to b, at which point the sign of μ changes. The boundary is T_inv = 2a/(Rb).

7. Seen from the Molecules’ Side

When the gas is throttled, it becomes dilute and the average distance between molecules increases.

When attraction is at work (dilute gas, lower temperature): work is required to pull molecules apart. No heat can be received from outside, so that energy has no choice but to be taken from the motion of the molecules themselves. The molecules slow down — that is, the gas cools.
When repulsion is at work (dense gas, high temperature): the molecules are already pushing against one another. Because the molecules compete for space, pv is larger than RT even at the same temperature. When throttling makes this extra work of pushing aside unnecessary, that portion goes into the motion of the molecules, because h = u + pv is constant. The molecules speed up — that is, the gas warms.

Which of the two wins is decided by the temperature and density at that point. That is why one cannot flatly say “this gas cools” or “this gas warms”: both occur, depending on conditions.

8. The Inversion Curve

The set of points where μ = 0, drawn on the (T, p) plane, is the inversion curve. Setting the numerator of (7) to zero gives a simple form for the van der Waals gas.

\frac{2 a}{v^{2}} = \frac{R T b}{{(v - b)}^{2}} \Rightarrow T_{inv} (v) = \frac{2 a}{R b} {(1 - \frac{b}{v})}^{2}

(9)

Combining this with p = p(T_inv, v) and sweeping v, we can draw a closed dome. The two ends at p → 0 are 0.75T_c and 6.75T_c, and the apex is (3T_c, 9p_c). Inside the dome is μ > 0 (cooling), and outside it — on the side hotter than the inversion curve and on the side at higher pressure than the inversion curve (above the apex value 9p_c, the point is outside whatever the temperature) — is μ < 0 (warming). The side at temperatures below 0.75T_c is also outside the dome, but there the inversion curve is a branch running off to negative pressure (p < 0), and on the low-pressure side, where the gas can exist stably, μ > 0 still holds.

Fig. 1: The sign of μ changes with the conditions. Left: μ versus temperature (one curve per pressure). The temperature at which a curve crosses μ = 0 is the inversion temperature. Right: the inversion curve (the dome), which shows on which side the current point lies. If you select He or H₂, you can see that room temperature is already outside the dome.

Gas Pressure p = 0.50 bar Temperature T = 365 K

Current μ—

μ_T = −C_pμ—

Inversion temperature (vdW, p→0)—

Inversion temperature (measured maximum)—

Critical temperature T_c—

Current verdict—

The sign of μ tells you only “the direction at the start of throttling” μ is the local slope at each point, so if the pressure is dropped by a large amount, the gas may cross the inversion curve along the way. Even if the inlet is outside the dome (μ < 0), if it is outside on the high-pressure side, then with a large enough drop the gas enters the inside partway through, and it may cool overall. However, if the inlet temperature itself is higher than the top of the dome (6.75T_c in vdW), the gas never enters the inside no matter how far the pressure is lowered, and it never cools. Conversely, if the inlet is inside, the gas stays inside to the end, so it always cools overall. Only in this direction can we say flatly that “if it is inside, it always cools.”

How far can van der Waals be trusted? This equation predicts an inversion temperature of 6.75T_c for every gas. For ordinary gases such as N₂ and O₂ the measured value is about 5T_c, so the prediction comes out roughly 30–40 % too high. However, He and H₂, where quantum effects matter, depart from this trend; for He the prediction is, if anything, on the low side.

Gas	vdW 2a/Rb	Measured maximum inversion temperature	Ratio
He	35 K	45 K	0.78
H₂	223 K	205 K	1.09
N₂	852 K	621 K	1.37
O₂	1043 K	761 K	1.37
CH₄	1287 K	939 K	1.37
CO₂	2052 K	1500 K	1.37

When the sign changes, which gases cool at room temperature and which warm — the logic of all this comes out correctly. The numbers, however, are not accurate enough to be used directly for design.

9. What Is It Good For?

(a) Liquefying gases

Air (N₂ and O₂) has an inversion temperature above 600 K, so if you throttle it straight from room temperature, it cools. The cooled gas precools the incoming gas, which is then throttled again — the apparatus that repeats this to liquefy air is the Linde–Hampson air liquefier. Liquid nitrogen is made this way.

