Joule-Thomson Expansion

Push a real gas slowly through a porous plug or a partly-opened valve, insulated so no heat leaks in, and its temperature changes even though it does no useful work and exchanges no heat. Nitrogen and air cool; hydrogen and helium, started at room temperature, warm up. This is the Joule–Thomson (or Joule–Kelvin) effect, and it is the workhorse of gas liquefaction: every household refrigerator, every air-separation plant, and the final stage of every helium liquefier relies on it. The puzzle is that an ideal gas throttled this way would not change temperature at all — so the entire effect is a direct, macroscopic readout of the intermolecular forces that make a gas real. The process conserves one specific quantity: the enthalpy $H=U+PV$ . Throttling is isenthalpic, and the temperature change per unit pressure drop is captured by a single response function, the Joule–Thomson coefficient $\mu_{JT}=(\partial T/\partial P)_H$ . A short thermodynamic argument reduces it to $\mu_{JT}=(1/C_P)\,[\,T(\partial V/\partial T)_P-V\,]$ , which vanishes identically for an ideal gas and changes sign at a characteristic inversion temperature for a real one. This article derives the isenthalpic condition, obtains $\mu_{JT}$ from a Maxwell relation, proves the ideal-gas result, and locates the van der Waals inversion curve with its clean prediction $T_{\text{inv}}^{\max}=2a/Rb=\tfrac{27}{4}T_c$ . The engineering cycles (Linde–Hampson, Collins) and the microscopic cluster expansion are named where they belong, not derived here.

Key takeaways

Throttling a gas through a plug or valve at constant enthalpy is the Joule–Thomson process; the temperature change per unit pressure drop is $\mu_{JT}=(\partial T/\partial P)_H$ .
A Maxwell relation reduces it to $\mu_{JT}=\dfrac{1}{C_P}\left[T\left(\dfrac{\partial V}{\partial T}\right)_P-V\right]=\dfrac{V}{C_P}(T\alpha-1)$ , with $\alpha$ the isobaric expansivity.
For an ideal gas $T(\partial V/\partial T)_P=V$ exactly, so $\mu_{JT}=0$ : no cooling, no heating. The whole effect is a real-gas phenomenon driven by intermolecular forces.
$\mu_{JT}>0$ means the gas cools as it expands (since $dP<0$ ); this happens only below the inversion temperature $T_{\text{inv}}$ , where $\mu_{JT}=0$ .
Van der Waals gives $T_{\text{inv}}^{\max}=2a/Rb=\tfrac{27}{4}T_c=6.75\,T_c$ : nitrogen cools at room temperature, but helium and hydrogen must be pre-cooled below their (low) inversion temperatures first.
Key idea: throttling holds enthalpy fixed, and whether a gas heats or cools is decided by the competition between the attractive $-a/V^2$ pull and the repulsive excluded volume $b$ — a purely real-gas contest with no ideal-gas counterpart.

The core insight

Throttling is isenthalpic, so the sign of the temperature change is set by $\mu_{JT}=(1/C_P)[T(\partial V/\partial T)_P-V]$ . This is exactly zero for an ideal gas; for a real gas it is positive (cooling) below the inversion temperature and negative (heating) above it, and the van der Waals model pins the maximum inversion temperature at $6.75$ times the critical temperature.

Where this lives today

The Joule–Thomson effect is the thermodynamic engine of the cryogenic and refrigeration industries. The Linde–Hampson cycle — throttle, then recycle the cooled gas through a counter-current heat exchanger so that each pass pre-cools the incoming stream — turns the modest per-pass cooling into full liquefaction, and is the basis of industrial air separation into oxygen, nitrogen, and argon. Every vapour-compression refrigerator and air conditioner throttles its working fluid through an expansion valve, exploiting the same $(\partial T/\partial P)_H$ . Helium and hydrogen liquefiers add an expansion-engine or turbine pre-cooling stage (the Collins cycle) to get below the inversion temperature before the final throttle. Beyond the van der Waals estimate, the quantitative inversion curve and the temperature dependence of $\mu_{JT}$ follow from the second virial coefficient $B(T)$ and the intermolecular potential, treated in the virial and cluster expansions; the connection $\mu_{JT}\propto T\,dB/dT-B$ makes throttling a sensitive thermometer of intermolecular forces.

