Liquid state — vapour pressure, viscosity, surface tension: Exercises

Before we start, one shared toolkit — the three formulas this whole chapter rests on. Each was derived in the parent note (parent topic); here we only use them.

Symbols, in plain words (so nothing is used before it is named):

$p$ = vapour pressure (push of the escaped gas). $T$ = absolute temperature in kelvin (never °C in these formulas).
$Δ H_{v a p}$ = heat needed to boil one mole (J/mol). $R = 8.314 J mol^{- 1} K^{- 1}$ = gas constant.
$γ$ = surface tension (N/m = J/m²). $θ$ = contact angle (how the liquid meets the glass). $ρ$ = density (kg/m³). $g = 9.8 m/s^{2}$ . $r$ = tube radius (m).
$η$ = viscosity (Pa·s). $E_{a}$ = activation energy to slip past a neighbour (J/mol). $A$ = a constant prefactor.

Level 1 — Recognition

L1.1

State the sign relationship: for a liquid with stronger intermolecular forces (IMF), does each of vapour pressure, surface tension, and viscosity go up or down?

Recall Solution

Stronger IMF = molecules hold each other harder. So:

Vapour pressure ↓ — harder to escape into the gas.
Surface tension ↑ — the inward pull on the skin is stronger.
Viscosity ↑ — layers grip each other harder, so flow is harder.

Mnemonic from the parent note: "Strong forces LOVE HIGH walls but HATE ESCAPE."

L1.2

The units of surface tension can be written two ways. Show that $N m^{- 1}$ and $J m^{- 2}$ are the same thing.

Recall Solution

Energy = force × distance, so $1 J = 1 N\cdotm$ . Divide both sides by $m^{2}$ : $\frac{J}{m ^{2}} = \frac{N\cdotm}{m ^{2}} = \frac{N}{m} .$ They are identical. This is the whole "force-per-length = energy-per-area" idea in one line.

Level 2 — Application

L2.1

Water has $p_{1} = 1.00$ atm at $T_{1} = 373 K$ and $Δ H_{v a p} = 40.7 kJ/mol$ . Find its vapour pressure at $T_{2} = 363 K$ ( $90^{\circ} C$ ).

Recall Solution

WHAT: use Clausius–Clapeyron to slide from $373 K$ down to $363 K$ . WHY: it links two $(p, T)$ points through $Δ H_{v a p}$ — exactly one unknown, $p_{2}$ . $ln \frac{p _{2}}{1.00} = - \frac{40700}{8.314} (\frac{1}{363} - \frac{1}{373})$ Inner bracket: $\frac{1}{363} = 2.7548 \times 1 0^{- 3}$ , $\frac{1}{373} = 2.6810 \times 1 0^{- 3}$ , difference $= + 7.385 \times 1 0^{- 5} K^{- 1}$ . $ln p_{2} = - (4895.4) (7.385 \times 1 0^{- 5}) = - 0.3615$ $p_{2} = e^{- 0.3615} = 0.697 atm .$ Sensible? Below the boiling point (373 K) the vapour pressure must be below 1 atm — and $0.70 < 1$ . ✓

L2.2

A capillary tube of radius $r = 0.30 mm$ is dipped in water: $γ = 0.072 N/m$ , $θ = 0^{\circ}$ , $ρ = 1000 kg/m^{3}$ , $g = 9.8$ . Find the rise $h$ .

Recall Solution

WHAT: balance the raised column's weight against the upward surface-tension pull (see figure). $h = \frac{2 γ c o s θ}{ρ g r} = \frac{2 ( 0.072 ) ( 1 )}{1000 \cdot 9.8 \cdot 3.0 \times 1 0 ^{- 4}}$ $h = \frac{0.144}{2.94} = 0.0490 m = 4.9 cm .$ Sensible? A wider tube than the 0.2 mm parent example, and $h \propto 1/ r$ , so this rise (4.9 cm) is lower than the parent's 7.3 cm. ✓

Figure — Liquid state — vapour pressure, viscosity, surface tension

Look at the figure: the green arrows are the surface-tension force pulling the water column up all around the circumference $2 π r$ ; the red arrow is the weight pulling it down. Equilibrium is where they cancel.

