Liquid state — vapour pressure, viscosity, surface tension: Question bank

Before we start, a short refresher so no symbol is used unexplained:

$γ$ = surface tension = force per unit length pulling a surface tight, in $N m^{- 1}$ .
$η$ = viscosity = internal friction resisting flow, in $Pa\cdots$ .
$Δ H_{v a p}$ = heat needed to turn one mole of liquid into vapour, in $J mol^{- 1}$ .
$θ$ = contact angle = the angle the liquid surface makes with the solid wall, measured inside the liquid.
$p$ = pressure, and specifically the vapour pressure of the liquid, in $atm$ or $Pa$ .
$R$ = universal gas constant $= 8.314 J mol^{- 1} K^{- 1}$ ; it appears wherever we relate energy per mole to temperature $T$ .
In the capillary-rise law $h = \frac{2 γ cos θ}{ρ g r}$ : $h$ = height risen (m), $ρ$ = liquid density ( $kg m^{- 3}$ ), $g$ = gravitational acceleration ( $9.8 m s^{- 2}$ ), $r$ = tube radius (m).
In $γ = F / ℓ$ : $F$ = force (N) needed to drag a wire, $ℓ$ = length of that wire (m), and $d x$ = a tiny distance the wire moves, sweeping out new surface area.

True or false — justify

Vapour pressure of a liquid increases if you use a wider dish (larger surface area).

False — a wider surface speeds up evaporation AND condensation equally, so the equilibrium pressure is unchanged; it depends only on temperature and IMF.

Adding more liquid to a closed flask raises its vapour pressure.

False — as long as some liquid remains, vapour pressure is fixed by

T

; extra liquid changes how much can evaporate but not the equilibrium pressure.

At the normal boiling point, vapour pressure equals exactly 1 atm.

True — "normal" boiling point is defined as the temperature where the liquid's vapour pressure reaches

1 atm

(the external pressure it must push against).

Surface tension has units of newtons (a force).

False — it is force per unit length,

N m^{- 1}

, equivalently energy per area

J m^{- 2}

; the "per length" is essential.

Heating a liquid lowers both its surface tension and its viscosity.

True — thermal motion weakens the inward pull (↓

γ

) and helps molecules hop past each other (↓

η

); both drop with rising

T

Heating a liquid lowers its vapour pressure because molecules calm down.

False — heating raises vapour pressure; more molecules exceed the escape energy set by the Boltzmann Distribution, so more escape into vapour.

A liquid with strong hydrogen bonding tends to have low vapour pressure but high viscosity.

True — strong Intermolecular Forces make escape hard (low

p

) and sliding hard (high

η

); the same forces push both effects.

Water rises in a thin glass tube and mercury rises too, just less.

False — mercury has contact angle

θ > 9 0^{\circ}

cos θ < 0

, giving

h < 0

; it is depressed, not merely raised less.

Doubling a capillary tube's radius doubles the rise height.

False —

h = \frac{2 γ cos θ}{ρ g r}

shows

h \propto 1/ r

, so doubling

r

halves the rise.

Two liquids at the same temperature always have the same vapour pressure.

False — vapour pressure depends on the liquid's IMF too; ether and water at

2 5^{\circ}

C differ hugely because ether's weak IMF let molecules escape easily.

Spot the error

"Viscosity follows

η = A e^{- E_{a} / R T}

, just like a reaction rate."

Wrong sign — viscosity uses

η = A e^{+ E_{a} / R T}

, so

η

decreases as

T

rises; flow is easier when molecules have energy to hop the barrier

E_{a}

"Since the Clausius–Clapeyron equation is

ln \frac{p _{2}}{p _{1}} = - \frac{Δ H _{v a p}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})

, raising

T

lowers

p

Wrong conclusion — for

T_{2} > T_{1}

the bracket

(\frac{1}{T _{2}} - \frac{1}{T _{1}})

is negative, and the leading minus makes the whole right side positive, so

p_{2} > p_{1}

; vapour pressure rises.

"Surface tension makes drops spherical because spheres have the largest surface area."

Backwards — a sphere has the smallest surface area for a given volume, and the surface-minimising tension drives the drop to that shape.

"In deriving

γ = F / ℓ

, we always use area

d A = ℓ d x

Incomplete — for a thin film with two liquid faces,

d A = 2 ℓ d x

, so

γ = F / (2 ℓ)

; forgetting the factor of two doubles the answer.

