Visual walkthrough — Liquid state — vapour pressure, viscosity, surface tension

Parent: Liquid state. Prerequisites: Intermolecular Forces, Capillary Action and Contact Angle.

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Step 1 — Why a surface pulls inward at all

WHAT: Look at two molecules — one buried deep in the liquid, one sitting on the top skin.

WHY: Everything about surface tension starts here. If we cannot see why the surface is special, the formula is just symbols. A molecule is held by intermolecular forces (the "stickiness" between molecules — see Intermolecular Forces). The question is: does that stickiness pull evenly or unevenly?

PICTURE:

• The deep molecule (lavender) has neighbours on all sides. Add up all the little pull-arrows → they cancel. Net force $=0$.
• The surface molecule (coral) has neighbours below and beside but none above. The downward/inward pulls don't cancel → there is a leftover pull pointing into the liquid.

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Step 2 — Turning "skin tension" into a number: defining $\gamma$

WHAT: Put a number on how strong that skin is. Call it $\gamma$ (Greek "gamma").

WHY: We need one measurable quantity for the tension so the rest is arithmetic. There are two honest ways to define it, and they turn out identical.

PICTURE:

We drag a sliding wire of length $\ell$ (the width of the skin it holds) a small distance $dx$, doing work $dW$. This uncovers new surface area $dA=\ell dx$.

$\gamma =\underset{\text{energy per new area}}{\underbrace{\frac{dW}{dA}}}=\frac{Fdx}{\ell dx}=\underset{\text{force per length}}{\underbrace{\frac{F}{\ell }}}$

Term by term:

• $F$ — the force you must pull with to hold the skin taut (newtons, N).
• $\ell$ — the length of skin edge that force is spread along (metres, m).
• $dx$ — the tiny slide distance; it appears on top and bottom and cancels.
• $\gamma$ — the survivor: force per unit length, units ${\text{N m}}^{-1}$, equal to energy per unit area ${\text{J m}}^{-2}$.

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Step 3 — Why the liquid meets the wall at an angle: $\theta$

WHAT: Where liquid touches the tube wall it doesn't stop flat — it curves up (or down). The angle it makes is the contact angle $\theta$.

WHY: The pull we found in Step 1 acts along the surface. Near a wall, the surface tilts, so the pull tilts too. Only the vertical part of that pull can lift liquid. We need $\theta$ to find how much of the pull points up. (Full detail: Capillary Action and Contact Angle.)

PICTURE:

• If the liquid likes the wall more than itself (water in glass) it climbs the wall → the surface curves up → small $\theta$ (near $0^{\circ }$). This is wetting.
• If the liquid likes itself more than the wall (mercury in glass) it pulls away → surface curves down → large $\theta$ (over $90^{\circ }$). This is non-wetting.

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Step 4 — Only the vertical slice of the pull lifts the liquid: $\cos \theta$

WHAT: Split the surface-tension pull at the wall into an up-part and a sideways-part. Keep the up-part.

WHY here does $\cos$ appear, not $\sin$ or $\tan$? $\cos \theta$ is the tool that asks: "of a force tilted by angle $\theta$ from the vertical, what fraction points straight up?" The cosine of an angle is the adjacent side over the hypotenuse of a right triangle — and "adjacent to $\theta$" here is precisely the vertical direction we care about. Sine would give the wasted sideways part; tangent isn't a fraction of the whole force, so it's the wrong question.

PICTURE:

The pull has size $\gamma$ per metre. Tilted by $\theta$ from vertical:

$\text{upward pull per metre}=\gamma \cos \theta$

• $\gamma$ — full pull per metre of edge (from Step 2).
• $\cos \theta$ — the fraction that points up (the vertical slice).
• When $\theta =0$ (perfect wetting): $\cos 0=1$ → the whole pull lifts. Maximum climb.
• When $\theta =90^{\circ }$: $\cos 90^{\circ }=0$ → nothing lifts. No rise.
• When $\theta >90^{\circ }$ (mercury): $\cos \theta$ is negative → the "up" pull is actually down. Keep this — it is the whole story of mercury in Step 8.

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Step 5 — Add up the pull all the way around the ring: the $2\pi r$ edge

WHAT: The liquid touches the wall along a circle of radius $r$. Total upward force = upward-pull-per-metre $\times$ length of that circle.

WHY: Force-per-length only becomes a real force once multiplied by the length it acts over. The wetted edge is the tube's inner circumference, a circle of radius $r$, whose length is $2\pi r$.

PICTURE:

$F_{\text{up}}=\underset{\text{up-pull per metre}}{\underbrace{(\gamma \cos \theta )}}\times \underset{\text{length of the wet ring}}{\underbrace{2\pi r}}=2\pi r\gamma \cos \theta$

• $r$ — inner radius of the tube (metres). Thin tube = small $r$.
• $2\pi r$ — the circumference; the total length of edge doing the lifting.
• $F_{\text{up}}$ — the whole upward force, in newtons.

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Step 6 — What weighs the column down: gravity on the raised liquid

WHAT: The lifted liquid is a cylinder: base area $\pi r^2$, height $h$. Its weight pulls down.