Level 3 — Analysis

L3.1

A liquid's vapour pressure is $0.100$ atm at $300 K$ and $0.500$ atm at $340 K$ . Find $Δ H_{v a p}$ (in kJ/mol).

Recall Solution

WHAT: two measured points, one unknown $Δ H_{v a p}$ — invert Clausius–Clapeyron. $ln \frac{0.500}{0.100} = - \frac{Δ H _{v a p}}{8.314} (\frac{1}{340} - \frac{1}{300})$ LHS $= ln 5 = 1.6094$ . Bracket: $\frac{1}{340} = 2.9412 \times 1 0^{- 3}$ , $\frac{1}{300} = 3.3333 \times 1 0^{- 3}$ , difference $= - 3.9216 \times 1 0^{- 4}$ . $1.6094 = - \frac{Δ H _{v a p}}{8.314} (- 3.9216 \times 1 0^{- 4}) = Δ H_{v a p} (4.717 \times 1 0^{- 5})$ $Δ H_{v a p} = \frac{1.6094}{4.717 \times 1 0 ^{- 5}} = 34120 J/mol \approx 34.1 kJ/mol .$ Sensible? A realistic value for a moderately volatile molecular liquid (water is ~40.7). ✓

L3.2

Two liquids at the same temperature obey $η = A e^{E_{a} / R T}$ with the same $A$ . Liquid X has $E_{a} = 15 kJ/mol$ , liquid Y has $E_{a} = 25 kJ/mol$ . At $T = 300 K$ , how many times more viscous is Y than X?

Recall Solution

WHAT: take the ratio so $A$ cancels. $\frac{η _{Y}}{η _{X}} = \frac{A e ^{E_{a, Y} / R T}}{A e ^{E_{a, X} / R T}} = e^{(E_{a, Y} - E_{a, X}) / R T}$ $= e^{(25000 - 15000) / (8.314 \cdot 300)} = e^{10000/2494.2} = e^{4.0093} = 55.1.$ Sensible? A higher barrier means far fewer molecules can hop past neighbours — so Y is dramatically thicker (~55×), which is why glycerol pours so much slower than water. ✓

Level 4 — Synthesis

L4.1

Mercury does not rise in a glass capillary — it is depressed. For mercury: $γ = 0.485 N/m$ , $θ = 14 0^{\circ}$ , $ρ = 13600 kg/m^{3}$ , $r = 0.30 mm$ , $g = 9.8$ . Compute $h$ and interpret the sign.

Recall Solution

WHAT: same capillary formula — but now $θ = 14 0^{\circ} > 9 0^{\circ}$ , so $cos θ$ is negative. $cos 14 0^{\circ} = - 0.766.$ $h = \frac{2 γ c o s θ}{ρ g r} = \frac{2 ( 0.485 ) ( - 0.766 )}{13600 \cdot 9.8 \cdot 3.0 \times 1 0 ^{- 4}}$ Numerator $= - 0.7430$ . Denominator $= 39.98$ . $h = - 0.01858 m = - 1.86 cm .$ Interpretation: the minus sign means a depression of 1.86 cm below the outside level — mercury is non-wetting (it beads up, pulls away from glass). See the figure: the meniscus bulges up (convex) for mercury, whereas water's dips down (concave).

L4.2

Show that if you double the temperature of a liquid whose viscosity follows $η = A e^{E_{a} / R T}$ , the viscosity does not simply halve. Derive the exact factor for $E_{a} = 20 kJ/mol$ going from $300 K$ to $600 K$ .

Recall Solution

WHAT: form the ratio to see the true (exponential) dependence. $\frac{η ( 600 )}{η ( 300 )} = e^{E_{a} / (R \cdot 600) - E_{a} / (R \cdot 300)} = e^{\frac{E _{a}}{R} (\frac{1}{600} - \frac{1}{300})}$ $= e^{\frac{20000}{8.314} (1.6667 \times 1 0^{- 3} - 3.3333 \times 1 0^{- 3})} = e^{2405.6 \cdot (- 1.6667 \times 1 0^{- 3})} = e^{- 4.009} = 0.0181.$ Meaning: viscosity drops to about 1.8 % of its value — a factor of ~55 decrease, not a factor of 2. Temperature acts inside an exponent, so its effect is far more dramatic than a linear "halving." This is why the honey-warming trick works so spectacularly.

