Visual walkthrough — Critical constants Tc, Pc, Vc; law of corresponding states

Prerequisites we lean on: the Van der Waals Equation of State, Real Gases and Compressibility Factor Z, Liquefaction of Gases and Andrews Isotherms, and the plain Ideal Gas Equation.

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Step 1 — Draw the thing we are studying: a $P$–$V$ isotherm

WHAT. Fix the temperature $T$ of one mole of gas. Slowly squeeze it: reduce the volume $V$ and watch the pressure $P$. Plot $P$ (up) against $V$ (across). One temperature gives one curve — an isotherm (iso = same, therm = heat).

WHY. Liquefaction is a shape event on this plot. A gas is a smoothly falling curve. A liquid forming shows up as a flat plateau (pressure stops changing while volume shrinks — gas is turning to liquid). So the whole story of "can we make a liquid?" is "does the curve have a flat plateau?".

PICTURE. Below, three curves at three temperatures.

• Bottom (blue) curve, $T<T_c$: a wide plateau — gas and liquid coexist here.
• Middle (yellow) curve, $T=T_c$: the plateau has shrunk to a single kink point.
• Top (pink) curve, $T>T_c$: no flat part at all — pressing only ever gives a dense gas, never a liquid.

The special middle curve, and the special point on it, is what we chase.

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Step 2 — Give the curve an equation: van der Waals, solved for $P$

WHAT. We need a formula for that curve. The Van der Waals Equation of State for one mole is $\left(P+\frac{a}{V^2}\right)(V-b)=RT.$ Solve for $P$ so we have "height as a function of across": $P=\underset{\text{repulsion / pressure push}}{\underbrace{\frac{RT}{V-b}}}-\underset{\text{attraction pull-down}}{\underbrace{\frac{a}{V^2}}}.$

Term by term:

• $R$ = gas constant (a fixed number, $0.0821{\text{L atm mol}}^{-1}{\text{K}}^{-1}$), $T$ = temperature — together $RT$ sets the overall "push".
• $b$ = the volume the molecules themselves occupy, so $V-b$ = the free space left to move in. Making the denominator $V-b$ (not just $V$) is what stops the pressure staying finite as we squeeze — see the wall.
• $a$ = strength of intermolecular attraction; the $-a/V^2$ term pulls the pressure down because molecules tugging each other inward hit the walls less hard.

WHY split it this way. The two terms fight. $\frac{RT}{V-b}$ shoots up as $V\to b$ (a wall). $-\frac{a}{V^2}$ dives down as $V$ gets small. When these two fights nearly balance, the curve can go momentarily flat — and that flatness is the plateau. So the plateau is literally a tug-of-war between $a$ and $b$.

PICTURE. The two pieces and their sum.

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Step 3 — Say "flat" and "straight" in symbols: the two derivative conditions

WHAT. We want the temperature and place where the plateau has shrunk to a single point. That point is where the curve does two things at once:

1. Its tangent is horizontal (the leftover of the flat plateau).
2. It changes from concave-down to concave-up — an inflection, i.e. it is momentarily straight.

WHY a derivative? A derivative $\frac{\partial P}{\partial V}$ is just the slope of the curve — "how much does height change per step across". A slope of $0$ means flat. The second derivative $\frac{{\partial }^{2}P}{\partial V^2}$ measures how the slope itself is changing — the bending (curvature). Curvature $0$ means not bending — momentarily straight. This is exactly the tool we need because we are describing a shape, and derivatives are the language of slope and bend. No other tool reads "flat + straight" off a curve so directly.

${\left(\frac{\partial P}{\partial V}\right)}_T=0\text{(flat)},{\left(\frac{{\partial }^{2}P}{\partial V^2}\right)}_T=0\text{(straight)}.$

PICTURE. Watch the tangent line and the curvature as $T$ rises toward $T_c$.

• $T<T_c$: the curve dips and rises — the slope goes negative then positive, so somewhere it's zero, but it's still bending (a real min/max, not straight).
• $T=T_c$: the dip flattens out completely — slope zero and no bend. Both conditions true at the same $V$. That is the critical point.

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Step 4 — Do the two derivatives

WHAT. Differentiate $P=\frac{RT}{V-b}-\frac{a}{V^2}$ with respect to $V$, twice, and set each to zero.

