Dalton's law of partial pressures: Visual walkthrough

This is the picture-first companion to Dalton's law of partial pressures. We derive its central result in seven numbered steps.

Step 1 — What "pressure" even is (the wall gets hit)

WHAT. Before any formula, we need to know what the word pressure means physically. Pressure is the push per unit area that gas molecules deliver to a wall by banging into it.

WHY this first. Every symbol later ( $p_{i}$ , $P_{total}$ ) is a pressure. If pressure is a mystery, the whole derivation is a mystery. So we anchor it to a picture of balls hitting a wall.

PICTURE. In the figure, black dots are gas molecules. Each red arrow is one molecule slamming into the right wall and bouncing back. More hits per second, or harder hits, means more pressure.

Figure — Dalton's law of partial pressures

Everything below is a story about these four letters.

Step 2 — One gas alone: the ideal gas law

WHAT. For a single pure gas, experiment and Kinetic Theory of Gases give one clean relationship tying all four letters together:

$push on walls P room V = how many n fixed constant R how hot T$

WHY this tool and not another. We want pressure. This is the only elementary equation that hands us $P$ directly in terms of the count $n$ , the room $V$ , and the heat $T$ . It comes straight from $PV=nRT$. Rearranged for pressure:

$p = \frac{n R T}{V}$

Read the right side like a sentence: more molecules ( $n ↑$ ) → more hits → more pressure; hotter ( $T ↑$ ) → faster hits → more pressure; bigger box ( $V ↑$ ) → hits spread thinner → less pressure (it's in the denominator).

PICTURE. Same box, same balls. The figure shows the three knobs — turn $n$ up, turn $T$ up, or turn $V$ up — and what each does to the pressure gauge.

Step 3 — The key idealisation: colours ignore each other

WHAT. Now put two kinds of gas in the same box — say red balls (gas 1) and grey balls (gas 2). The central assumption of an ideal gas is: molecules take up no room of their own and exert no forces on each other.

WHY it matters. If a red ball can't feel a grey ball, then the red balls behave exactly as if the grey balls were not there — and vice versa. This single sentence is the whole engine of Dalton's law. It is also where the law will later break for real gases, because real molecules do attract each other.

PICTURE. The figure overlays two "ghost" boxes onto one real box: the red gas acting alone, and the grey gas acting alone. Their collision patterns don't interfere — a red hit and a grey hit are separate events.

Step 4 — Give each colour its own pressure

WHAT. Because each gas ignores the other, we may apply the Step 2 formula to each colour separately, using the shared box volume $V$ and the shared temperature $T$ :

$p_{1} = \frac{n _{1} R T}{V}, p_{2} = \frac{n _{2} R T}{V}$

Term by term:

$p_{1}$ = the pressure the red gas would make if it were alone in the box — this is its partial pressure.
$n_{1}$ = number of moles of red gas.
$V, T, R$ = same for both colours, because they physically share the box.

WHY the shared $V$ . This is subtle and important: we do not use whatever volume the red gas occupied before mixing. It now roams the whole box, so its room is the full $V$ .

PICTURE. Two gauges — a red one reading $p_{1}$ , a grey one reading $p_{2}$ — each computed as if its gas owned the entire box alone.

Step 5 — Add the collisions: pressures sum

WHAT. The real gauge on the box measures every hit — red hits and grey hits. Since the hits are independent, the total hit-rate is simply the sum. So the total pressure is the sum of the partials:

$P_{total} = p_{1} + p_{2} = \frac{n _{1} R T}{V} + \frac{n _{2} R T}{V}$

Now factor out the common piece $\frac{R T}{V}$ — it's the same in both terms because $R$ , $T$ , $V$ are shared:

$P_{total} = \frac{( n _{1} + n _{2} ) R T}{V} = \frac{n _{total} R T}{V}$

where $n_{total} = n_{1} + n_{2}$ is just the total count of all molecules.

