IIT-JAM Liquid State — Vapour Pressure, Surface Tension and Viscosity

By Aniket Bhardwaj · 16 September 2026 · IIT-JAM Chemistry

The liquid state is a small unit, and that is exactly why it is worth doing properly. Three measurable properties — vapour pressure, surface tension and viscosity — carry almost all the questions, each has one main equation, and all three are numerically simple once you keep the units straight. Students who skip this unit because it "looks descriptive" then lose easy marks to a capillary-rise calculation that takes ninety seconds. This guide gives the equations, the meaning of every symbol, and four fully computed examples.

Why liquids behave the way they do

A liquid sits between a gas and a solid: the molecules are as close together as in a solid, but they are free to move past one another. Every liquid property in this unit is a consequence of that single picture. Molecules at the surface have fewer neighbours than molecules in the bulk, so the surface costs energy — that is surface tension. Molecules must be dragged past one another for the liquid to flow — that is viscosity. Molecules at the surface with enough kinetic energy escape — that is vapour pressure. Stronger intermolecular forces raise surface tension and viscosity and lower vapour pressure, and that one sentence answers a large share of the conceptual questions in this unit.

The three equations

Clausius–Clapeyron (vapour pressure vs temperature):
ln (P₂/P₁) = − (ΔH_vap/R) (1/T₂ − 1/T₁)

Capillary rise (surface tension): h = 2γ cos θ / (r ρ g)
Excess pressure: inside a drop ΔP = 2γ/r · inside a soap bubble ΔP = 4γ/r (two surfaces)

Poiseuille flow: V/t = π ΔP r⁴ / (8 η L)
Ostwald viscometer comparison: η₁/η₂ = (ρ₁ t₁) / (ρ₂ t₂)
Temperature dependence: η = A e^E_a/RT (note the positive exponent)

What each symbol means, and in what unit

γ — surface tension, in N m⁻¹, which is identical to J m⁻². Both readings are correct: force per unit length along the surface, or energy per unit new area.
θ — contact angle. For water on clean glass it is close to 0°, so cos θ ≈ 1 and the liquid rises. For mercury on glass it is obtuse, cos θ is negative, and the liquid is depressed. The formula handles both automatically if you keep the sign.
r — radius of the capillary (or of the drop), in metres. Note the fourth power of r in Poiseuille's equation: halving the tube radius cuts the flow rate to one sixteenth.
η — coefficient of viscosity. SI unit Pa s. The older CGS unit is the poise: 1 P = 0.1 Pa s, so 1 centipoise = 1 mPa s. Water is very close to 1 mPa s at 293 K, which makes it a convenient mental reference.
ΔH_vap — molar enthalpy of vaporisation, in J mol⁻¹ if you are using R = 8.314 J K⁻¹ mol⁻¹. Mixing kJ and J here is the single most common arithmetic failure in this unit.

Worked example 1 — capillary rise of water

Water rises in a glass capillary of radius 0.20 mm. Take γ = 0.0728 N m⁻¹, θ = 0°, ρ = 998 kg m⁻³, g = 9.81 m s⁻². Find the height.

r = 0.20 mm = 2.0 × 10⁻⁴ m

Numerator: 2γ cos θ = 2 × 0.0728 × 1 = 0.1456 N m⁻¹

Denominator: r ρ g = 2.0 × 10⁻⁴ × 998 × 9.81
998 × 9.81 = 9790.38 → 9790.38 × 2.0 × 10⁻⁴ = 1.9581

h = 0.1456 / 1.9581 = 0.07436 m = 7.44 cm

Worked example 2 — the same tube with mercury

γ(Hg) = 0.485 N m⁻¹, θ = 140°, ρ = 13534 kg m⁻³, same capillary.

cos 140° = −0.7660, so 2γ cos θ = 2 × 0.485 × (−0.7660) = −0.7431

r ρ g = 2.0 × 10⁻⁴ × 13534 × 9.81 = 2.0 × 10⁻⁴ × 132768.5 = 26.554

h = −0.7431 / 26.554 = −0.02798 m

Mercury is depressed by 2.80 cm. The negative sign is the answer, not an error — it tells you the meniscus is convex and the liquid stands below the outside level. A JAM question that gives you an obtuse contact angle is testing exactly this.

Worked example 3 — vapour pressure by Clausius–Clapeyron

Water boils at 373.15 K under 1.00 atm and ΔH_vap = 40.7 kJ mol⁻¹. Estimate its vapour pressure at 353.15 K (80 °C).