First derivative (slope): $\frac{\partial P}{\partial V}=\underset{\text{push slope, negative}}{\underbrace{-\frac{RT}{(V-b{)}^2}}}+\underset{\text{pull slope, positive}}{\underbrace{\frac{2a}{V^3}}}=0\Rightarrow \frac{RT}{(V-b{)}^2}=\frac{2a}{V^3}.(1)$

Second derivative (curvature): $\frac{{\partial }^{2}P}{\partial V^2}=\underset{\text{curls up}}{\underbrace{\frac{2RT}{(V-b{)}^3}}}-\underset{\text{curls down}}{\underbrace{\frac{6a}{V^4}}}=0\Rightarrow \frac{2RT}{(V-b{)}^3}=\frac{6a}{V^4}.(2)$

WHY. Each derivative just applies the power rule to the two terms of Step 2. The signs tell the story: in $(1)$ the push pulls the slope negative and the pull pushes it positive — they cancel at the flat spot. In $(2)$ the two curvatures cancel at the straight spot.

PICTURE. The two conditions as two curves crossing zero.

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Step 5 — Divide equation (2) by (1): kill $R,T,a$, harvest $V_c$

WHAT. Divide the second condition by the first. Almost everything cancels:

\;\Rightarrow\; \frac{2}{V-b} = \frac{3}{V}.$$ **Term by term:** on the left $R,T$ cancel and one power of $(V-b)$ survives; on the right $a$ cancels and one power of $V$ survives. The *physics constants vanish* — what's left is **pure geometry** of where the curve straightens. Cross-multiply: $$2V = 3(V-b) \;\Rightarrow\; 2V = 3V - 3b \;\Rightarrow\; \boxed{V_c = 3b}.$$ **WHY divide?** Because dividing is the cleanest way to annihilate the messy shared factors ($R$, $T$, $a$) and isolate the *location* $V$. The critical volume is **three times the excluded volume $b$** — not equal to it. (See the [!mistake] in the parent: molecules can't be crushed down to $b$.) **PICTURE.** The critical point sits at $V_c=3b$, comfortably to the right of the "wall" at $V=b$. ![[deepdives/dd-chemistry-2.4.09-d2-s05.png]] --- ## Step 6 — Back-substitute $V_c=3b$ into (1): harvest $T_c$ **WHAT.** Put $V_c=3b$ into condition $(1)$. Then $V_c-b = 3b-b = 2b$: $$\frac{RT_c}{(2b)^2} = \frac{2a}{(3b)^3} \;\Rightarrow\; \frac{RT_c}{4b^2} = \frac{2a}{27b^3}.$$ Multiply both sides by $4b^2$: $$RT_c = \frac{2a}{27b^3}\cdot 4b^2 = \frac{8a}{27b} \;\Rightarrow\; \boxed{T_c = \frac{8a}{27Rb}}.$$ **Term by term:** $(2b)^2 = 4b^2$ is the free-space term squared; $(3b)^3=27b^3$ is the full volume cubed. The clean numbers $4, 27, 8$ are just what these powers of $2$ and $3$ produce. **WHY.** With the *place* $V_c$ known, condition $(1)$ turns into a plain equation for the one thing left — the *temperature* $T_c$. Bigger attraction $a$ ⇒ higher $T_c$ (harder-attracting gases liquefy more easily, up to a higher ceiling). Bigger $b$ ⇒ lower $T_c$. **PICTURE.** $T_c$ is the temperature of the one isotherm whose kink is the critical point. ![[deepdives/dd-chemistry-2.4.09-d2-s06.png]] --- ## Step 7 — Plug $V_c, T_c$ into the equation of state: harvest $P_c$ **WHAT.** Read the *height* of the critical point straight off the curve $P = \dfrac{RT}{V-b} - \dfrac{a}{V^2}$ using $V=V_c=3b$, $T=T_c=\tfrac{8a}{27Rb}$: $$P_c = \frac{R\cdot \frac{8a}{27Rb}}{2b} - \frac{a}{(3b)^2} = \frac{8a}{27b\cdot 2b} - \frac{a}{9b^2} = \frac{8a}{54b^2} - \frac{a}{9b^2}.$$ Common denominator $27b^2$: $$P_c = \frac{4a}{27b^2} - \frac{3a}{27b^2} = \boxed{\frac{a}{27b^2}}.$$ **Term by term:** $R$ cancels inside the first fraction (an $R$ up top, an $R$ hidden in $T_c$), $\frac{8a}{54b^2}=\frac{4a}{27b^2}$, and $\frac{a}{9b^2}=\frac{3a}{27b^2}$. The subtraction leaves a single $a$ over $27b^2$. **WHY.** $V_c$ was the *where*, $T_c$ was the *which curve*; $P_c$ is simply the *height* of the point once both are pinned. Nothing new — just evaluation. **PICTURE.** All three constants marked on the one critical isotherm. ![[deepdives/dd-chemistry-2.4.09-d2-s07.png]] --- ## Step 8 — The degenerate check: what happens for $T>T_c$ and $T<T_c$? **WHAT.