Liquid state — vapour pressure, viscosity, surface tension: Worked examples

This page is the drill ground for the parent topic. We will not learn new theory here — instead we hunt down every kind of problem the three liquid properties can throw at you, and solve one example for each. If any symbol feels unfamiliar, it was built in the parent note and in Clausius-Clapeyron Equation, Capillary Action and Contact Angle and Intermolecular Forces.

Before we compute anything, let us name the symbols once, in plain words, so no line surprises you:

Definition The cast of symbols

$p$ — vapour pressure: the push of the gas sitting above a liquid once escaping = returning. Measured in atm, kPa, or mmHg.
$T$ — absolute temperature in kelvin (K). Always kelvin in these formulas — never °C — because the maths counts energy from true zero.
$Δ H_{v a p}$ — heat of vaporisation: energy (J or kJ per mole) needed to rip a mole of molecules out of the liquid into gas.
$R = 8.314 J mol^{- 1} K^{- 1}$ — the gas constant, a fixed conversion between energy and (temperature × amount).
$γ$ (gamma) — surface tension: the inward pull of the surface skin, in N/m (= J/m²).
$θ$ (theta) — contact angle: the angle the liquid's edge makes with the tube wall. Small $θ$ = "loves the wall" (wetting); $θ > 9 0^{\circ}$ = "hates the wall" (non-wetting).
$η$ (eta) — viscosity: internal friction resisting flow, in Pa·s (SI) or poise (CGS).
$E_{a}$ — activation energy for flow: the little energy hill a molecule must climb to slip past its neighbours.

The scenario matrix

Every problem in this topic is one of the cells below. The worked examples that follow are labelled with the cell they hit.

#	Cell (case class)	What makes it special	Example
A	Clausius–Clapeyron: find $Δ H_{v a p}$	two $(p, T)$ pairs given	Ex 1
B	Clausius–Clapeyron: find $p_{2}$ at new $T$	solve for a pressure	Ex 2
C	Clausius–Clapeyron: find boiling $T$	set $p = p_{e x t}$ , solve for $T$	Ex 3
D	Surface tension: capillary RISE ( $θ < 9 0^{\circ}$ , $h > 0$ )	wetting liquid	Ex 4
E	Surface tension: capillary DEPRESSION ( $θ > 9 0^{\circ}$ , $h < 0$ )	sign flip on $cos θ$	Ex 5
F	Surface tension: degenerate $θ = 9 0^{\circ}$	$cos θ = 0 \Rightarrow h = 0$	inside Ex 5
G	Surface tension: work / energy per area	droplet-splitting, $γ = d W / d A$	Ex 6
H	Viscosity: Arrhenius, find $η$ ratio	sign of exponent, $T$ up ⇒ $η$ down	Ex 7
I	Viscosity: find $E_{a}$ from two viscosities	solve the exponent	Ex 8
J	Real-world word problem	translate story → equation	Ex 9
K	Exam twist / limiting behaviour	$T \to \infty$ , $r \to 0$ , unit trap	Ex 10

Example 1 — Cell A: find $Δ H_{v a p}$ from two pressures

Worked example

A liquid has vapour pressure $p_{1} = 0.500 atm$ at $T_{1} = 320 K$ and $p_{2} = 1.00 atm$ at $T_{2} = 350 K$ . Find $Δ H_{v a p}$ .

Forecast: Doubling the pressure over a 30 K jump — is $Δ H_{v a p}$ closer to 40 or to 100 kJ/mol? Guess before reading.

Step 1. Write the integrated Clausius–Clapeyron equation. $ln \frac{p _{2}}{p _{1}} = - \frac{Δ H _{v a p}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ Why this step? This is the only law relating two $(p, T)$ points to $Δ H_{v a p}$ , assuming $Δ H_{v a p}$ stays constant over the range.

Step 2. Compute the left side. $ln \frac{1.00}{0.500} = ln 2 = 0.6931$ Why this step? The ratio of pressures is all that matters on the left — units of pressure cancel, so atm is fine.

Step 3. Compute the temperature bracket. $\frac{1}{350} - \frac{1}{320} = 0.0028571 - 0.0031250 = - 2.679 \times 1 0^{- 4} K^{- 1}$ Why this step? Going to higher $T$ makes $1/ T$ smaller, so the bracket is negative — it will cancel the leading minus sign and leave $Δ H_{v a p}$ positive, as physical sense demands.