The physical picture

Imagine gas forced steadily from a high-pressure chamber, through a porous plug that throttles the flow, into a low-pressure chamber, with the walls thermally insulating so that no heat is exchanged. Upstream, the surroundings push a volume $V_1$ of gas through the plug at pressure $P_1$ , doing work $P_1V_1$ on the gas; downstream, the gas pushes out a volume $V_2$ against pressure $P_2$ , doing work $P_2V_2$ . With no heat flow, the first law says the internal energy changes by exactly the net work done on the gas, and — as the derivation below shows — this forces the enthalpy to be the same on both sides. Throttling is an isenthalpic process. It is also strongly irreversible: entropy is produced as the gas expands turbulently through the plug, so this is nothing like a reversible adiabatic expansion, and the temperature change is generally different. Why should the temperature change at all? In an ideal gas the internal energy depends only on temperature, and $PV=RT$ makes $H=U+RT$ a function of $T$ alone; holding $H$ fixed therefore holds $T$ fixed, and nothing happens. In a real gas two things break this. First, the molecules attract one another, so pulling them apart during expansion costs potential energy that is paid for out of the kinetic energy — the gas cools. Second, the molecules have finite size and repel at short range; that excluded-volume effect works the other way and tends to warm the gas on expansion. Which wins depends on temperature: at low temperature the slow molecules feel the attraction most and the gas cools; at high temperature the fast, hard collisions dominate and the gas warms. The crossover is the inversion temperature.

Figure

Van der Waals inversion curve on the temperature–pressure plane: cooling inside, heating outside

There is one decisive prediction to hold onto. The inversion temperature is not a small correction: for the van der Waals gas the maximum inversion temperature is $6.75$ times the critical temperature. For nitrogen ( $T_c\approx126$ K) that puts the ceiling far above room temperature, so ambient nitrogen throttles cold and air-liquefaction plants work directly. For helium ( $T_c\approx5.2$ K) the inversion ceiling is only around $40$ K, so room-temperature helium heats on throttling and cannot be liquefied by a valve until it has first been pre-cooled below that ceiling. The same tiny formula, $2a/Rb$ , explains both why air liquefiers are commonplace and why helium liquefiers need an extra pre-cooling stage.

Figure

Isenthalps on the T–P plane: μ_JT is the slope, and the inversion curve threads their maxima

The core equations

Throttling is defined by constant enthalpy, and the temperature response is a single derivative of the equation of state; everything below flows from these two facts.

H \equiv U + PV, \qquad H_1 = H_2 \quad\text{(throttling is isenthalpic)} \qquad (1)

where: $H$ is the enthalpy, $U$ the internal energy, $P$ the pressure and $V$ the volume. The enthalpy is the natural potential for a constant-pressure flow process; that the two sides of the plug share the same $H$ (not the same $U$ or $T$ ) is what the flow-work bookkeeping in the first derivation establishes. The enthalpy itself is treated in Helmholtz Free Energy, Enthalpy, and Gibbs Free Energy.

\mu_{JT} \equiv \left(\frac{\partial T}{\partial P}\right)_H \qquad (2)

where: $\mu_{JT}$ is the Joule–Thomson coefficient, the temperature change per unit pressure change at fixed enthalpy. Because throttling lowers the pressure ( $dP<0$ ), a positive $\mu_{JT}$ produces a temperature drop (cooling); a negative $\mu_{JT}$ produces heating. Its units are K/Pa (often quoted in K/bar).

\mu_{JT} = \frac{1}{C_P}\left[\,T\left(\frac{\partial V}{\partial T}\right)_P - V\,\right] = \frac{V}{C_P}\,(T\alpha - 1) \qquad (3)

where: $C_P=(\partial H/\partial T)_P$ is the isobaric heat capacity (a Response Function), and $\alpha=\tfrac{1}{V}(\partial V/\partial T)_P$ is the isobaric thermal expansivity. This is the master formula: it turns the abstract derivative $(\partial T/\partial P)_H$ into something computable from the equation of state alone. The bracket vanishes exactly when $T\alpha=1$ .

