Joule-Thomson Expansion

Picture a rigid, thermally insulated tube divided by a porous plug or a narrow throttle valve. Gas at pressure $p_1$ and temperature $T_1$ sits upstream; gas emerges downstream at a lower pressure $p_2$ and some new temperature $T_2$. No heat crosses the insulated walls, and no shaft work is extracted — the only work done is the flow work of pushing gas through the plug. This is fundamentally different from the free Joule expansion into a vacuum, where the prerequisite result is that an ideal gas undergoes no temperature change because no external work is done and no attractive forces exist to overcome. In throttling, the gas does do net work on itself, and that work is done against intermolecular forces if any exist. Consider a fixed amount of gas, one mole, pushed steadily through the plug. Upstream, a piston-like action does work $p_1V_1$ pushing the gas in; downstream, the gas pushes with work $p_2V_2$ expanding into the low-pressure region. Since the tube is insulated and there is no other heat or work exchange, the First Law applied to this steady-flow segment gives conservation of the sum $U + pV$, which is enthalpy. The process is isenthalpic: $H_1 = H_2$, even though pressure and volume both change. Because $H$ is conserved rather than $T$, any temperature change reflects how enthalpy is partitioned between $U$ and $pV$ as pressure drops. For an ideal gas, $H$ depends only on $T$, so constant $H$ forces constant $T$: no temperature change at all. For a real gas, $H$ depends on both $T$ and $p$ because intermolecular potential energy contributes to $U$. Dropping the pressure changes the average intermolecular separation, which changes the potential energy contribution to $U$, and the temperature must shift to keep $H$ fixed. Physically: if attractive forces dominate (gas well below its Boyle-like regime), pulling molecules apart during expansion costs energy, which is drawn from kinetic energy — the gas cools. If repulsive forces dominate (gas already very compressed, or at high temperature where kinetic energies swamp attractive wells), expansion releases stored repulsive potential energy into kinetic energy — the gas warms.

where: $H_1, H_2$ are the molar (or total) enthalpies upstream and downstream of the plug.

where: $\mu_{JT}$ is the Joule-Thomson coefficient, the slope of an isenthalp on a $T$–$p$ diagram, typically in K atm⁻¹ or K bar⁻¹.

where: $C_p$ is the constant-pressure heat capacity, always positive, so the sign of $\mu_{JT}$ is opposite the sign of $(\partial H/\partial p)_T$.

where: $V$ is molar volume; this form uses only $C_p$ and the equation of state, both experimentally accessible.

where: $a, b$ are van der Waals constants; this predicts $\mu_{JT} = 0$ at the inversion temperature $T_{inv} = 2a/(Rb)$.

Example 1 — Foundational: computing μ_JT from tabulated data for nitrogen. At 300 K and 1 atm, nitrogen gas has $C_{p,m} = 29.1\ \text{J K}^{-1}\text{mol}^{-1}$. Experimental data show that near this state, throttling nitrogen from 20 atm to 1 atm at an average temperature of 300 K produces a temperature drop of $\Delta T = -21.5$ K per 19 atm drop (a standard tabulated result consistent with $\mu_{JT}(N_2, 300\text{ K}) \approx 0.25$ K atm⁻¹). Compute the expected temperature change for a smaller, controlled throttle from $p_1 = 5.00$ atm to $p_2 = 4.00$ atm at 300 K, assuming $\mu_{JT}$ is roughly constant over this narrow interval. Step 1. Use the defining relation in finite-difference form, valid since $\Delta p$ is small compared to the pressure range over which $\mu_{JT}$ varies appreciably:

Step 2. Insert numbers, with $\Delta p = p_2 - p_1 = 4.00 - 5.00 = -1.00$ atm:

Step 3. Sanity check. The pressure drop is small (1 atm out of 5), so linear extrapolation of $\mu_{JT}$ is reasonable, and the sign is correct: nitrogen at room temperature is well below its inversion temperature (621 K, from Table data with $a = 1.408$ atm L² mol⁻², $b = 0.03913$ L mol⁻¹, giving $T_{inv} = 2a/(Rb) = 2(1.408)/(0.08206 \times 0.03913) \approx 876$ K by the crude van der Waals estimate — real $N_2$'s measured inversion temperature is closer to 621 K because the low-density van der Waals approximation overestimates $T_{inv}$). Either way, $T \ll T_{inv}$, so cooling on expansion, $\mu_{JT} > 0$, is expected. The predicted $-0.25$ K cooling for a 1 atm drop is small and physically sensible — throttling refrigerators use multi-atmosphere pressure drops precisely because the per-atmosphere effect is modest.