Step 4. Solve. $Δ H_{v a p} = - \frac{R l n 2}{( 1/ T _{2} - 1/ T _{1} )} = - \frac{8.314 \times 0.6931}{- 2.679 \times 1 0 ^{- 4}} \approx 21, 510 J/mol \approx 21.5 kJ/mol$

Verify: Two negatives → positive $Δ H_{v a p}$ ✓ (heat is always absorbed to vaporise). ~21.5 kJ/mol is a modest value — sensible for a fairly volatile liquid (weak IMF, low boiling point). The forecast "40 vs 100" was a trap: it's below both.

Example 2 — Cell B: find the vapour pressure at a new temperature

Worked example

Ethanol has $Δ H_{v a p} = 38.6 kJ/mol$ and $p_{1} = 1.00 atm$ at its normal boiling point $T_{1} = 351.5 K$ . What is its vapour pressure at $T_{2} = 300 K$ (room-ish)?

Forecast: We are cooling below the boiling point, so $p_{2}$ must be less than 1 atm. How much less — 0.1 atm? 0.3?

Step 1. Same law, now solving for $p_{2}$ . $ln \frac{p _{2}}{p _{1}} = - \frac{Δ H _{v a p}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ Why this step? Everything except $p_{2}$ is known; we isolate the unknown by exponentiating at the end.

Step 2. Temperature bracket. $\frac{1}{300} - \frac{1}{351.5} = 0.0033333 - 0.0028450 = 4.883 \times 1 0^{- 4} K^{- 1}$ Why this step? Cooling means smaller $T_{2}$ , larger $1/ T_{2}$ , so the bracket is positive now — the opposite sign from Ex 1. Watch how it drives $p_{2}$ down.

Step 3. Plug in. $ln \frac{p _{2}}{1.00} = - \frac{38600}{8.314} \times 4.883 \times 1 0^{- 4} = - (4642.8) (4.883 \times 1 0^{- 4}) = - 2.267$ Why this step? The negative log means $p_{2} < p_{1}$ — consistent with our forecast that cooling lowers vapour pressure.

Step 4. Exponentiate. $p_{2} = e^{- 2.267} = 0.1036 atm$

Verify: $0.104 atm \approx 79 mmHg$ . Real ethanol at 300 K is ~0.09 atm — our constant- $Δ H$ answer is close ✓. And it is well below 1 atm, matching "cooling lowers $p$ ." ✓

Example 3 — Cell C: find the boiling temperature at reduced pressure

Worked example

On a mountain the outside pressure is $0.60 atm$ . Using water's $Δ H_{v a p} = 40.7 kJ/mol$ and normal boiling point $T_{1} = 373 K$ (at $p_{1} = 1.00 atm$ ), find the boiling temperature $T_{2}$ .

Forecast: Water boils when its vapour pressure equals the external pressure. Lower outside pressure ⇒ boils easier ⇒ at a temperature below 373 K. Above or below 90 °C?

Step 1. Set the boiling condition: $p_{2} = p_{e x t} = 0.60 atm$ . Why this step? Boiling happens exactly when internal vapour pressure can push against the atmosphere — so the target vapour pressure is the outside pressure.

Step 2. Rearrange the law for $1/ T_{2}$ . $ln \frac{p _{2}}{p _{1}} = - \frac{Δ H _{v a p}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ $\frac{1}{T _{2}} = \frac{1}{T _{1}} - \frac{R}{Δ H _{v a p}} ln \frac{p _{2}}{p _{1}}$ Why this step? $T_{2}$ sits inside a reciprocal; we solve for $1/ T_{2}$ first, then invert.

Step 3. Numbers. $ln \frac{0.60}{1.00} = - 0.5108, \frac{R}{Δ H _{v a p}} = \frac{8.314}{40700} = 2.043 \times 1 0^{- 4}$ $\frac{1}{T _{2}} = \frac{1}{373} - (2.043 \times 1 0^{- 4}) (- 0.5108) = 0.0026810 + 1.0436 \times 1 0^{- 4} = 0.0027854$ Why this step? The $ln$ is negative, so subtracting it adds to $1/ T_{2}$ , making $T_{2}$ smaller — exactly the "boils cooler up high" effect.