\left(P + \frac{a}{v^2}\right)(v-b) = RT \qquad (4)

where: $v=V/n$ is the molar volume, $a$ measures the intermolecular attraction, $b$ the excluded volume (finite molecular size), and $R$ the gas constant. This is the Van der Waals Equation of State; feeding it into equation (3) produces the inversion curve. The two parameters encode exactly the two competing effects sketched above.

T_{\text{inv}}^{\max} = \frac{2a}{Rb} = \frac{27}{4}\,T_c, \qquad T_c=\frac{8a}{27Rb} \qquad (5)

where: $T_{\text{inv}}^{\max}$ is the highest temperature at which a van der Waals gas can still be cooled by throttling (the zero-pressure top of the inversion curve), and $T_c$ its critical temperature. The ratio $27/4=6.75$ is a pure number, independent of $a$ , $b$ , and the gas — a corresponding-states prediction.

Derivation AThrottling conserves enthalpy

The goal is to prove equation (1): that steady flow through a thermally insulated plug leaves the enthalpy unchanged, even though the gas does net work and its internal energy, pressure, and temperature all change. The strategy is to follow a fixed parcel of gas from the high-pressure side to the low-pressure side and account for every scrap of work done on it and by it, using the fact that no heat crosses the insulated walls. The one subtlety is that the "work" here is flow work — the work the upstream gas does to shove the parcel through and the work the parcel does to make room downstream — and these two flow-work terms are what convert internal energy into enthalpy.

Step 1. Isolate a parcel that starts entirely upstream, occupying volume $V_1$ at pressure $P_1$ , and ends entirely downstream, occupying $V_2$ at $P_2$ . The walls are insulated, so $Q=0$ ; by the first law, $\Delta U = U_2-U_1 = W$ , the net work done on the parcel.

Step 2. Count the work. On the upstream side the rest of the gas acts like a piston at constant pressure $P_1$ , sweeping the parcel's volume from $V_1$ to $0$ as it passes the plug; the work done on the parcel is $+P_1V_1$ . On the downstream side the parcel expands from $0$ to $V_2$ against constant pressure $P_2$ , doing work $P_2V_2$ on its surroundings, i.e. work $-P_2V_2$ is done on the parcel. The net work on the parcel is $W = P_1V_1 - P_2V_2$ .

Step 3. Combine. $U_2 - U_1 = P_1V_1 - P_2V_2$ , which rearranges to $U_2 + P_2V_2 = U_1 + P_1V_1$ , i.e. $H_2 = H_1$ . The enthalpy is conserved across the plug.

Step 4. Note what is not conserved and what this is not. The internal energy changes, the temperature generally changes, and — because the expansion through the plug is turbulent and uncontrolled — entropy is produced: this is an irreversible process, not a reversible adiabat. Only $H$ is pinned. That is why the relevant coefficient is $(\partial T/\partial P)_H$ and not $(\partial T/\partial P)_S$ . Note: the flow-work bookkeeping $\Delta U = P_1V_1-P_2V_2 \Rightarrow H_1=H_2$ can be confirmed symbolically, and the ideal-gas special case $H=H(T)$ confirms that constant $H$ then forces constant $T$ .

Go deeper · Derivation B · 1 minThe Joule–Thomson coefficient from a Maxwell relation

Derivation BThe Joule–Thomson coefficient from a Maxwell relation

Here the aim is to derive the master formula (3), $\mu_{JT}=(1/C_P)[T(\partial V/\partial T)_P-V]$ , starting from the definition $\mu_{JT}=(\partial T/\partial P)_H$ and reducing it to derivatives of the equation of state. The strategy is a standard two-step manoeuvre: first use a triple-product (cyclic) identity to trade the awkward constant- $H$ derivative for two ordinary ones, then evaluate $(\partial H/\partial P)_T$ using the fundamental relation $dH=T\,dS+V\,dP$ together with a Maxwell relation. The subtlety worth flagging is the Maxwell relation itself: $(\partial S/\partial P)_T=-(\partial V/\partial T)_P$ comes from the Gibbs free energy and is what injects the equation of state into the answer. The Maxwell relations themselves are established in Maxwell Relations.

