Liquefaction of gases — Linde, Claude processes (concept): Visual walkthrough

Step 1 — What "throttling" physically is

WHAT. Picture a horizontal insulated tube. In the middle sits a porous plug (imagine cotton wool). On the left, a piston pushes gas in at high pressure $P_{1}$ ; on the right, a piston lets gas out at low pressure $P_{2}$ , with $P_{2} < P_{1}$ .

WHY this setup. We want the gas to expand without pushing any machinery on the outside and without touching a hot or cold object. The plug does exactly that: it just resists flow. So whatever temperature change appears cannot be blamed on heat or on running an engine — it must come from inside the gas. That is what makes throttling the clean experiment.

PICTURE. Look at the two pistons squeezing gas through the grey plug — left side red-hot pressure, right side blue low pressure.

Figure — Liquefaction of gases — Linde, Claude processes (concept)

Here $P$ is pressure (push per unit area), $V$ is volume (space the gas fills), $T$ is temperature (a measure of how fast molecules jiggle). Subscript $1$ = before the plug, subscript $2$ = after.

Step 2 — Bookkeeping the energy: why enthalpy is constant

WHAT. Push the whole left-side blob of gas (volume $V_{1}$ ) through the plug. The left piston does work on the gas; the right piston has work done on it by the gas as the gas swells out to $V_{2}$ .

WHY. The First law of thermodynamics says $Δ U = q + w$ : change in internal energy = heat in + work done on the gas. The tube is insulated, so $q = 0$ , and (per the assumptions above) there is no bulk kinetic-energy term to carry. Now add up the work.

Left piston pushes at constant $P_{1}$ , sweeping volume $V_{1}$ to zero: work done on the gas $= + P_{1} V_{1}$ .
Right piston is shoved out at constant $P_{2}$ , gas grows from $0$ to $V_{2}$ : work done on the gas $= - P_{2} V_{2}$ .

$Δ U = U_{2} - U_{1} = left pushes in P_{1} V_{1} gas pushes out - P_{2} V_{2}$

Rearrange so "before" sits on one side, "after" on the other:

$U_{1} + P_{1} V_{1} = U_{2} + P_{2} V_{2}$

PICTURE. The green area on the left ( $P_{1} V_{1}$ ) is the work fed in; the blue area on the right ( $P_{2} V_{2}$ ) is the work spent. Watch the left area collapse as the right area opens up.

The boxed equation is exactly $H_{1} = H_{2}$ . See the linked idea Enthalpy and constant-H processes.

Step 3 — The question we actually want to answer

WHAT. We do not care about $H$ for its own sake — we care about temperature. So we ask: as the pressure falls by a little bit $d P$ (which is negative, since $P$ drops), while keeping $H$ fixed, how much does $T$ change?

WHY a derivative, and why this specific one. A derivative answers "for a tiny nudge in one thing, how much does another thing respond?" — it is the rate of response. We want the rate at which $T$ responds to $P$ , but only along the special path where $H$ stays constant (because that is the path throttling actually follows). That constraint is written as a subscript $H$ .

PICTURE. On a temperature–pressure graph, draw a curve of constant $H$ (an "isenthalp"). $μ_{J T}$ is just its slope. Where the slope is positive, sliding left (lower $P$ ) means sliding down (lower $T$ ) → cooling.

Step 4 — Rewriting the slope into measurable pieces

WHAT. The definition $(\partial T / \partial P)_{H}$ is honest but useless in the lab — nobody has a "hold $H$ constant" knob. We convert it into quantities we can measure: how volume swells with temperature, and heat capacity.