However, H₂ has an inversion temperature of only 205 K and He only 45 K. If you throttle them at room temperature, they warm up instead. Therefore, to liquefy hydrogen, it must first be cooled with liquid nitrogen to around 80 K, bringing it below the inversion temperature, and only then throttled. Helium must be precooled further, with liquid hydrogen or liquid neon — and it is for this reason that low-temperature physics has built up its staircase of “what do we cool with first?”

(b) Air conditioners and refrigerators

Inside a household air conditioner is a restriction called the expansion valve, and what happens there is exactly a throttling expansion at constant H. The difference from the experiment in tab ① of the simulator, however, is that the refrigerant enters the region inside the saturation line.

Throttling a gas and throttling a liquid differ by orders of magnitude in effect When only a gas is throttled, the temperature drop is about 20 K for a 100 bar pressure drop. When a saturated liquid is throttled, however, the moment the pressure drops a portion evaporates all at once (flashing), taking that heat of vaporization from the rest of the liquid. The temperature sticks to the saturation temperature at the lower pressure, and with a pressure drop of only about ten to a few tens of bar, it falls by 40 K or even 50 K in one go.
Whether latent heat can be used is what makes this difference. This is why air conditioners use a refrigerant that undergoes a phase change, and not plain air.

On a p–h diagram, the expansion valve appears as a straight downward line, because h does not change. Look at that vertical line in tab ② of the simulator.

Component	What it does	In thermodynamic terms
Compressor	Compresses low-pressure vapor to high pressure	Adiabatic compression (uses electricity)
Condenser (outdoor unit)	Rejects heat and returns the vapor to liquid	Constant-pressure heat rejection
Expansion valve	Throttles the refrigerant down to low pressure	Isenthalpic change (Joule–Thomson)
Evaporator (indoor unit)	Absorbs heat and returns the refrigerant to vapor	Constant-pressure heat absorption

What is switched between cooling and heating is only the direction in which the refrigerant circulates (the reversing valve), and the roles of the indoor and outdoor units are exchanged. That is why the same machine can do both.

The performance of this device is measured by the coefficient of performance (COP), the heat moved divided by the work put into the compressor. Per 1 kg of refrigerant, let q_e be the heat received in the evaporator, q_k the heat rejected in the condenser, and w the work put into the compressor. Then the COP for cooling is q_e/w and the COP for heating is q_k/w. In steady operation q_k = q_e + w, so for the same cycle the COP for heating is greater than the COP for cooling by exactly 1.

Exceeding 1 does not violate the first law Because the machine is not making heat but only moving it from the low-temperature side to the high-temperature side. The upper limit on how much can be moved is set by a reversible engine (Carnot) operating between the same two temperatures: with the evaporating temperature T_e and the condensing temperature T_k, the limit is T_e/(T_k − T_e) for cooling and T_k/(T_k − T_e) for heating. The more the temperature differences across the heat exchangers are narrowed, bringing T_e and T_k closer to the indoor and outdoor temperatures, the closer one gets to this limit, but the heat exchangers then have to be made larger — this is why the outdoor unit is as big as it is.

10. Summary

Question	Answer
Why does H not change when a gas is throttled?	Because the process is adiabatic, Q = 0, and the difference between the work of pushing in and the work of pushing out becomes ΔU (Eq. 4)
What is μ?	The temperature change when the pressure is changed at constant H, (∂T/∂p)_H. From the chain relation, (∂H/∂p)_T = −μC_p (Eq. 3)
Why does it not occur in an ideal gas?	Because T(∂V/∂T)_p = V holds exactly, μ = 0
What decides whether the gas cools or warms?	Which wins: the attraction 2a/RT or the size b. The boundary is the inversion curve (Eqs. 8, 9)
Why does hydrogen warm up at room temperature?	Its inversion temperature is only 205 K, so room temperature is already outside the dome
Where is it in an air conditioner?	The expansion valve. It is the straight downward line on the p–h diagram

In three lines ① When a gas is throttled adiabatically, only the temperature moves, while the enthalpy stays unchanged.
② The direction of the movement is decided by a contest between the attraction between molecules (cooling) and the size of molecules (warming).
③ The boundary of that contest is the inversion curve, and inside it the gas always cools — both air liquefaction and air conditioners operate inside it.

The van der Waals constants a and b are tabulated values fitted to the critical point, and C_p is the ideal-gas part obtained from the partition function with a correction for the real gas added. The measured inversion temperatures are standard literature values from cryogenic engineering. Related pages: Derivation of the ideal gas, Derivation of the Carnot cycle, The simulator.