Step 1. Apply the cyclic identity to the three variables $T,P,H$ : $\left(\frac{\partial T}{\partial P}\right)_H\left(\frac{\partial P}{\partial H}\right)_T\left(\frac{\partial H}{\partial T}\right)_P=-1$ . Solving for the first factor and using $(\partial H/\partial T)_P=C_P$ gives $\mu_{JT}=-\dfrac{1}{C_P}\left(\dfrac{\partial H}{\partial P}\right)_T$ .

Step 2. Evaluate $(\partial H/\partial P)_T$ . From $H=U+PV$ and the fundamental relation, $dH=T\,dS+V\,dP$ . Differentiating at constant $T$ : $\left(\frac{\partial H}{\partial P}\right)_T = T\left(\frac{\partial S}{\partial P}\right)_T + V$ .

Step 3. Use the Maxwell relation from $dG=-S\,dT+V\,dP$ , namely $\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$ . Substituting, $\left(\frac{\partial H}{\partial P}\right)_T = V - T\left(\frac{\partial V}{\partial T}\right)_P$ .

Step 4. Assemble. $\mu_{JT}=-\dfrac{1}{C_P}\Big[V-T\big(\tfrac{\partial V}{\partial T}\big)_P\Big]=\dfrac{1}{C_P}\Big[T\big(\tfrac{\partial V}{\partial T}\big)_P-V\Big]$ . Writing $\alpha=\tfrac1V(\partial V/\partial T)_P$ gives the compact form $\mu_{JT}=\tfrac{V}{C_P}(T\alpha-1)$ . Cooling requires $T\alpha>1$ ; the inversion condition is $T\alpha=1$ . Note: $(\partial H/\partial P)_T=V-T(\partial V/\partial T)_P$ can be confirmed symbolically, and the ideal-gas relation $V=RT/P$ makes the bracket vanish identically.

Go deeper · Derivation C · 1 minThe van der Waals inversion temperature and the 6.75 ratio

Derivation CThe van der Waals inversion temperature and the 6.75 ratio

The target is equation (5): to show that a van der Waals gas has a well-defined inversion temperature, that its low-density value is $2a/Rb$ , and that the maximum inversion temperature is exactly $27/4$ times the critical temperature — a parameter-free number. The strategy is to evaluate the bracket $T(\partial V/\partial T)_P-V$ for the van der Waals equation of state to leading order in the density, set it to zero, and then compare with the known critical temperature $T_c=8a/27Rb$ . The subtlety is that the full inversion curve is a closed loop in the $T$ – $P$ plane; the single number $2a/Rb$ is its zero-pressure ceiling, the highest temperature at which any cooling is possible.

Step 1. Work to first order in density. Write the van der Waals equation as $P=\dfrac{RT}{v-b}-\dfrac{a}{v^2}$ and expand for large $v$ : $\dfrac{Pv}{RT}=1+\dfrac{1}{v}\Big(b-\dfrac{a}{RT}\Big)+\dots$ , so the second virial coefficient is $B(T)=b-a/RT$ .

Step 2. For a gas with $B(T)$ , the Joule–Thomson bracket becomes $T(\partial V/\partial T)_P-V \to T\dfrac{dB}{dT}-B$ per mole to this order. Compute $T\,dB/dT-B = T\cdot\dfrac{a}{RT^2}-\big(b-\dfrac{a}{RT}\big)=\dfrac{a}{RT}-b+\dfrac{a}{RT}=\dfrac{2a}{RT}-b$ .

Step 3. The inversion condition sets this to zero: $\dfrac{2a}{RT_{\text{inv}}}=b$ , i.e. $T_{\text{inv}}=\dfrac{2a}{Rb}$ . Above it the bracket is negative (heating); below it, positive (cooling). This low-density value is the maximum inversion temperature, the top of the inversion dome.