WHY — the cyclic relation, explained. On the constant- $H$ curve, both $T$ and $P$ move together. Walk a tiny closed loop on the $H$ -surface: nudge $P$ a bit, then $T$ a bit, so that $H$ returns to its start ( $d H = 0$ ). Writing " $H$ changed by zero" in terms of its two response rates, $d H = (\frac{\partial H}{\partial P})_{T} d P + (\frac{\partial H}{\partial T})_{P} d T = 0.$ Each symbol: $(\partial H / \partial P)_{T}$ = how $H$ reacts to a pressure nudge at fixed $T$ ; $(\partial H / \partial T)_{P}$ = how $H$ reacts to a temperature nudge at fixed $P$ . Setting the sum to zero and solving for $d T / d P$ (the constant- $H$ slope) gives exactly: $(\frac{\partial T}{\partial P})_{H} = - \frac{( \partial H / \partial P ) _{T}}{( \partial H / \partial T ) _{P}}$ This is just "the two ways $H$ can change must cancel."

The denominator is the heat capacity at constant pressure: $(\partial H / \partial T)_{P} = C_{p}$ — "how much enthalpy you must pour in to warm the gas by one degree at fixed $P$ ." So: $μ_{J T} = - \frac{1}{C _{p}} (\frac{\partial H}{\partial P})_{T}$

Now the numerator — algebraic walkthrough. Start from the exact differential of enthalpy (combining the first and second laws), $d H = T d S + V d P$ , where $S$ is entropy. Hold $T$ fixed and divide by $d P$ : $(\frac{\partial H}{\partial P})_{T} = T (\frac{\partial S}{\partial P})_{T} + V .$ The awkward entropy term is removed by a Maxwell relation — one of four bridges that say two mixed second-derivatives of a state function are equal. The relevant one is $(\partial S / \partial P)_{T} = - (\partial V / \partial T)_{P}$ (measurable! it is just thermal expansion). Substituting: $(\frac{\partial H}{\partial P})_{T} = V - T (\frac{\partial V}{\partial T})_{P} .$

PICTURE. Two competing quantities on a bar chart: the "thermal-expansion push" $T (\partial V / \partial T)_{P}$ versus the "current size" $V$ . Whichever bar is taller wins; their difference (scaled by $1/ C_{p}$ ) is $μ_{J T}$ .

Step 5 — The ideal gas: the "control experiment" that gives zero

WHAT. Plug an ideal gas into the bracket. From $P V = n R T$ (with $n$ = moles and $R$ = gas constant, defined at the top) we get $V = n R T / P$ , so at fixed $P$ : $(\frac{\partial V}{\partial T})_{P} = \frac{n R}{P}$

WHY do this first. An ideal gas has no intermolecular attractions. It is our baseline: whatever the ideal gas does is not the interesting real-gas physics. Put it in the bracket: $T \cdot \frac{n R}{P} - V = T \cdot \frac{n R}{P} - \frac{n R T}{P} = 0$

Every term cancels exactly. So:

WHY this is the whole point. The cooling that liquefies real air is not "gases cool when they expand." It is a defect — the fact that real molecules attract each other (the van der Waals $a$ ; see van der Waals equation and Intermolecular forces). Ideal $\Rightarrow$ zero is the proof.

PICTURE. The two bars from Step 4, now exactly equal height — they cancel to a flat zero line. The isenthalp is perfectly horizontal.

Step 6 — The real gas: where the attractions tip the balance

WHAT. For a real gas the molecules pull on each other. When you warm it at fixed pressure, it expands more eagerly than an ideal gas would (attractions relax), so $T (\partial V / \partial T)_{P}$ can grow larger than $V$ . The bracket goes positive.

WHY — expanding the van der Waals equation at low pressure. The van der Waals equation for one mole is $(P + \frac{a}{V ^{2}}) (V - b) = R T,$ where $a$ measures attraction and $b$ the molecular volume. At low pressure the gas is nearly ideal, so $V \approx R T / P$ is large; then $b / V ≪ 1$ and $a / (V^{2})$ is small. Multiplying out and keeping only the leading correction gives $V \approx \frac{R T}{P} + b - \frac{a}{R T} .$ Differentiate at fixed $P$ (note $b$ is constant and $a / R T$ depends on $T$ ): $(\frac{\partial V}{\partial T})_{P} \approx \frac{R}{P} + \frac{a}{R T ^{2}} .$ Now build the bracket $T (\partial V / \partial T)_{P} - V$ : $T (\frac{R}{P} + \frac{a}{R T ^{2}}) - (\frac{R T}{P} + b - \frac{a}{R T}) = \frac{a}{R T} - b + \frac{a}{R T} = \frac{2 a}{R T} - b .$ (The two $R T / P$ terms cancel — the ideal part vanishes, exactly as Step 5 promised.)

