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JEE Mains Chemistry · Solutions

Henry's Law and Solubility of Gases

The partial pressure of a gas above a solution equals its Henry constant times its mole fraction in the solution, p = KH·x, and the constant depends on the gas, the solvent and the temperature.

Why this matters

Eight PYQs, six of them multiple choice, and three from 2026. Four apply p = KH·x to find a mole fraction, a number of millimoles or a mass of dissolved gas; four ask what the Henry constant depends on and why warm water holds less gas. The arithmetic is short, so the marks go on the partial pressure and the units.

Concept 1 of 2: Henry's law, p = KH·x

A gas dissolves until the pressure it exerts from inside the liquid matches its partial pressure outside. Double the partial pressure and twice as much dissolves. The Henry constant KHK_H is the pressure needed per unit mole fraction, so a LARGE KHK_H means a gas that dissolves poorly.

Definition

  • p=KH xp = K_H\,x: pp is the PARTIAL pressure of the gas, xx its mole fraction in the solution.
  • Partial pressure from air: p=(mole fraction in air)×Ptotalp = (\text{mole fraction in air}) \times P_{\text{total}}.
  • Units of KHK_H are those of pressure; convert pp to the same unit first (1 atm=760 mmHg1\ \text{atm} = 760\ \text{mmHg}, 1 kbar=1000 bar1\ \text{kbar} = 1000\ \text{bar}).
  • Moles dissolved in 1 L of water: water is 1000/18=55.561000/18 = 55.56 mol, and xx is tiny, so n≈55.56 xn \approx 55.56\,x.
  • Log form: log⁡p=log⁡KH+log⁡x\log p = \log K_H + \log x, a straight line of slope 1 and intercept log⁡KH\log K_H.

Henry's law

p=KH xngas≈x×55.56 mol per litre of waterp = K_H\,x \qquad n_{\text{gas}} \approx x \times 55.56 \text{ mol per litre of water}

Worked example

Air at a total pressure of 2 atm contains 20 mol% O2\mathrm{O_2}. For O2\mathrm{O_2} in water, KH=4.0×104K_H = 4.0 \times 10^{4} atm. Find the mole fraction of O2\mathrm{O_2} in the water and the millimoles dissolved in 1 L.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q43Moderate

Example 1 · Solutions · Henry's Law and Solubility of Gases

At 298 K , the mole percentage of N2( g)N_{2}(\text{ }g) in air is 80%80\%. Water is in equilibrium with air at a pressure of 10 atm . What is the mole fraction of N2( g)N_{2}(\text{ }g) in water at 298 K ? (KHK_{H} for N2N_{2} is 6.5×107 mmHg6.5 \times10^{7}\text{ }mmHg)

Use the partial pressure, not the total pressure

Henry's law uses the gas's own partial pressure. For a gas that is 20% of air at 5 atm, p=1p = 1 atm. Putting 5 atm into p=KHxp = K_H x gives a mole fraction five times too large.

Match the pressure unit to KH

If KHK_H is in mmHg, convert the partial pressure from atm to mmHg (×760\times 760) before dividing. Dividing atm by mmHg gives an answer 760 times too small, and that wrong value is usually an option.

Concept 2 of 2: What the Henry constant depends on

KHK_H belongs to a gas-solvent pair at a given temperature. It does not change with concentration while the solution stays dilute. Dissolving a gas gives out heat, so warming the liquid pushes gas out: KHK_H rises with temperature and solubility falls.

Definition

  • Constant with concentration over the ideally dilute range.
  • Different for the same gas in different solvents.
  • Rises with temperature for most gases in water, so solubility falls as the water warms.
  • Larger KHK_H at the same partial pressure means a smaller mole fraction dissolved.
  • Cold water holds more dissolved oxygen than hot water; boiling drives dissolved gases out.
GasTemperatureHenry constant in water (kbar)What it shows
He293 K144.97The largest constant here, so the least soluble gas
H2\mathrm{H_2}293 K69.16About half of helium's constant, so about twice as soluble
N2\mathrm{N_2}293 K76.48Less soluble than oxygen at the same temperature
N2\mathrm{N_2}303 K88.84The constant rises on warming by 10 K, so solubility falls
O2\mathrm{O_2}293 K34.86About 2.2 times as soluble as nitrogen at 293 K
O2\mathrm{O_2}303 K46.82Warmer water holds less oxygen
The same gas at two temperatures: the constant is not fixed for a gas.
CO2\mathrm{CO_2}298 K1.67A small constant: very soluble, which is why soda water holds so much
A larger constant means a less soluble gas; every gas listed at two temperatures has the larger constant at the higher one.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q45Moderate

Example 2 · Solutions · Henry's Law and Solubility of Gases

Given below are two statements: Statement I : The Henry's law constant KHK_{H} is constant with respect to variations in solution's concentration over the range for which the solutions is ideally dilute. Statement II: KHK_{H} does not differ for the same solute in different solvents. In the light of the above statements, choose the correct answer from the options.

KH is not a property of the gas alone

The statement 'KH does not differ for the same gas in different solvents' is false. The constant measures how readily the gas dissolves in that particular liquid, so it changes with the solvent and with temperature.

Warm water holds less gas

Dissolving a gas is exothermic, so raising the temperature drives gas out. The Henry constant rises with temperature and the solubility falls; water near 4 °C holds more oxygen than boiling water.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

  • Henry's law, p = KH·x

    Henry's law

    p=KH xngas≈x×55.56 mol per litre of waterp = K_H\,x \qquad n_{\text{gas}} \approx x \times 55.56 \text{ mol per litre of water}

Reference tables (1)

What the Henry constant depends on7 rows
GasTemperatureHenry constant in water (kbar)What it shows
He293 K144.97The largest constant here, so the least soluble gas
H2\mathrm{H_2}293 K69.16About half of helium's constant, so about twice as soluble
N2\mathrm{N_2}293 K76.48Less soluble than oxygen at the same temperature
N2\mathrm{N_2}303 K88.84The constant rises on warming by 10 K, so solubility falls
O2\mathrm{O_2}293 K34.86About 2.2 times as soluble as nitrogen at 293 K
O2\mathrm{O_2}303 K46.82Warmer water holds less oxygen
The same gas at two temperatures: the constant is not fixed for a gas.
CO2\mathrm{CO_2}298 K1.67A small constant: very soluble, which is why soda water holds so much
A larger constant means a less soluble gas; every gas listed at two temperatures has the larger constant at the higher one.

Watch out for (4)

Test yourself on Solutions

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.