Step 4. Compare to the critical temperature. Van der Waals gives $T_c=8a/27Rb$ . Hence $\dfrac{T_{\text{inv}}^{\max}}{T_c}=\dfrac{2a/Rb}{8a/27Rb}=\dfrac{2\cdot27}{8}=\dfrac{27}{4}=6.75$ . The $a$ and $b$ cancel: this is a corresponding-states prediction, the same for every van der Waals gas. Note: the results $B(T)=b-a/RT$ , the bracket $2a/RT-b$ , the root $T_{\text{inv}}=2a/Rb$ , and the exact ratio $T_{\text{inv}}^{\max}/T_c=27/4$ can all be confirmed symbolically, along with the N₂ and He numerical values.

Limiting cases

Each limit of equation (3) is a physical check on the master formula.

Case	Result	Meaning
Ideal gas	$T(\partial V/\partial T)_P=V\Rightarrow\mu_{JT}=0$	No temperature change on throttling; the effect is intrinsically a real-gas one.
Low temperature, $T<T_{\text{inv}}$	$\mu_{JT}>0$	Attraction dominates; pulling molecules apart costs kinetic energy — the gas cools.
High temperature, $T>T_{\text{inv}}$	$\mu_{JT}<0$	Repulsion / excluded volume dominates; the gas warms on expansion.
At the inversion temperature	$T\alpha=1,\ \mu_{JT}=0$	The two effects balance; throttling neither heats nor cools to leading order.

Common trapThe Joule–Thomson expansion is not a reversible adiabatic expansion

It is tempting to picture throttling as "adiabatic expansion, so it must cool like a gas pushing a piston." That is wrong on two counts. First, throttling is irreversible: the gas expands turbulently through the plug, producing entropy, so it is nothing like the reversible adiabat $PV^\gamma=$ const that always cools. Second, throttling conserves enthalpy, not entropy, and an ideal gas throttled at constant enthalpy does not change temperature at all — whereas the same ideal gas expanding reversibly and adiabatically against a piston does cool (it does external work). The two "adiabatic" processes are physically distinct: one does useful work and is reversible; the other does no net useful work, is irreversible, and only changes temperature because the gas is real. Confusing them predicts cooling for an ideal gas, which is simply false.

Worked examples

Example 1 — Does nitrogen cool at room temperature, and by how much? Take nitrogen as a van der Waals gas, $a=0.137\ \mathrm{Pa\,m^6/mol^2}$ , $b=3.87\times10^{-5}\ \mathrm{m^3/mol}$ , with $C_P=29.1\ \mathrm{J/(mol\,K)}$ , at $T=300$ K.

Step 1. Find the inversion temperature ceiling: $T_{\text{inv}}^{\max}=2a/Rb=2(0.137)/[(8.314)(3.87\times10^{-5})]=\mathbf{852\ K}$ (the experimental maximum is $\approx621$ K — van der Waals overestimates, but the point stands). Since $300\ \mathrm{K}<T_{\text{inv}}$ , we are inside the dome: nitrogen cools.

Step 2. Estimate the coefficient with the low-density formula $\mu_{JT}\approx\dfrac{1}{C_P}\Big(\dfrac{2a}{RT}-b\Big)$ . Compute $\dfrac{2a}{RT}=\dfrac{0.274}{(8.314)(300)}=1.099\times10^{-4}\ \mathrm{m^3/mol}$ ; subtract $b$ : $1.099\times10^{-4}-3.87\times10^{-5}=7.12\times10^{-5}\ \mathrm{m^3/mol}$ .

Step 3. Divide by $C_P$ : $\mu_{JT}=7.12\times10^{-5}/29.1=2.45\times10^{-6}\ \mathrm{K/Pa}=\mathbf{0.245\ K/bar}$ , positive (cooling). The measured value near room temperature is about $0.22$ K/bar — the simple model lands within ten percent. The takeaway: a $200$ -bar throttling of room-temperature nitrogen drops its temperature by roughly $200\times0.22\approx44$ K in one pass — enough that, recycled through a counter-current heat exchanger, it cascades all the way to liquefaction.

Example 2 — Why must helium be pre-cooled? Take helium, $a=0.00346\ \mathrm{Pa\,m^6/mol^2}$ , $b=2.38\times10^{-5}\ \mathrm{m^3/mol}$ , and ask what happens at room temperature.

