General Chemistry · Solutions · GAS SOLUBILITY · HENRY'S LAW
Gas Solubility and Pressure — Henry's Law Sg = kPg
The concentration of dissolved gas is not an input. It settles by itself from the balance of particles going in and coming out at the surface.
First predict, then push the piston and observe, and finally check yourself.
Estimated time 15–20 min · Prerequisites: partial pressure, molarity, dynamic equilibrium
Learning objectives After this activity you can …
explain that dissolving a gas is a dynamic equilibrium of particles going in and coming out at the surface.
predict how the solubility Sg at the new equilibrium changes when the partial pressure Pg changes, and confirm it by measurement.
find out, by changing conditions and measuring, which conditions change the Henry's law constant k.recognize that the Henry's law constant k is set by the gas, the solvent and the temperature, not by the pressure.
1. Gas molecules keep entering the liquid at the surface (dissolving) and leaving it (escaping). Equilibrium is reached when the two rates are equal.
2. The concentration of dissolved gas at equilibrium is the solubility Sg under those conditions. The gas side is described by the partial pressure Pg. 3. The Henry's law constant k links Sg and Pg. In the activity you find out yourself which change alters k.3. The Henry's law constant k is the proportionality constant between Sg and Pg. It depends on the kind of gas, the solvent and the temperature.
The result summary (what happens to Sg and k when the pressure is doubled) opens after you lock in a prediction.
Result: Sg = kPg — when Pg doubles, S_g also doubles at the new equilibrium, and k (the slope) stays the same. When the temperature rises, k for a gas becomes smaller.
Entry check Before you start — prerequisites and key terms
At constant temperature, you compress the same amount of gas to half its volume. How does the number of gas molecules hitting the liquid surface per unit time change?
It increases (about 2 times)It decreasesIt does not change
Key terms
Partial pressure Pg: the pressure exerted by one gas in a gas mixture. ·
Solubility Sg: the concentration of dissolved gas at equilibrium (mM = mmol/L). ·
Henry's law constant k: the constant in Sg = kPg; it depends on the solute, the solvent and the temperature. ·
Dynamic equilibrium: rate of dissolving = rate of escaping.
Predict Think first, before you observe
At 25 ℃ you push the piston and reduce the gas volume to half, so Pg doubles. What is true after the new equilibrium is reached?
Sg doubles and k stays the samek doubles, so Sg also doublesSg is a fixed property of the substance, so it does not changeSg is proportional to P², so it becomes 4 times larger
ConfidenceMedium
S_g = —Henry's law S_g = k·P_g · compared with the exact value k(T)·P_g
Change the piston, the temperature or the gas. The measured solubility is compared live with the exact Henry's law value.
The explanation opens after you lock in a prediction
At equilibrium, gas molecules enter the solution as fast as they leave it. When the piston halves the volume,
the gas pressure doubles and more molecules hit the surface (going in). The solubility then rises until the dissolved concentration is high enough for the escape rate to match again.
The same law applies to O₂ and CO₂ dissolved in paddy-field water and in nutrient solutions for hydroponics.
Sg = k Pg
Henry's law · k depends on the solute, the solvent and the temperature
A straight line through the origin
A plot of solubility against partial pressure is a straight line through the origin, and its slope is k. In water at 25 ℃ the slopes are in the order
O₂ > CO > N₂ > He. Example: N₂ at 79.03 kPa gives 4.75×10⁻⁴ M, k = 6.1×10⁻⁶ mol/m³·Pa.
k = Sg / Pg
Changing the pressure does not change the slope (k) · temperature ↑ → k of a gas ↓
Limits Limits of the model
This is a 2D particle model. In real water, the dissolved O₂ is only a few % (or less) of the molecules in the gas phase, so this model ignores depletion of the gas phase
and draws the dissolved particles on a different scale (1 particle = a fixed number of mM). The mM value is the measured number of dissolved particles × the scale; the exact value k(T)·Pg is calculated from literature constants.
The reaction of CO₂ with water (forming H₂CO₃) and deviations from Henry's law at high pressure or in concentrated solutions are not included.
Check Formative check
The check and reflection open after you lock in a prediction. (They are hidden before the prediction so that the questions do not hint at the result.)
Check Formative check — pass all items to complete
Q1The temperature stays the same, and only the partial pressure of the gas above the solution is doubled. Which statement is correct once equilibrium is reached again?
The Henry's law constant has doubledThe Henry's law constant has halvedThe number of gas molecules in the liquid has doubledThe number of gas molecules in the liquid has not changed
Hint
On the graph, look separately at the direction in which the measured point moves and at whether the slope of the line changes.
Q2 · Hands-on taskWith O₂ selected (25 ℃), use the piston to make Pg about 2 times (180–225 kPa). Wait for the new equilibrium, then press [Check].
Now: P_g — · S_g — · equilibrium reached —
Q3 · Calculation—
Hint
S_g = kP_g. Convert the units first: 1 mol/m³ = 1 mM, 1 MPa = 10⁶ Pa.
Q4 · Fill in the blankFill in the blank: The Henry's law constant k depends on the solute, the solvent and the [_____], and it does not change when the gas pressure changes.
Hint
Move the second slider in the panel: the slope of the line on the graph changes.
Q5 · Apply (higher-order, not required for completion)At 25 ℃, which dissolves more in water (in mM): O₂ at 50 kPa or He at 100 kPa? Decide from the slopes on the graph.
O₂ at 50 kPaHe at 100 kPa (the pressure is 2 times higher)The same
Reflect Reflect — metacognition · completion
My prediction: Not predicted yet
When the pressure is doubled, the solubility doubles. In one sentence, explain why k stays the same:
Apply Use the idea in a new context (not required for completion)
When you open a bottle of soda, bubbles rush up. A diver who comes up too fast from deep water can get decompression sickness from N₂ bubbles in the blood.