WHY: The liquid stops rising when the upward pull exactly matches the downward weight. So we need the weight in symbols before we can balance.

PICTURE:

Weight $=$ mass $\times g=$ (density $\times$ volume) $\times g$:

$F_{\text{down}}=\underset{\text{density}}{\underbrace{\rho }}\times \underset{\text{volume of column}}{\underbrace{\pi r^{2}h}}\times \underset{\text{gravity}}{\underbrace{g}}=\rho gh\pi r^2$

• $\rho$ — density of the liquid, mass per volume (${\text{kg m}}^{-3}$); water $\approx 1000$.
• $\pi r^2$ — area of the circular base.
• $h$ — height the column has risen (what we are ultimately solving for).
• $g$ — gravitational pull, $9.8{\text{m s}}^{-2}$.

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Step 7 — Balance the two forces and solve for $h$

WHAT: Set up-pull equal to down-weight and isolate $h$.

WHY: At equilibrium the column is still — no net force. That is exactly $F_{\text{up}}=F_{\text{down}}$. Solving this single equation delivers the boxed result.

PICTURE:

$\underset{F_{\text{up}}\text{ (Step 5)}}{\underbrace{2\pi r\gamma \cos \theta }}=\underset{F_{\text{down}}\text{ (Step 6)}}{\underbrace{\rho gh\pi r^2}}$

Divide both sides by $\pi r$ (a $\pi r$ lives in each side):

$2\gamma \cos \theta =\rho ghr$

Now divide by $\rho gr$ to leave $h$ alone:

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Step 8 — Every case, including the degenerate ones

WHAT: Sweep $\theta$ through all its regimes and check zero/limiting inputs.

WHY: The contract: the reader must never meet a scenario we skipped. The sign of $\cos \theta$ controls everything.

PICTURE:

Case	$\theta$	$\cos \theta$	$h$	What you see
Perfect wetting (water/glass)	$0^{\circ }$	$+1$	max positive	climbs highest
Partial wetting	$0^{\circ }$–$90^{\circ }$	positive	positive	climbs some
Neutral	$90^{\circ }$	$0$	$0$	flat, no rise
Non-wetting (mercury/glass)	$90^{\circ }$–$180^{\circ }$	negative	negative	depresses — sinks below outside level

Degenerate / limiting inputs:

• $r\to 0$ (infinitely thin tube): $h\to \infty$. Real tubes never reach zero radius, so no paradox — just very tall rise.
• $r\to \infty$ (wide vessel): $h\to 0$. In a bucket you see no capillary rise — matches life.
• $\gamma =0$ (no skin): $h=0$. No tension, no lift. Sanity ✓.
• $g\to 0$ (weightlessness): $h\to \infty$ — in orbit liquids creep along walls indefinitely. Astronauts confirm this.

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The one-picture summary

Read it left to right: the surface skin (Step 1–2) pulls up along the wet ring $2\pi r$ (Step 5), only the $\cos \theta$ slice counts (Step 4), and it climbs until the column's weight $\rho gh\pi r^2$ (Step 6) matches — giving $h=2\gamma \cos \theta /(\rho gr)$.

Recall Feynman retelling — the whole walk in plain words

A molecule on the top skin has no friends above, so it's yanked inward — that inward yank is what we measure as $\gamma$, the skin's pull per metre of edge. Slip a thin tube into water. Where the water hugs the glass, its skin tilts; the upward share of the skin's pull is $\gamma \cos \theta$ — full pull if it hugs perfectly ($\theta =0$), zero if it meets the wall square-on, and downward if the liquid hates the wall (mercury). That upward pull runs all the way around the wet ring, a circle of length $2\pi r$, so the total lift is $2\pi r\gamma \cos \theta$. The water rises, dragging up a little cylinder of weight $\rho gh\pi r^2$, until lift equals weight. Cancel a $\pi r$, divide by $\rho gr$, and out drops $h=2\gamma \cos \theta /\rho gr$: thinner tube, taller climb; heavier liquid or fatter tube, shorter climb; and if the liquid won't wet the wall, it sinks instead of climbing.

Recall Quick self-test

Why does $\cos$ (not $\sin$) appear in the lift? ::: Because $\cos \theta$ is the fraction of the tilted pull that points straight up (adjacent-over-hypotenuse of the tilt triangle); $\sin$ would give the wasted sideways part. Why is the wetted length $2\pi r$ and not $\pi r^2$? ::: The pull acts along the edge where liquid meets wall — a circle (circumference $2\pi r$) — not over an area. A tube of half the radius gives what rise? ::: Twice the rise, because $h\propto 1/r$. Mercury has $\theta \approx 140^{\circ }$; what does the formula predict? ::: $\cos \theta <0\Rightarrow h<0$: mercury is depressed below the outside level, not raised.

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Connections

• Capillary Action and Contact Angle — deeper look at what sets $\theta$
• Intermolecular Forces — the root cause of the inward pull and of $\gamma$
• Boiling Point and Phase Diagrams — surface effects and the liquid state