Step 1. Inversion ceiling: $T_{\text{inv}}^{\max}=2a/Rb=2(0.00346)/[(8.314)(2.38\times10^{-5})]=\mathbf{35\ K}$ (experiment $\approx40$ K). At $300$ K we are far above the inversion temperature.

Step 2. Sign of $\mu_{JT}$ : the bracket $2a/RT-b$ at $300$ K is $2(0.00346)/[(8.314)(300)]-2.38\times10^{-5}=2.77\times10^{-6}-2.38\times10^{-5}<0$ , so $\mu_{JT}<0$ : room-temperature helium warms on throttling.

Step 3. Conclusion: a helium liquefier cannot start with a plain valve at room temperature; the gas must first be pre-cooled below $\sim40$ K (historically with liquid hydrogen, in modern machines with an expansion engine — the Collins cycle) so that throttling finally cools rather than heats. The takeaways: the inversion temperature is decisive — the same formula $2a/Rb$ explains why air liquefiers throttle directly (N₂: ceiling far above room T) and why helium liquefiers need a pre-cooling stage (He: ceiling only ≈40 K).

Origins · 1 mininversion temperature and how this was discovered

inversion temperature

concept · The temperature that decides whether throttling cools or heats a real gas

The inversion temperature $T_{\text{inv}}$ is the temperature at which the Joule–Thomson coefficient changes sign: below it a throttled gas cools, above it the gas warms, and at it $\mu_{JT}=0$ . The effect was discovered by James Prescott Joule and William Thomson (later Lord Kelvin) in their porous-plug experiments of the 1850s, undertaken to test whether real gases departed from the ideal-gas law that Joule's earlier free-expansion work had seemed to support. The van der Waals model gives the clean prediction $T_{\text{inv}}^{\max}=2a/Rb=\tfrac{27}{4}T_c$ , a parameter-free multiple of the critical temperature. That most common gases have inversion temperatures far above room temperature — while hydrogen and helium do not — is precisely why room-temperature air can be liquefied by throttling but hydrogen and helium must first be pre-cooled below their inversion temperatures.

History

1845

— Joule's free-expansion experiment finds no temperature change for a near-ideal gas

1852

— Joule and William Thomson begin the porous-plug (throttling) experiments, published through 1862

1862

— Joule and Thomson report the small cooling of air and the warming of hydrogen

1895

— Carl von Linde and, independently, William Hampson patent air liquefaction by Joule–Thomson throttling with counter-current heat exchange

1898

— James Dewar liquefies hydrogen, pre-cooling it below its inversion temperature

1908

— Heike Kamerlingh Onnes liquefies helium, using a Joule–Thomson final stage after hydrogen pre-cooling

Going further

References. Reif, Fundamentals of Statistical and Thermal Physics, §5.10 (Joule–Thomson expansion) · Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed., Ch. 6 · Zemansky & Dittman, Heat and Thermodynamics, Ch. 11 · Joule & Thomson, "On the Thermal Effects of Fluids in Motion," Phil. Trans. R. Soc. (1853–1862). The key results of this article are: the flow-work identity $\Delta U=P_1V_1-P_2V_2\Rightarrow H_1=H_2$ ; $(\partial H/\partial P)_T=V-T(\partial V/\partial T)_P$ from the Maxwell relation; the ideal-gas result $T(\partial V/\partial T)_P-V=0\Rightarrow\mu_{JT}=0$ ; the van der Waals second virial coefficient $B=b-a/RT$ and bracket $2a/RT-b$ ; $T_{\text{inv}}^{\max}=2a/Rb$ and the exact ratio $T_{\text{inv}}^{\max}/T_c=27/4$ ; and the numerical values $T_{\text{inv}}^{\max}(\mathrm{N_2})=852$ K, $T_c(\mathrm{N_2})=126$ K, $T_{\text{inv}}^{\max}(\mathrm{He})=35$ K, and $\mu_{JT}(\mathrm{N_2},300\,\mathrm{K})=0.245$ K/bar.

Builds on: Helmholtz Free Energy, Enthalpy, and Gibbs Free Energy · Maxwell Relations · Van der Waals Equation of State. Leads to: Adiabatic Demagnetization and Magnetic Cooling · Linde–Hampson liquefaction cycle · Helium cryogenics and the Collins cycle.