** We *assumed* a point where slope and curvature both vanish exists. Does it always? No — only exactly at $T=T_c$. - **$T<T_c$:** the isotherm has a real wiggle. $\partial P/\partial V$ is zero at **two** places (a local min and a local max of the van der Waals loop). The physical plateau (from an [[Liquefaction of Gases and Andrews Isotherms|Andrews isotherm]]) replaces this wiggle. Liquefaction is possible. - **$T=T_c$:** the two zeros of the slope **merge** into one — the inflection with horizontal tangent. Exactly our derivation. This merging is *why* we needed *both* derivatives to be zero: a double root. - **$T>T_c$:** $\partial P/\partial V$ is **never** zero — the curve is monotonically falling. No flat spot, no plateau, no liquid. You get a [[Supercritical Fluids|supercritical fluid]] instead. **WHY it matters.** The condition "first *and* second derivative zero" is precisely the algebraic fingerprint of "two turning points colliding into one." That collision *is* the critical point. Below, they haven't collided; above, they never existed. **PICTURE.** The two turning points sliding together as $T\to T_c$, then vanishing above. ![[deepdives/dd-chemistry-2.4.09-d2-s08.png]] > [!mistake] "Any horizontal tangent is the critical point." > **Why it feels right:** the plateau is flat, so flatness seems enough. **Fix:** below $T_c$ there are *two* flat-ish turning points on the vdW loop and neither is the critical point. Only when they **merge** (adding the curvature-zero condition) do you get the true critical point. That's why one derivative isn't enough — you need both. --- ## The one-picture summary ![[deepdives/dd-chemistry-2.4.09-d2-s09.png]] One isotherm at $T_c$, one inflection point, three read-offs: $$V_c = 3b \;\;(\text{where}), \qquad T_c=\frac{8a}{27Rb}\;\;(\text{which curve}), \qquad P_c=\frac{a}{27b^2}\;\;(\text{how high}),$$ all born from the single geometric statement "**flat and straight at the same spot**": $$\frac{\partial P}{\partial V}=0 \quad\text{and}\quad \frac{\partial^2 P}{\partial V^2}=0.$$ > [!recall]- Feynman retelling (say it in plain words) > Picture pressing on a gas at a fixed temperature and drawing how the pressure climbs as the volume drops. Cold enough, and the line goes dead flat for a while — that flat stretch is the gas quietly turning into a liquid. Warm it up and that flat stretch gets shorter. At one magic temperature the flat stretch shrinks to a single dot: the curve is flat *and* dead straight there for just an instant. "Flat" is the slope being zero; "straight" is the bending being zero. Slope and bending are exactly what the two derivatives measure, so we set both to zero. Van der Waals gives a formula for the curve, so we differentiate it twice, set both to zero, and divide one equation by the other — all the messy constants ($R$, $T$, $a$) evaporate, leaving a tidy geometric fact: the magic volume is three times the molecules' own size, $V_c=3b$. Feed that back in and out pop the magic temperature and pressure. Hotter than that temperature, the flat dot never forms — no liquid, just a supercritical haze. Colder, and there are two little turning points instead of one; the critical point is the exact instant those two collide. > [!recall]- Quick self-check > Why two derivative conditions, not one? ::: One gives horizontal tangent (flat); the second gives zero curvature (straight/inflection). The critical point needs both, because it's where two turning points merge into a single double point. > Why divide equation (2) by (1)? ::: To cancel $R$, $T$ and $a$ at once, leaving a pure geometric relation that yields $V_c=3b$. > What is $V_c$ in terms of $b$, and why not $V_c=b$? ::: $V_c=3b$; molecules cannot be crushed to their own excluded volume $b$ even at the critical point. --- Related: [[Real Gases and Compressibility Factor Z]] uses these constants to build $Z_c=\tfrac{3}{8}$; [[Supercritical Fluids]] is the $T>T_c$ region of Step 8; [[Intermolecular Forces]] is the physical meaning of $a$.