PICTURE. A tug-of-war: the yellow attraction term $2 a / R T$ against the red size term $b$ . Cold on the left (attraction wins, cooling), hot on the right (size wins, heating).

This is precisely the connection to Real gases and compressibility factor Z: the same $a$ -vs- $b$ competition that makes $Z \neq = 1$ is what makes $μ_{J T} \neq = 0$ .

Step 7 — The turning point: the inversion temperature

WHAT. The exact border between cooling and heating is where the bracket equals zero: $\frac{2 a}{R T _{i}} - b = 0 ⟹ T_{i} = \frac{2 a}{R b}$

WHY it matters — the degenerate cases.

$N_{2}$ and $O_{2}$ : $T_{i}$ is well above room temperature, so throttling at 300 K already cools them.
$H_{2}$ ( $T_{i} \approx 200$ K) and He ( $T_{i} \approx 40$ K): room temperature is above $T_{i}$ , so a naive throttle heats them. You must pre-cool below $T_{i}$ first (e.g. with liquid $N_{2}$ ), then throttling starts cooling. This is the single most-missed edge case.

PICTURE. The $μ_{J T}$ -versus- $T$ curve crossing zero at $T_{i}$ : green "cooling" region on the left, red "heating" region on the right, with $N_{2}$ (real constants) and $H_{2}$ (its own small $a, b$ ) marked at 300 K.

The one-picture summary

Everything above collapses into a single decision chart: insulated leak ⇒ constant $H$ ⇒ compare $T (\partial V / \partial T)_{P}$ with $V$ ⇒ sign of the bracket ⇒ cool or heat, gated by $T_{i} = 2 a / R b$ .

Recall Feynman retelling of the whole walkthrough

We shove gas through a plug in an insulated tube. Because no heat sneaks in, the flow is slow, and the pistons' work exactly cancels into $H_{1} = H_{2}$ , the enthalpy rides across unchanged. We then ask: on that constant- $H$ path, does temperature fall? Calculus turns that hard question into a fair fight between two numbers — how eagerly the gas expands when warmed, $T (\partial V / \partial T)_{P}$ , versus how big it already is, $V$ . For a perfect gas these tie exactly, so nothing happens. But real molecules are sticky (the $a$ in van der Waals), which makes the expansion term win when it's cold — the gas steals energy from itself to pull its sticky molecules apart, and cools. Warm it too much and molecular bulkiness (the $b$ ) takes over and it heats instead. The exact crossover is $T_{i} = 2 a / R b$ . That is why nitrogen throttles cold at room temperature but hydrogen and helium must be pre-chilled first — and it is the tiny self-cooling that the Linde machine stacks over and over until air drips out as liquid.

Recall

Why is throttling a constant-enthalpy process? ::: Insulated ⇒ $q = 0$ ; slow flow ⇒ no bulk kinetic-energy term; the work in ( $P_{1} V_{1}$ ) minus work out ( $P_{2} V_{2}$ ) gives $U_{1} + P_{1} V_{1} = U_{2} + P_{2} V_{2}$ , i.e. $H_{1} = H_{2}$ . What is $μ_{J T}$ for an ideal gas, and why? ::: Exactly $0$ ; the bracket $T (\partial V / \partial T)_{P} - V = T \cdot n R / P - n R T / P$ cancels because there are no attractions. Below or above $T_{i}$ do you get cooling? ::: Below $T_{i}$ (bracket positive, $μ_{J T} > 0$ , so $d P < 0$ gives $d T < 0$ ). What are the two competing terms in the real-gas bracket? ::: $2 a / R T$ (attraction, favours cooling) versus $b$ (molecular size, favours heating).

Parent: 2.4.10 Liquefaction of gases — Linde, Claude processes (concept) (Hinglish) · see also Critical constants ($T_c, P_c, V_c$).