Dalton's law of partial pressures: Question bank — Chemistry deep dive

Before we start, one anchor so no symbol is used unexplained:

True or false — justify

Every prompt: decide T/F, then give the reason. A bare "true" earns nothing.

Partial pressure depends on the volume the gas occupied before mixing.

False.

p_{i} = n_{i} R T / V

uses the final shared

V

. The molecules forgot their old container the moment they filled the new one — only present-volume collisions count.

Dalton's law holds even if the gases chemically react.

False. Reaction changes each

n_{i}

, so the

p_{i}

you'd compute keeps shifting. The law needs fixed, non-reacting mole counts.

If you double the volume at fixed

T

and fixed moles, every partial pressure halves but the mole fractions stay the same.

True. Each

p_{i} \propto 1/ V

so all halve together; but

x_{i} = n_{i} / n_{total}

has no

V

in it, so ratios are untouched.

Adding an inert gas at constant volume and temperature raises the partial pressure of the gases already present.

False. Each original

p_{i} = n_{i} R T / V

is unchanged since its own

n_{i}

V

T

are unchanged. Only

P_{total}

rises. (Their mole fractions drop, though.)

Adding an inert gas at constant volume lowers the mole fraction of the gases already present.

True.

n_{total}

grows while each

n_{i}

is fixed, so each

x_{i} = n_{i} / n_{total}

shrinks — even though the partial pressures themselves don't move.

The mole fraction of a component can exceed 1 if that gas is at very high pressure.

False.

x_{i} = n_{i} / n_{total} \leq 1

always, because

n_{i}

is part of the sum

n_{total}

. Pressure is irrelevant to a count ratio.

For gas collected over water, the barometer reading equals the dry gas pressure.

False. The barometer reads the total, which is dry gas + water vapour. You must subtract aqueous tension:

p_{dry} = P_{total} - p_{water}

. See Vapour Pressure.

Two different ideal gases at the same

p

V

T

contain the same number of moles.

True.

n = p V / R T

has no term for the gas's identity — an ideal gas is "invisible" to its own chemistry, so equal

p, V, T

forces equal

n

Dalton's law follows from molecules not feeling each other.

True. Independence means each gas's collision rate (hence its pressure) adds without interference — exactly why the Kinetic Theory of Gases gives additive pressures.

Real gas mixtures obey Dalton's law perfectly at very high pressure.

False. At high pressure intermolecular forces and finite molecular size matter, so pressures no longer add cleanly — see Real Gases and van der Waals Equation.

Spot the error

Each line states a flawed worked step. Reveal names the mistake and the fix.

"Gas A came from a 2 L flask, gas B from a 3 L flask; mix them, so I add the pressures as measured."

Error: each

p_{i}

was defined in its old volume. Fix: recompute both at the new combined volume

V

using

p_{i} = n_{i} R T / V

, then add.

"The mixture is 40 g O₂ and 60 g N₂, so

x_{O_{2}} = 0.40

Error: that's a mass fraction. Fix: convert to moles first,

n = m / M

;

n_{O_{2}} = 40/32 = 1.25

n_{N_{2}} = 60/28 \approx 2.14

, so

x_{O_{2}} \approx 0.37

, not

0.40

p_{i} = x_{i} P_{total}

, and since I want

x_{i}

I'll use the pressure fraction

p_{i} / P

— oh wait that's circular, so I'll use the volume the gas took before mixing as its fraction."

Error: substituting pre-mix volume for mole fraction. Fix: for ideal gases the mole fraction equals the pressure fraction and equals the volume fraction at the same $p, T$ — but never the pre-mix volume in a different container.

"H₂ over water: total 760 mmHg, so

n_{H_{2}} = (760/760) V / R T

Error: used total instead of dry pressure. Fix: subtract aqueous tension first, then plug the dry

p_{H_{2}}

into

n = p V / R T

"Since

\sum x_{i} = 1

, the partial pressures also sum to 1 atm."

Error: confusing the dimensionless fractions with pressures. Fix:

\sum x_{i} = 1

(unitless) forces

\sum p_{i} = P_{total}

, which is whatever the total is — not automatically 1 atm.