Step 4. Invert. $T_{2} = \frac{1}{0.0027854} = 359.0 K = 85.9^{\circ} C$

Verify: 85.9 °C < 100 °C ✓ — food takes longer to cook on mountains, exactly as observed. Below 90 °C, matching the forecast. ✓

Example 4 — Cell D: capillary rise (wetting, $h > 0$ )

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

Step 1. Write the capillary formula (derived in the parent from force balance). $h = \frac{2 γ c o s θ}{ρ g r}$ Why this step? It equates the weight of the raised column ( $ρ g h π r^{2}$ ) with the upward pull of surface tension around the circumference ( $2 π r γ cos θ$ ). Look at figure s01: the red arrows are the surface pull, the blue arrow is gravity.

Step 2. Since $θ = 0^{\circ}$ , $cos θ = 1$ . Why this step? Perfect wetting means the surface pull is fully vertical — nothing is wasted sideways, so the rise is maximal for this liquid.

Step 3. Plug in ( $r = 1.0 \times 1 0^{- 4} m$ ). $h = \frac{2 ( 0.072 ) ( 1 )}{( 1000 ) ( 9.8 ) ( 1.0 \times 1 0 ^{- 4} )} = \frac{0.144}{0.98} = 0.1469 m \approx 14.7 cm$

Verify: Halving $r$ from 0.2 mm to 0.1 mm doubled the height (7.3 → 14.7 cm) ✓ — that is $h \propto 1/ r$ confirmed. Units: $\frac{N/m}{( kg/m ^{3} ) ( m/s ^{2} ) ( m )} = \frac{N/m}{N/m ^{3}} = m$ ✓.

Example 5 — Cells E & F: capillary depression and the degenerate angle

Step 1. Same formula — the physics did not change, only $θ$ . $h = \frac{2 γ c o s θ}{ρ g r}$ Why this step? The formula is universal; the sign of $h$ falls out of $cos θ$ automatically. No new equation needed.

Step 2. Evaluate $cos 14 0^{\circ}$ . $cos 14 0^{\circ} = - 0.766$ Why this step? Because $θ > 9 0^{\circ}$ , cosine is negative. This is the whole story of a non-wetting liquid — see the downward-curving red arrows in figure s02.

Step 3. Plug in ( $r = 5.0 \times 1 0^{- 4} m$ ). $h = \frac{2 ( 0.485 ) ( - 0.766 )}{( 13600 ) ( 9.8 ) ( 5.0 \times 1 0 ^{- 4} )} = \frac{- 0.7430}{66.64} = - 0.01115 m \approx - 1.11 cm$ Why this step? Negative $h$ means the liquid inside the tube sits below the outside level — a depression, not a rise.

Step 4 (Cell F, degenerate). If $θ = 9 0^{\circ}$ , then $cos 9 0^{\circ} = 0$ , so $h = 0$ . Why this step? At exactly $9 0^{\circ}$ the surface pull is purely horizontal — no vertical component — so the liquid neither rises nor falls. This is the knife-edge case dividing rise from depression.

Verify: $h = - 1.11 cm$ < 0 ✓ — mercury is depressed, exactly why barometer corrections push mercury down. And $θ = 9 0^{\circ}$ gives $h = 0$ ✓, the clean boundary between Cells D and E.

Example 6 — Cell G: work to break a drop into a mist (energy per area)

Worked example

A single water drop of radius $R = 1.0 mm$ is sprayed into $n = 1000$ identical tiny droplets. With $γ = 0.072 N/m$ , how much energy does this cost?

Forecast: Splitting one drop into 1000 makes lots of new surface. Is the energy in microjoules or millijoules?

Step 1. Conserve volume to find the small radius $r$ . $\frac{4}{3} π R^{3} = n \cdot \frac{4}{3} π r^{3} \Rightarrow r = \frac{R}{n ^{1/3}} = \frac{1.0 mm}{100 0 ^{1/3}} = \frac{1.0}{10} = 0.10 mm$ Why this step? Water isn't created or destroyed, so total volume is fixed. Since $100 0^{1/3} = 10$ , each droplet is 10× smaller in radius.

Step 2. New surface area created $= (total small area) - (original area)$ . $Δ A = n \cdot 4 π r^{2} - 4 π R^{2} = 4 π (1000 (1.0 \times 1 0^{- 4})^{2} - (1.0 \times 1 0^{- 3})^{2})$ $= 4 π (1000 \times 1 0^{- 8} - 1 0^{- 6}) = 4 π (1 0^{- 5} - 1 0^{- 6}) = 4 π (9.0 \times 1 0^{- 6}) = 1.131 \times 1 0^{- 4} m^{2}$ Why this step? $γ$ is energy per unit area ( $γ = d W / d A$ ), so we need the increase in area, not the total.