Level 5 — Mastery

L5.1

A liquid boils at $350 K$ under $1.00 atm$ and has $Δ H_{v a p} = 30.0 kJ/mol$ . A mountaineer is where atmospheric pressure is only $0.60 atm$ . At what temperature will this liquid boil there?

Recall Solution

WHAT: boiling happens when vapour pressure equals the surrounding pressure. So set $p_{1} = 1.00$ atm at $T_{1} = 350$ K (the known boiling point), and $p_{2} = 0.60$ atm, solve for $T_{2}$ . WHY Clausius–Clapeyron: it is the only equation linking the two boiling conditions through $Δ H_{v a p}$ . $ln \frac{0.60}{1.00} = - \frac{30000}{8.314} (\frac{1}{T _{2}} - \frac{1}{350})$ LHS $= ln 0.60 = - 0.5108$ . Constant $\frac{Δ H}{R} = 3608.4$ . $- 0.5108 = - 3608.4 (\frac{1}{T _{2}} - 2.8571 \times 1 0^{- 3})$ $\frac{1}{T _{2}} - 2.8571 \times 1 0^{- 3} = \frac{0.5108}{3608.4} = 1.4155 \times 1 0^{- 4}$ $\frac{1}{T _{2}} = 2.9987 \times 1 0^{- 3} \Rightarrow T_{2} = 333.5 K .$ Sensible? Lower external pressure ⇒ molecules need less energy to break out ⇒ lower boiling point. $333.5 < 350$ . ✓ (Same reason water boils below $100^{\circ}$ C on a mountain.)

L5.2

A capillary experiment for an unknown liquid gives rise $h = 3.2 cm$ in a tube of radius $r = 0.25 mm$ , with $θ = 0^{\circ}$ and density $ρ = 800 kg/m^{3}$ . Find its surface tension $γ$ . Then predict the rise in a tube of half the radius.

Recall Solution

WHAT (part 1): invert the capillary formula for $γ$ . $γ = \frac{h ρ g r}{2 c o s θ} = \frac{( 0.032 ) ( 800 ) ( 9.8 ) ( 2.5 \times 1 0 ^{- 4} )}{2 ( 1 )}$ Numerator $= 0.032 \cdot 800 \cdot 9.8 \cdot 2.5 \times 1 0^{- 4} = 0.06272$ . Divide by 2: $γ = 0.03136 N/m \approx 0.0314 N/m .$ WHAT (part 2): $h \propto 1/ r$ . Halving $r$ doubles $h$ : $h_{n e w} = 2 \times 3.2 = 6.4 cm .$ Sensible? $γ \approx 0.031$ N/m is typical of an organic liquid (less than water's 0.072, consistent with weaker IMF). And a thinner tube climbs higher — the $1/ r$ law again. ✓

Quick self-test

Answer :::- means reveal.

What sign sits in the viscosity exponent, and what does it imply?

Positive:

η = A e^{+ E_{a} / R T}

, so viscosity falls as

T

rises.

When does capillary

h

come out negative?

When

θ > 9 0^{\circ}

(non-wetting, e.g. mercury), so

cos θ < 0

— a depression.

Boiling occurs when the vapour pressure equals _____ ?

The external (surrounding) pressure.

Doubling

T

changes viscosity by a factor

e^{\frac{E _{a}}{R} (1/2 T - 1/ T)}

— is this bigger or smaller than a factor of 2?

Far more dramatic — it can be tens of times smaller, because

T

acts inside an exponent.

Connections

Clausius-Clapeyron Equation — engine behind L2.1, L3.1, L5.1
Capillary Action and Contact Angle — the geometry of L2.2, L4.1, L5.2
Intermolecular Forces — the root cause of every trend used here
Boiling Point and Phase Diagrams — where L5.1's pressure-dependence lives
Boltzmann Distribution — why escape and hopping fractions grow exponentially with $T$
← back to the topic note