"Bigger molecules always evaporate faster, so larger molar mass means higher vapour pressure."

Wrong — larger molecules usually have stronger dispersion forces, making escape harder, so vapour pressure typically falls with molar mass in a homologous series.

"The Clausius–Clapeyron derivation is exact."

Not exact — it assumes ideal vapour,

V_{g a s} ≫ V_{l i q}

, and constant

Δ H_{v a p}

; these approximations make it accurate only over modest temperature ranges.

Why questions

Why does a molecule inside the liquid feel no net pull, but a surface molecule feels one?

An interior molecule is surrounded on all sides, so pulls cancel; a surface molecule has neighbours below and beside but none above, leaving a net inward pull that tightens the surface.

Why does vapour pressure depend on temperature but not on the amount of liquid?

The escaping fraction is set by the Boltzmann Distribution at temperature

T

; equilibrium balances rates per unit area, so total quantity cancels and only

T

(and IMF) survives.

Why does the capillary force act around the circumference

2 π r

rather than over the area

π r^{2}

Surface tension is a force per unit length acting along the contact line where liquid meets glass — that line is the circumference

2 π r

, not the cross-sectional area.

Why does viscosity of gases rise with temperature while liquids fall?

In gases, faster molecules transfer more momentum between layers (more collisions ⇒ more drag); in liquids, the limiting factor is escaping IMF traps, which heat helps, so drag drops.

Why is the same underlying cause — intermolecular forces — able to raise viscosity and surface tension yet lower vapour pressure?

Strong forces make it hard to slide past (↑

η

), hard to break the surface (↑

γ

), and hard to escape into gas (↓

p

); each property measures resistance to a different kind of separation.

Why does honey pour more easily when warmed?

Warming supplies energy to overcome the activation barrier

E_{a}

η = A e^{+ E_{a} / R T}

, so more molecules can slip past neighbours and viscosity falls sharply.

Why must we assume the vapour is ideal to reach the Clausius–Clapeyron form?

Only for an ideal gas can we write

V_{g a s} = R T / p

, which is the substitution that converts

d p / d T

into the clean integrable

d ln p / d T = Δ H_{v a p} / R T^{2}

Edge cases

What happens to capillary height when

θ = 9 0^{\circ}

exactly?

cos 9 0^{\circ} = 0

h = 0

— the liquid neither rises nor falls; the wall is perfectly "neutral" toward the liquid.

What does

θ > 9 0^{\circ}

(like mercury on glass) predict for

h

cos θ < 0

makes

h < 0

, a depression — the liquid surface dips below the outside level and forms a convex (bulging) meniscus.

What is the vapour pressure of a pure solid, and does Clausius–Clapeyron still apply?

Solids also have a (small) vapour pressure via sublimation; the same equation applies with

Δ H_{s u b}

in place of

Δ H_{v a p}

As temperature approaches the critical point, what happens to surface tension?

It falls toward zero — at the critical point liquid and vapour become indistinguishable, so there is no surface to have tension.

If a liquid perfectly wets the glass (

θ = 0

), what is

cos θ

and the rise?

cos 0^{\circ} = 1

, giving the maximum possible rise

h = 2 γ / (ρ g r)

for that tube; any wetting less perfect reduces it.

What would happen to boiling point if external pressure were lowered (e.g. on a mountain)?

Boiling occurs when vapour pressure equals external pressure; lower external pressure is reached at a lower vapour pressure, so the liquid boils at a lower temperature (see Boiling Point and Phase Diagrams).

In the limit

r \to 0

(an infinitely thin tube), what does

h = 2 γ cos θ / ρ g r

predict, and is it physical?

It predicts

h \to \infty

; physically the rise is enormous but finite, since the idealised thin-cylinder model breaks down and evaporation limits it in reality.

Recall One-line survival summary

Strong IMF ⇒ ↑ $γ$ , ↑ $η$ , ↓ $p$ . Heat ⇒ ↓ $γ$ , ↓ $η$ , ↑ $p$ . Capillary $h \propto cos θ / r$ : sign of $cos θ$ decides rise vs depression.

Connections

Parent topic
Intermolecular Forces · Clausius-Clapeyron Equation · Boltzmann Distribution
Capillary Action and Contact Angle · Boiling Point and Phase Diagrams · Gibbs Free Energy