Step 3. Energy $= γ Δ A$ . $W = 0.072 \times 1.131 \times 1 0^{- 4} = 8.14 \times 1 0^{- 6} J = 8.14 μ J$

Verify: Positive work ✓ — you must spend energy to make surface (nature prefers minimum area, so it fights you). Order of magnitude ~micro-joules ✓ for a millimetre drop. This uses the energy-per-area face of $γ$ , not force-per-length — same $γ$ , different question.

Example 7 — Cell H: viscosity ratio between two temperatures (sign check)

Worked example

A liquid has flow activation energy $E_{a} = 15 kJ/mol$ . By what factor does its viscosity change when heated from $T_{1} = 300 K$ to $T_{2} = 350 K$ ?

Forecast: Heating honey makes it thinner. So $η_{2} / η_{1}$ should be less than 1. Roughly 0.9? 0.4?

Step 1. Write the Arrhenius law for viscosity (note the + sign in the exponent). $η = A e^{+ E_{a} / R T}$ Why this step? Unlike gases, liquid flow needs molecules to hop over a barrier $E_{a}$ ; the $+$ sign means higher $T$ (bigger denominator) gives a smaller exponent and thus lower $η$ — see Boltzmann Distribution for why the barrier-beating fraction grows with $T$ .

Step 2. Take the ratio so the unknown constant $A$ cancels. $\frac{η _{2}}{η _{1}} = exp [\frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]$ Why this step? We never know $A$ , but in a ratio it disappears — the classic trick, mirroring Clausius–Clapeyron.

Step 3. Numbers. $\frac{1}{350} - \frac{1}{300} = 0.0028571 - 0.0033333 = - 4.762 \times 1 0^{- 4}$ $\frac{η _{2}}{η _{1}} = exp [\frac{15000}{8.314} \times (- 4.762 \times 1 0^{- 4})] = exp (- 0.8592) = 0.4235$ Why this step? The negative bracket (heating) makes the exponent negative, so the ratio is below 1 — the viscosity drops.

Verify: $η_{2} / η_{1} = 0.42$ ✓ — viscosity fell to ~42% of its cold value. Heating thinned the liquid, matching the honey experience and the parent-note mistake (b). ✓

Example 8 — Cell I: find $E_{a}$ from two measured viscosities

Worked example

A liquid's viscosity drops from $η_{1} = 1.20 mPa\cdots$ at $T_{1} = 298 K$ to $η_{2} = 0.80 mPa\cdots$ at $T_{2} = 320 K$ . Find $E_{a}$ for flow.

Forecast: A modest viscosity drop over 22 K — expect $E_{a}$ of a few kJ/mol to a couple of tens.

Step 1. Start from the ratio form and solve for $E_{a}$ . $ln \frac{η _{2}}{η _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) \Rightarrow E_{a} = \frac{R l n ( η _{2} / η _{1} )}{( \frac{1}{T _{2}} - \frac{1}{T _{1}} )}$ Why this step? Two data points fix the one unknown $E_{a}$ ; the constant $A$ has already cancelled.

Step 2. Left side (viscosity ratio — units cancel, so mPa·s is fine). $ln \frac{0.80}{1.20} = ln 0.6667 = - 0.4055$ Why this step? Only the ratio matters, so we needn't convert to SI.

Step 3. Temperature bracket. $\frac{1}{320} - \frac{1}{298} = 0.0031250 - 0.0033557 = - 2.307 \times 1 0^{- 4} K^{- 1}$ Why this step? Heating shrinks $1/ T$ , giving a negative bracket — which, divided into a negative log, yields a positive $E_{a}$ .

Step 4. Combine. $E_{a} = \frac{8.314 \times ( - 0.4055 )}{- 2.307 \times 1 0 ^{- 4}} = \frac{- 3.371}{- 2.307 \times 1 0 ^{- 4}} \approx 14, 620 J/mol \approx 14.6 kJ/mol$

Verify: Positive $E_{a}$ ✓ (a real energy barrier). ~15 kJ/mol is a typical low-viscosity organic-liquid value ✓, matching the forecast range.

Example 9 — Cell J: real-world word problem (perfume in a warm room)

Worked example

A perfume with $Δ H_{v a p} = 42 kJ/mol$ has vapour pressure $12.0 mmHg$ at $T_{1} = 293 K$ (cool morning). A guest turns the heating up to $T_{2} = 308 K$ . What is the new vapour pressure, and why does the room "smell stronger"?

