Account for each gas in a mixture

Add a component without changing the other component’s amount

The lab fixes the shared vessel at 10 L and 300 K. With 1 mol A and 0.5 mol B, the partial pressures are 249.42 and 124.71 kPa, giving 374.13 kPa total. Increase B to 1 mol while retaining A: B’s partial pressure becomes 249.42 kPa and the total becomes 498.84 kPa. A’s partial pressure stays 249.42 kPa.

Yet A’s fraction of the pressure falls from two thirds to one half. A component can retain the same absolute contribution while becoming a smaller share of the total. Capture both states and distinguish the contribution from its fraction rather than saying only that A “decreases.”

Both gases occupy the whole vessel

For an ideal component i, its partial pressure is Pi = niRT/V. It is the pressure its amount would exert alone at the same temperature in the same full volume. V is not divided into a compartment for each species. The drawing uses labeled inventory markers inside one shared box; their positions do not show a measured distribution or imply physical partitions.

For nonreacting ideal components at equilibrium, Ptotal = PA + PB. This is Dalton’s pressure-addition relation in the scope used here. Since R, T and V are common, total pressure also equals (nA + nB)RT/V. The independent component ledger and the total-amount calculation must agree.

At 300 K in 10 L, each mole contributes 8.314 × 300/10 = 249.42 kPa under the rounded constant used here. Half a mole contributes half as much. The substance labels A and B are placeholders: the model does not assign molecular masses, reaction chemistry or condensation properties.

Fractions need a nonzero denominator

A mole fraction compares a component amount with the total amount. For A, xA = nA/(nA + nB). Divide the corresponding pressure equations and the common RT/V factor cancels, leaving PA/Ptotal = xA for this ideal mixture. A and B mole fractions sum to one when total amount is positive.

If A = 1 mol and B = 0.5 mol, xA = 1/1.5 = 2/3 and xB = 1/3. If B is absent, xA = 1. If both are absent, the model pressure is zero, but xA would require 0/0 and is undefined. Assigning both fractions zero would falsely suggest a defined composition whose fractions sum to zero. The interface keeps that case explicit.

Mole fraction is not generally mass fraction

Suppose, in a separate hypothetical example, A has molar mass 20 g/mol and B has molar mass 40 g/mol. One mole of each gives equal mole and partial-pressure fractions, one half each, but masses of 20 g A and 40 g B. A’s mass fraction is 20/60 = one third. An equal number of moles need not weigh the same.

Conversely, 20 g of each hypothetical gas gives 1 mol A and 0.5 mol B. Their mass fractions are equal, but A supplies two thirds of the ideal total pressure. Convert masses to moles before using the pressure model. These hypothetical molar masses are a worked example, not hidden properties of the lab’s A/B selectors.

Account for additional components before inferring amount

Gas collected over water may include water vapor. A total pressure reading then includes both the target gas and the vapor, under the applicable mixture assumptions. For an original teaching case, suppose total absolute pressure is 100.0 kPa and the supplied equilibrium water-vapor contribution is 3.2 kPa. The target-gas partial pressure is 96.8 kPa, not 100.0 kPa.

Using that 96.8 kPa value in n = PV/(RT) requires the actual shared volume and absolute temperature. The vapor value must match the conditions and equilibrium assumptions. The present two-component model does not calculate saturation, humidity, gas dissolution or water-vapor pressure; it only adds stipulated ideal-gas amounts.

If components react, their individual mole amounts can change. If a component condenses, it no longer all contributes as gas. Intermolecular interactions can also invalidate ideal additivity at relevant densities. The same algebraic ledger does not eliminate the need to establish which physical state and model apply.

Transfer: a vessel at fixed temperature and volume has 0.5 mol A and 1.5 mol B. Its total pressure is 400 kPa. Find each partial pressure. Then state the result if all A is removed while B, temperature and volume remain unchanged.

Check the component-removal case

Total amount is 2 mol, so A supplies 0.5/2 = one quarter and B supplies three quarters. Their partial pressures are 100 and 300 kPa. Removing A leaves B at 300 kPa, now the entire pressure. B's amount and partial pressure remain unchanged, while its fraction rises to one. This is an idealized removal with T and V fixed, not a claim about a particular selective-separation apparatus.

OpenStax gas mixtures and partial pressures provides reference equations. Moles and mass reviews the distinction between amount and mass.