Account for pressure in an ideal-gas mixture

Both gases occupy the whole vessel

The mixture investigation contains nonreacting ideal gases A and B in the same 10 L vessel at 300 K. Changing one amount does not allocate it a separate fraction of the vessel’s volume. Each component’s molecules can explore the full shared space.

The circle and square markers distinguish inventories. Each mark represents 0.25 mol, not one molecule or a molecular size. Marker positions are not observations or a molecular simulation. Use the pressure ledger to reason about amount rather than interpreting spacing between these schematic marks as a density measurement.

Define partial pressure

For component i, Pi = niRT/V. Pi is its partial pressure, ni its amount in moles, and T and V are the common mixture temperature and whole-vessel volume. Partial pressure is what that amount would exert if it alone occupied this same volume at this temperature.

For an ideal nonreacting mixture, Ptotal = PA + PB. With one mole of each gas, each contributes 249.420 kPa, so total pressure is 498.840 kPa. These are absolute pressures. A pressure sensor responds to the combined momentum transfer; it does not provide two separate readings unless additional information about composition is available.

The OpenStax molecular gas discussion provides the kinetic basis for mixture pressure. Interactions, reactions, condensation or unequal nonequilibrium temperatures can require different treatment; this model does not predict them.

A changing fraction need not mean a changing amount

Hold A at 1 mol and set B to 0.5 mol. PA is 249.420 kPa, PB is 124.710 kPa and total is 374.130 kPa. A’s mole fraction is xA = nA/(nA+nB) = 2/3.

Increase B to 1 mol while retaining A, T and V. PA stays 249.420 kPa. Total pressure rises to 498.840 kPa, so A’s fraction falls to 1/2. There is less A as a fraction of the mixture but no less A in the vessel. This is a useful distinction whenever percentages change because the denominator changes.

In this ideal model Pi = xiPtotal. The pressure fraction equals the mole fraction. It does not generally equal mass fraction. For example, equal moles of species with molar masses 20 and 80 g/mol have equal partial pressures but mass fractions 20% and 80%. Their molecular speeds differ, while their average translational energy per molecule is the same at common T.

Handle boundary cases without inventing fractions

Set B to zero and keep A nonzero. B contributes zero pressure, A supplies all pressure and xA = 1. Reverse the settings for pure B.

Set both amounts to zero. The ideal model gives zero pressure, but composition fractions are undefined because the total amount in the denominator is zero. Do not assign 0%, 50% or any other composition to an empty inventory. A real evacuated container can still contain residual gas or vapor; the model’s empty state is an ideal limit.

At fixed volume and temperature, doubling both amounts doubles total pressure but preserves composition. This is different from adding only one component. The temperature control is absent here deliberately: the common temperature is fixed, rather than a hidden outcome of adding gas.

Think about what measurements identify

A total pressure reading, known volume and known temperature determine total ideal amount n = PV/(RT). They do not uniquely determine how that amount is split among species. Many compositions can have the same total pressure. An additional composition measurement or known component amount is needed.

No gas-handling procedure is required for this model exercise. For real recorded measurements, report pressure reference, temperature, volume, sample identity and calibration uncertainty before using a partial-pressure calculation. Do not treat these fictitious gas labels or high-pressure values as instructions for a physical apparatus.

Transfer to a changed composition

An ideal mixture has 0.5 mol A and 1.5 mol B at fixed T and V. Its initial total pressure is 240 kPa. Half of B is removed while all A remains, and the original temperature is restored. Find each initial partial pressure, the final total pressure and A’s final mole fraction.

Check the mixture transfer

Initially A contributes 0.5/2 × 240 = 60 kPa and B contributes 180 kPa. Removing half of B halves its partial pressure to 90 kPa while A remains at 60 kPa. Final total is 150 kPa and xA = 60/150 = 0.4, also 0.5/(0.5+0.75). A's fraction rises without any added A. Holding T and V fixed is essential to these proportional comparisons.