SCIENCE · INVESTIGATE AND EXPLAIN

Keep wet totals separate from dry-gas composition

Subtract a supplied vapor contribution before assigning dry product amount and mole fractions.

Learning outcome: Compare water contributions0 and 3 at total100 kPa; retain dry amount, species pressures and the naive overcount. Check that all partial pressures reconstruct the wet total.

Starting recap

Review amount dimensions, chemical formulas and charge, signed differences, ratio bounds and the distinction between a theoretical constraint and measured conversion.

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Try

At total100 kPa with water partial pressure 3 kPa, dry pressure is 97; a 2:1 dry mixture has partial pressures64.6667 and 32.3333 kPa.

Open the experiment

Predict

Does a dry A:B ratio 2:1 imply A is two-thirds of all particles including water vapor?

Your prediction and reason

Experiment

Compare water contributions0 and 3 at total100 kPa; retain dry amount, species pressures and the naive overcount. Check that all partial pressures reconstruct the wet total.

Separate dry-gas amount from water vapor

Worked teaching model · synthetic values · Hypothetical wet gas in1.000 L at300 K contains dry A:B=2:1 in mole ratio plus supplied water vapor pressure. Ideal mixture, no dissolution/reaction/condensation loss beyond the stipulated vapor contribution. R=8.314 kPa L/(mol K). Not a real collection procedure or water-vapor table.

Measured total absolute pressure: 100 kPa Water vapor partial pressure: 3 kPa

Reset keeps your prediction, captures and practice history.

Calculated evidence under the stated assumptions
Quantity	Value	Unit
Dry-gas pressure	97	kPa
Dry-gas total amount	0.03889023	mol
Dry species A partial pressure	64.66667	kPa
Dry species B partial pressure	32.33333	kPa
Water partial pressure	3	kPa
Total amount including vapor	0.04009302	mol
Naive overcount of dry amount	0.00120279	mol

Mole fractions and partial pressures require a clearly named wet or dry total.

The supplied2:1 ratio applies only to the dry gases, not the full wet mixture. Water vapor adds pressure but is not product A or B. Dissolution, chemical reaction or incomplete saturation would require more evidence; a supplied vapor pressure is not a measured water correction.

Retain the reaction/measurement boundary with every number. This mathematical packet is not a chemical procedure or evidence of actual laboratory technique.

Mole fractions and partial pressures require a clearly named wet or dry total.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Identify the denominator of a mole fraction

For an ideal mixture, each component’s partial pressure is its mole fraction times the total pressure defined over the same mixture. A dry composition ratio excludes water vapor. A wet total includes it. Confusing these two totals can preserve a superficially plausible ratio while assigning the wrong absolute amount .

Correct the virtual collection packet

The activity stipulates1.000 L at300 K, total absolute pressure 100 kPa and water partial pressure 3 kPa. The dry-gas pressure is 97 kPa, and its amount is 97/(8.314×300)≈.03889 mol. Using100 kPa instead gives about .04009 mol, overcounting dry material by approximately.001203 mol.

Partition the dry gas

For dry A:B=2:1, A has partial pressure(2/3)×97≈64.6667 kPa and B about 32.3333. Adding the 3 kPa water contribution reconstructs100. The wet mole fraction of A is 64.6667/100, not2/3; the latter is its dry mole fraction . State which is meant before multiplying by a total amount .

Preserve the assumptions

The supplied water pressure is a hypothetical input, not a measured saturation table. The model omits dissolution of A or B, chemical reaction, unaccounted condensation and collection losses. A real gas could be undersaturated with water, and a dissolved fraction would not be recovered simply by subtracting vapor pressure . Those are distinct evidence questions.

Transfer with a changed pressure

At total105 kPa and water5 kPa, dry pressure is 100 kPa and the same dry ratio gives 66.6667 and 33.3333 kPa. The extra measured pressure has not all become extra dry gas; some belongs to vapor. A sound inference retains the pressure ledger, wet/dry denominators and missing loss mechanisms. This is a supplied calculation, not instructions for collecting or releasing a gas.

Sources and evidence boundary

Conceptual sources checked 30 October 2026: Gas-mixture stoichiometry. All prose, cases, code and diagrams are original. The supplied models are not measured structures, a performed laboratory investigation or certification of independent mastery. No chemical handling, heating, pressure vessel or cryogenic experiment is requested. Retain predictions, contrasting evidence and a changed-case explanation.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Which dissolution, saturation or chemical-reaction assumptions are absent from this correction?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Infer an amount from a gas slope rather than a biased reading Next: Compare molecular speed without confusing it with bulk transport

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