Explain each effect in one sentence using partial pressure and Henry's law.
Next up: gas solubility and temperature (solubility–temperature graph) → colligative properties (vapor-pressure lowering, Raoult's law)
COMPLETE · Completed — all checks passed.
Worksheet — Gas solubility: Henry's law Sg = kPg
Name ________ Class ________ Date ________
1. Predict — When the volume is halved, what happens to Sg at the new equilibrium? And to k? ( doubles / stays the same / halves )
2. Observe — gas ______ , 25 ℃ — V=100%: P=____ kPa, S=____ mM / V=50%: P=____ kPa, S=____ mM / S÷P = ____ , ____
3. Graph — Plot the two measured points on the empty axes above and join them to the origin. What is the slope of the line?
4. Fill in the blank — k depends on the solute, the solvent and the [_____].
5. Calculate — At 25 ℃, k(CO₂) = 3.4×10⁻⁴ mol/m³·Pa (= 3.4×10⁻⁴ mol L⁻¹ kPa⁻¹) and P = 0.41 MPa. S = ______ M
Answer key (teacher only, separate) 1. S doubles, k stays the same · 2. S/P constant · 3. k · 4. temperature · 5. 0.14 M
Simulation: Gas solubility and Henry's law (print again from [Print worksheet] on the same page)
Teacher notes — open the lesson plan
Lesson plan · 5E phase and time
Explore · Explain. 15–20 min. Whole-class demo (?mode=teacher) or one device per student.
Demo in 3 steps
(1) O₂, 25 ℃, V=100%: equilibrium with similar numbers of dissolve and escape arrows, S ≈ 1.32 mM (k·101.3 kPa).
(2) V=50%: P doubles at once to 202.7 kPa → transient state (dissolving > escaping) → after about 5 s, S ≈ 2.63 mM (2 times), S/P unchanged.
(3) Raise the temperature to 40 ℃ or switch to He → the slope of the line (k) changes. Temperature and the kind of gas change k; pressure does not.
3 key questions
"The moment you push the piston, does S double at once? What changes first?"
"When you doubled the pressure, the point on the graph moved. Did the line move?"
"Why does a warm soft drink go flat faster?"
Common misconceptions
MC-sol-01: "Doubling the pressure doubles the Henry's law constant k" → k is the slope; the pressure is the position on the line. Changing P only moves the point along the line.
MC-sol-02: "Solubility is a fixed property of the substance and does not depend on pressure" → roughly true for solids and liquids, but gas solubility is directly proportional to the partial pressure.
Answer key
Prediction: S doubles, k stays the same · Q1 (c) · Q2 V 50–55% at equilibrium · Q4 temperature · Q5 O₂. Q3 by level is shown in the teacher bar.
Sources of values
N₂ k = 6.1×10⁻⁶ mol/m³·Pa (25 ℃), CO₂ k = 3.4×10⁻⁴ mol/m³·Pa (= 3.4×10⁻⁴ mol L⁻¹ kPa⁻¹) (25 ℃).
Literature values: O₂ 1.3×10⁻⁵, CO 9.7×10⁻⁶, He 3.8×10⁻⁶ mol/m³·Pa (25 ℃); temperature dependence k(T) = k₂₉₈·exp[C(1/T − 1/298.15)],
C = O₂ 1500, CO 1300, N₂ 1300, He 92, CO₂ 2400 K. Sources: R. Sander, Compilation of Henry's law constants for water as solvent — henrys-law.org (v5.0.0: Atmos. Chem. Phys. 23, 10901–12440 (2023); earlier edition 15, 4399–4981 (2015)) and the NIST Chemistry WebBook (SRD 69) Henry's law tables.
O₂ 1500 K = recommended by henrys-law.org (Burkholder et al. 2019, type L) and NIST (method L); He 92 K = NIST (method L, k°H 0.00038 mol/(kg·bar)) and Wilhelm et al. (1977) as listed in henrys-law.org. An earlier He value of 230 K had no literature support and was replaced (correction of k at 0 ℃ from +7.3% to +2.9%).
(At 100 kPa these give O₂ ≈1.3, CO ≈1.0, N₂ ≈0.65, He ≈0.38 mM. For N₂, Sander gives 6.4×10⁻⁶, about 5% different — the reference value is used.)
Model description
150 gas particles (fixed) hit the surface and form dissolved particles with probability p_in ∝ k(T)·T; dissolved particles escape at the surface with probability 0.5.
The equilibrium number of dissolved particles emerges in proportion to the collision frequency (∝ P). In external checks (long headless averages) it agrees with the exact value kP within ±3% on average, and a single snapshot agrees within about ±5% (mission tolerance: Basic 20%, Standard 15%, Advanced 10%). Revision r2 (2026-10-03): the mission is not judged from a snapshot. After compression, 1200 steps are discarded; the mission succeeds when the whole 95% confidence interval of the mean of 8–10 blocks of 600 steps lies within kP ± tolerance (E1 confidence-interval test, false-negative test with 200 seeds).
Note on the N₂ constant
The often-quoted relation "(4.75×10⁻⁴ mol/L)/79.03 kPa = 6.1×10⁻⁶ mol/m³·Pa" is slightly off: 0.475 mol/m³ ÷ 79,030 Pa = 6.0×10⁻⁶ (6.01×10⁻⁶). It probably comes from converting 6.1×10⁻⁴ M/atm. The simulation uses 6.1×10⁻⁶, so 79.03 kPa gives 0.482 mM (1.5% above 0.475 mM). A student who uses 0.475 is still within the tolerance for Basic Q3.
The literature N₂ value is 6.4×10⁻⁶ (henrys-law.org).
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