Forecast: Warmer ⇒ more molecules escape ⇒ higher vapour pressure ⇒ stronger smell. By what factor — 1.5×? 2×?

Step 1. Translate the story: "smells stronger" = higher vapour pressure = more perfume in the air. We want $p_{2}$ . Why this step? Vapour pressure is literally the count of scent molecules floating above the liquid — the physical meaning of the number we compute.

Step 2. Use Clausius–Clapeyron for $p_{2}$ . $ln \frac{p _{2}}{p _{1}} = - \frac{Δ H _{v a p}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$

Step 3. Bracket and exponent. $\frac{1}{308} - \frac{1}{293} = 0.0032468 - 0.0034130 = - 1.662 \times 1 0^{- 4}$ $ln \frac{p _{2}}{p _{1}} = - \frac{42000}{8.314} \times (- 1.662 \times 1 0^{- 4}) = - (5051.7) (- 1.662 \times 1 0^{- 4}) = 0.8396$ Why this step? Warming gives a negative bracket, and the leading minus flips it positive — so $p_{2} > p_{1}$ , matching "stronger smell."

Step 4. Solve. $p_{2} = 12.0 \times e^{0.8396} = 12.0 \times 2.315 = 27.8 mmHg$

Verify: $p_{2} / p_{1} = 2.32$ — the vapour pressure more than doubled for only a 15 K rise ✓. This is the exponential sensitivity of vapour pressure to temperature (Boltzmann Distribution), and exactly why warm rooms smell stronger. ✓

Example 10 — Cell K: exam twist & limiting behaviour

Worked example

Three quick conceptual-numeric traps. (a) In the viscosity law $η = A e^{E_{a} / R T}$ , what does $η$ approach as $T \to \infty$ ? (b) In $h = \frac{2 γ cos θ}{ρ g r}$ , what happens to $h$ as $r \to 0$ ? (c) A student writes water's surface tension as $γ = 72 N/m$ . Spot and fix the unit error.

Forecast: Each is a "what does the formula say in the extreme" check — try to answer all three before reading.

Step 1 (a). As $T \to \infty$ , the exponent $E_{a} / R T \to 0$ , so $e^{0} = 1$ , giving $η \to A$ . Why this step? Infinite temperature means every molecule can trivially clear the barrier — friction bottoms out at the constant $A$ , the "no-barrier" viscosity. It does not go to zero.

Step 2 (b). As $r \to 0$ , $h = \frac{2 γ cos θ}{ρ g r} \to \infty$ . Why this step? An infinitely thin tube would (idealised) lift the liquid infinitely high — $h \propto 1/ r$ blows up. In reality other effects intervene, but the formula's limit is $+ \infty$ for a wetting liquid.

Step 3 (c). Real water is $γ = 0.072 N/m = 72 mN/m$ , not $72 N/m$ . Why this step? The student mixed up milli-newtons and newtons. $72 N/m$ is 1000× too big — you could stand on such a skin! Correct value: $7.2 \times 1 0^{- 2} N/m$ .

Numeric check: $72 mN/m = 72 \times 1 0^{- 3} N/m = 0.072 N/m$ ✓.

Verify: (a) $η \to A$ , finite ✓. (b) $h \to \infty$ , consistent with $h \propto 1/ r$ ✓. (c) $0.072 N/m$ is the textbook value ✓. All three limits behave exactly as the formulas predict.

Recall Self-test: name the cell, then solve in your head

Which cell does each belong to? "Mercury in a glass tube, find the level change" ::: Cell E — capillary depression ( $cos θ < 0$ ) "Two viscosities given, find the flow barrier" ::: Cell I — solve Arrhenius for $E_{a}$ "At what temperature does water boil under 0.6 atm" ::: Cell C — set $p = p_{e x t}$ , solve for $T$ "Break one drop into a thousand, find the energy" ::: Cell G — $W = γ Δ A$ "What is $h$ when $θ = 9 0^{\circ}$ " ::: Cell F — degenerate case, $h = 0$

Connections

Parent topic
Clausius-Clapeyron Equation — the engine behind Ex 1–3 and 9
Capillary Action and Contact Angle — the physics of Ex 4–5
Boiling Point and Phase Diagrams — why Ex 3 works
Boltzmann Distribution — the exponential in Ex 7–9
Intermolecular Forces — the root cause behind every trend
Gibbs Free Energy — why $Δ G = 0$ starts the whole vapour-pressure derivation