"Adding argon raises

P_{total}

, so by

p_{i} = x_{i} P_{total}

the oxygen's partial pressure rises too."

Error: treating

x_{i}

as constant while

P_{total}

climbs. Fix:

x_{O_{2}}

drops by exactly the factor

P_{total}

rose, so

p_{O_{2}}

stays put — consistent with

p_{O_{2}} = n_{O_{2}} R T / V

being unchanged.

Why questions

Why do partial pressures add rather than average?

Because pressure is a rate of momentum delivery to the walls, and independent molecules deliver momentum simultaneously and separately — rates from independent sources add, they don't average.

Why does the identity of the gas not appear in

p_{i} = n_{i} R T / V

The ideal model strips away size and forces, leaving only how many molecules hit the wall — identity carries no term because ideal molecules are treated as featureless points.

Why must the gases be non-reacting for the law to apply as usually stated?

A reaction converts moles of one species into another, so the

n_{i}

values used in

p_{i} = n_{i} R T / V

are not fixed — the additive sum you compute keeps changing until equilibrium.

Why does collecting a gas over water introduce a correction but collecting over mercury does not?

Water has a significant vapour pressure at room temperature, so it saturates the gas; mercury's vapour pressure is negligibly tiny, so it contributes essentially nothing to subtract.

Why is the pressure fraction

p_{i} / P_{total}

equal to the mole fraction

x_{i}

Dividing

p_{i} = n_{i} R T / V

P_{total} = n_{total} R T / V

cancels the identical

R T / V

, leaving the pure count ratio

n_{i} / n_{total} = x_{i}

Why does Dalton's law break down for real gases whereas it's exact for ideal ones?

Real molecules attract and repel each other, so gas A does feel gas B; the collision rates are no longer independent, so the pressures no longer add exactly.

Edge cases

What is the partial pressure of a component whose mole fraction is exactly 0?

Zero —

p_{i} = x_{i} P_{total} = 0

. A species truly absent contributes no molecules and therefore no wall collisions.

If a mixture is a single pure gas, what is its partial pressure?

It equals the total pressure, since

x_{i} = 1

gives

p_{i} = 1 \cdot P_{total} = P_{total}

. "Partial" collapses to "total" for one component.

Two gases have equal moles; how do their partial pressures compare regardless of their molar masses?

They are equal.

p_{i} = n_{i} R T / V

depends only on

n_{i}

V

T

— molar mass never enters, so identical moles give identical partial pressures.

Collect a gas over water at a temperature so high the aqueous tension equals the atmospheric pressure. What happens to the dry gas pressure?

It drops to zero:

p_{dry} = P_{atm} - p_{water} = 0

. Physically the water boils, and vapour fills the space — you can collect no dry gas.

T \to 0

(approaching absolute zero, ideal-gas limit), what happens to every partial pressure at fixed

n

and

V

Each

p_{i} = n_{i} R T / V \to 0

. With no thermal motion there are no wall collisions, so both the parts and the total vanish together.

Can a gas with the largest mole fraction ever have a smaller partial pressure than another component in the same mixture?

No — in one mixture all components share the same

P_{total}

, so

p_{i} = x_{i} P_{total}

orders partial pressures exactly by mole fraction; larger

x_{i}

always means larger

p_{i}

Recall One-line survival summary

Partial pressure is set by your own moles in the final volume ( $p_{i} = n_{i} R T / V$ ); mole fractions rescale the total ( $p_{i} = x_{i} P_{total}$ ); over water, subtract aqueous tension; and none of it holds if the gases react or get squeezed until they feel each other.

Connections

Dalton's law of partial pressures — the parent this bank stress-tests.
Ideal Gas Equation — every "why" here traces back to $P V = n R T$ .
Mole Fraction and Concentration Terms — the mass-vs-mole trap lives here.
Kinetic Theory of Gases — the additivity-of-collisions justification.
Real Gases and van der Waals Equation — where the additivity fails.
Vapour Pressure — aqueous tension background for the over-water traps.