Example 2 — Applied: van der Waals estimate and liquefaction feasibility for CO₂ versus H₂. Use the low-density van der Waals formula to estimate $\mu_{JT}$ and $T_{inv}$ for carbon dioxide and hydrogen, and decide whether room-temperature throttling can cool each gas. Data: for CO₂, $a = 3.640$ atm L² mol⁻², $b = 0.04267$ L mol⁻¹, $C_{p,m} \approx 37.1$ J K⁻¹mol⁻¹ near 300 K. For H₂, $a = 0.2476$ atm L² mol⁻², $b = 0.02661$ L mol⁻¹, $C_{p,m} \approx 28.8$ J K⁻¹mol⁻¹ near 300 K. Use $R = 0.08206$ L atm K⁻¹ mol⁻¹ for the van der Waals terms and $R = 8.314$ J K⁻¹mol⁻¹ for energy units. Step 1. Inversion temperatures:

Step 2. Interpretation at $T = 300$ K. For CO₂, $300\ \text{K} \ll 2079\ \text{K}$, so CO₂ sits deep in its cooling regime; $\mu_{JT} > 0$ strongly. For H₂, $300\ \text{K} > 227\ \text{K}$, so room-temperature hydrogen sits above its inversion temperature; $\mu_{JT} < 0$ and hydrogen warms on throttling at 300 K. Step 3. Quantify $\mu_{JT}$ at 300 K for each gas. Convert $a$ from atm L² mol⁻² to SI (J m³ mol⁻²) is unnecessary if we keep consistent units: work in L·atm for the bracket, then convert to J using $1\ \text{L atm} = 101.325$ J. For CO₂:

Convert to J mol⁻¹ atm⁻¹... actually the bracket has units of volume (L mol⁻¹); dividing by $C_p$ (J K⁻¹mol⁻¹) requires converting the bracket to J atm⁻¹ mol⁻¹ using $1\ \text{L atm} = 101.325$ J:

This is close to the accepted experimental value near 1.1 K atm⁻¹ at 300 K (the low-density van der Waals estimate is approximate, but the sign and order of magnitude are correct). For H₂:

Convert: $-0.00649 \times 101.325 = -0.658\ \text{J atm}^{-1}\text{mol}^{-1}$.

Step 4. Sanity check and conclusion. Both signs match qualitative prediction: CO₂ cools substantially (about 0.7-1 K per atmosphere dropped, consistent with why CO₂ fire extinguishers frost heavily on discharge), while hydrogen very slightly warms per atmosphere at 300 K, consistent with the historical fact that hydrogen and helium must first be pre-cooled below their (low) inversion temperatures before Joule-Thomson throttling can liquefy them — the technological puzzle the next article resolves in detail.

James Prescott Joule had already established, with his free-expansion experiment, that an ideal gas shows no cooling when expanded without doing external work — a crucial but frustratingly inconclusive test for real gases, since the heat capacity of the surrounding water bath was far larger than that of the gas itself, masking any small real-gas effect. Joule then collaborated with William Thomson (Lord Kelvin), who had been developing an absolute temperature scale and needed a more sensitive way to detect deviations from ideal-gas behavior. Together they redesigned the experiment: instead of free expansion into vacuum, they forced gas continuously through a porous plug under steady pressures on each side, measuring the small but reproducible temperature change directly in the flowing gas rather than in a bath. This porous-plug experiment, reported in the 1850s, gave the first clean, quantitative evidence of intermolecular forces in real gases and directly enabled the practical liquefaction technology that followed within decades.

Ask the tutor: How does the inversion curve of a real gas look on a full T-p diagram, and why does it eventually close at high pressure?