1/T₂ = 1/353.15 = 2.83166 × 10⁻³ K⁻¹
1/T₁ = 1/373.15 = 2.67989 × 10⁻³ K⁻¹
(1/T₂ − 1/T₁) = 1.5177 × 10⁻⁴ K⁻¹

ΔH_vap/R = 40700 / 8.314 = 4895.4 K

ln(P₂/P₁) = −4895.4 × 1.5177 × 10⁻⁴ = −0.7430

P₂ = 1.00 × e^−0.7430 = 0.476 atm ≈ 362 mmHg

Honest note on accuracy. The measured vapour pressure of water at 80 °C is about 355 mmHg, so this estimate is roughly 2% high. That is expected: the derivation assumes ΔH_vap is constant over the temperature range, the vapour is ideal, and the volume of the liquid is negligible next to the vapour. In a JAM answer, stating those assumptions is worth as much as the number.

Worked example 4 — viscosity from an Ostwald viscometer, and its activation energy

(a) In the same viscometer, an unknown liquid of density 0.879 g cm⁻³ takes 156 s to flow; water (ρ = 0.998 g cm⁻³, η = 1.002 mPa s) takes 98 s. Find η of the liquid.

η = 1.002 × (0.879 × 156) / (0.998 × 98)

0.879 × 156 = 137.124 · 0.998 × 98 = 97.804

137.124 / 97.804 = 1.4020, so η = 1.002 × 1.4020 = 1.405 mPa s

(b) Water's viscosity falls from 1.002 mPa s at 293.15 K to 0.798 mPa s at 303.15 K. Find the activation energy of viscous flow.

ln(η₁/η₂) = ln(1.002 / 0.798) = ln 1.2556 = 0.22764

1/T₁ − 1/T₂ = 3.41122 × 10⁻³ − 3.29870 × 10⁻³ = 1.1253 × 10⁻⁴

E_a = R × 0.22764 / 1.1253 × 10⁻⁴ = 8.314 × 2023.0 = 16819 J mol⁻¹ ≈ 16.8 kJ mol⁻¹

Note the direction: viscosity decreases as temperature rises, which is the opposite of a gas, whose viscosity increases with temperature. JAM has asked that comparison as a conceptual question.

Extra numbers worth being able to produce

Excess pressure inside a water droplet of radius 1.0 mm: ΔP = 2γ/r = 2 × 0.0728 / 1.0 × 10⁻³ = 145.6 Pa. A soap bubble of the same radius with γ = 0.025 N m⁻¹ has two surfaces, so ΔP = 4γ/r = 4 × 0.025 / 1.0 × 10⁻³ = 100 Pa. Forgetting the factor of 4 for a bubble is a classic trap.

Common mistakes that cost marks

Mixing kJ with J. If R is 8.314 J K⁻¹ mol⁻¹, then ΔH_vap must be in J mol⁻¹. Using 40.7 instead of 40700 makes the exponent a thousand times too small and the answer indistinguishable from 1 atm.
Sign errors in Clausius–Clapeyron. Write it once carefully and check the physics: raising the temperature must raise the vapour pressure. If your answer says otherwise, the sign is wrong, not the physics.
Dropping cos θ. It only equals 1 for a perfectly wetting liquid. Mercury needs the negative cosine.
Radius vs diameter. Capillary questions often quote the bore diameter. Halving it before substituting is part of the question.
Poise and Pa s. 1 poise = 0.1 Pa s and 1 cP = 1 mPa s. Answers a factor of 10 or 1000 out are almost always this.
Writing viscosity's Arrhenius form with a minus sign. It is η = A e^+E_a/RT, because viscosity falls as T rises — the opposite of a rate constant.
Confusing vapour pressure with boiling point. A liquid boils when its vapour pressure equals the external pressure, which is why water boils below 100 °C at altitude.

How to prepare this unit for JAM

Property	Main equation	Effect of raising temperature	What you should be able to do
Vapour pressure	Clausius–Clapeyron	Increases	Find P at a second T, or find ΔH_vap from two P–T pairs
Surface tension	h = 2γ cos θ / rρg	Decreases	Capillary rise or depression; excess pressure in drops and bubbles
Viscosity	Ostwald comparison; Poiseuille	Decreases (liquids)	Compare two liquids in one viscometer; extract E_a from two temperatures

Treat this table as a study plan, not as a prediction of the paper. For the authoritative syllabus and the current pattern, read the official IIT-JAM notification for your year.

Check your vapour-pressure working instantly. The Clausius–Clapeyron calculator takes two of the four quantities (P₁, P₂, T₁, T₂) with ΔH_vap and returns the missing one, so you can verify a whole problem set instead of one hand calculation.

Open the Clausius–Clapeyron Calculator →

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