Why is throttling isenthalpic rather than adiabatic-reversible?

Both processes are adiabatic in the sense that no heat crosses the walls, but they differ in reversibility and in what is conserved. Reversible adiabatic expansion (a gas smoothly pushing a piston) conserves entropy and always cools, because the gas does external work. Throttling forces the gas turbulently through a plug: it is irreversible, produces entropy, and does no net useful work — the upstream flow work $P_1V_1$ is spent making room downstream as $P_2V_2$ . Accounting for exactly those flow-work terms shows the internal energy plus $PV$ is conserved, i.e. the enthalpy. So the constraint is constant $H$ , not constant $S$ , and that is why an ideal gas cools in the reversible case but not when throttled.

Physically, why does a real gas cool when it expands through the valve?

Below the inversion temperature the intermolecular attraction dominates. As the gas expands, the average distance between molecules grows, so the (negative) potential energy of attraction increases toward zero — work must be done against the attractive forces. Because the process is isenthalpic and does no external work to draw on, that energy is taken from the molecules' kinetic energy, and lower kinetic energy means lower temperature. Above the inversion temperature the fast molecules spend most of their time in hard, repulsive near-collisions; expanding then releases repulsive potential energy into kinetic energy, and the gas warms. The inversion temperature is where these two contributions cancel.

Why does the effect vanish for an ideal gas?

An ideal gas has no intermolecular forces and no molecular volume, so its internal energy depends only on temperature and its enthalpy $H=U+PV=U+RT$ is a function of $T$ alone. Holding $H$ fixed then holds $T$ fixed — no temperature change is possible. Equivalently, in the master formula $\mu_{JT}=(1/C_P)[T(\partial V/\partial T)_P-V]$ , the ideal-gas law $V=RT/P$ gives $T(\partial V/\partial T)_P=T\cdot R/P=V$ , so the bracket is exactly zero. The Joule–Thomson effect is therefore a direct measurement that a real gas is not ideal.

What sets the inversion temperature, and why is it so different for helium?

To leading order in density the inversion condition is $2a/RT_{\text{inv}}=b$ , i.e. $T_{\text{inv}}=2a/Rb$ , so it is set by the ratio of the attraction parameter $a$ to the excluded-volume parameter $b$ . Helium is a tiny, weakly interacting atom: its $a$ is minuscule (weak van der Waals attraction), so $2a/Rb$ is small — only about $40$ K experimentally. Hydrogen is similar (inversion near $200$ K). Nitrogen and oxygen have much larger $a$ , pushing their inversion temperatures well above room temperature. That is why air liquefiers throttle directly, while helium and hydrogen liquefiers must first pre-cool the gas below its inversion temperature.

Is the van der Waals prediction

T_{\text{inv}}^{\max}=6.75\,T_c

accurate?

The ratio is a clean corresponding-states prediction and gets the right order of magnitude, but van der Waals is only a qualitative model. For nitrogen it gives $T_{\text{inv}}^{\max}\approx852$ K against a measured maximum near $621$ K — an overestimate of about a third. Real gases are better described by the measured second virial coefficient $B(T)$ and its temperature derivative, since $\mu_{JT}\propto T\,dB/dT-B$ ; the full experimental inversion curve is a closed loop that van der Waals reproduces only roughly. The lesson to keep is structural: there is an inversion temperature, it is several times $T_c$ , and it decides the sign of the cooling.

How does a real liquefier turn a fraction of a kelvin per bar into liquefaction?

A single throttle drops the temperature only modestly — a fraction of a kelvin per bar. The trick is regeneration: the cooled, low-pressure gas leaving the throttle is sent back through a counter-current heat exchanger to pre-cool the incoming high-pressure stream before it reaches the valve. Each cycle therefore starts colder than the last, and the temperature cascades downward until part of the stream liquefies (the Linde–Hampson cycle). For gases whose inversion temperature is below ambient, such as helium, an expansion engine or turbine provides the initial cooling to get under the inversion temperature before the Joule–Thomson stage takes over.