WHY factoring is legal. Addition of fractions with the same denominator: $\frac{a}{V} + \frac{b}{V} = \frac{a + b}{V}$ . Nothing deeper — but geometrically it says "counting red hits plus grey hits is the same as counting all hits at once."

PICTURE. Two skinny bars (red $p_{1}$ , grey $p_{2}$ ) stack into one tall bar $P_{total}$ . The heights add because the collisions don't interfere.

Step 6 — Scale by the crowd: the mole-fraction form

WHAT. We now ask a new question: what fraction of the total pressure does one colour contribute? Divide the red gas's own equation by the total equation:

$\frac{p _{1}}{P _{total}} = \frac{\frac{n _{1} R T}{V}}{\frac{n _{total} R T}{V}}$

The whole factor $\frac{R T}{V}$ is identical on top and bottom, so it cancels, leaving pure counting:

$\frac{p _{1}}{P _{total}} = \frac{n _{1}}{n _{total}} = x_{1}$

Here $x_{1} = \frac{n _{1}}{n _{total}}$ is the mole fraction — see Mole Fraction and Concentration Terms — the share of red balls in the whole crowd (a number between 0 and 1).

WHY divide. Division is the tool for "what portion of the whole?" And because $R T / V$ is shared, dividing wipes out all the physics and leaves only the headcount ratio. That's why partial pressure depends only on the fractions, not on temperature or volume.

PICTURE. A pie: if red balls are one-fifth of all balls, the red slice of the pressure pie is one-fifth. The figure shows a crowd split $x_{1}$ red / rest grey mapped onto a pressure bar of the same split.

Step 7 — The edge cases (never leave the reader stranded)

WHAT. We test the corners of the formula so no scenario surprises you.

Case A — one gas missing ( $n_{2} = 0$ ). If there are no grey balls, $x_{2} = 0$ so $p_{2} = 0$ , and $P_{total} = p_{1}$ . The formula gracefully collapses to a single pure gas. ✔

Case B — equal amounts. Two gases with $n_{1} = n_{2}$ give $x_{1} = x_{2} = \frac{1}{2}$ , so each carries half the pressure regardless of which gases they are (a helium atom and a huge $SF_{6}$ molecule contribute equally — pressure counts molecules, not mass).

Case C — gas collected over water. Water always adds its own vapour (see Vapour Pressure). The measured pressure is dry gas plus water vapour, so: $p_{dry} = P_{atm} - p_{water}$ where $p_{water}$ = aqueous tension. Forgetting this is the classic trap.

Case D — reacting gases (the law's limit). If the colours react, the counts $n_{i}$ change, so the whole derivation (which assumed fixed $n_{i}$ ) no longer applies. Dalton's law is for non-reacting mixtures only.

PICTURE. Four mini-panels, one per case, each a small pressure bar showing what happens.

The one-picture summary

This single figure compresses all seven steps: two independent gases (red + grey), each computed alone via $p_{i} = n_{i} R T / V$ , their pressure bars stacking into $P_{total}$ , and the same stack re-read as mole fractions $p_{i} = x_{i} P_{total}$ .

Recall Feynman retelling — the whole walkthrough in plain words

Pressure is just balls banging on a wall (Step 1). For one kind of ball, how hard they bang depends on how many there are, how hot, and how big the box — that's $p = n R T / V$ (Step 2). Now mix two colours of balls that can't feel each other (Step 3). Because red ignores grey, we can pretend each colour is alone and give it its own "banging number," its partial pressure (Step 4). The wall feels all the bangs, so we just add the two numbers — and the shared $R T / V$ lets them factor into one total (Step 5). If we ask "what share does red own?", the physics cancels and we're left with pure counting: red's share of the pressure equals red's share of the balls, $p_{i} = x_{i} P$ (Step 6). Finally we poke the corners: no grey balls → back to one gas; equal counts → equal shares no matter the mass; over water → subtract water's own bangs; reacting balls → the counts change, so the whole thing is off-limits (Step 7). That's Dalton's law, drawn.

Recall

Why can we factor $R T / V$ out of $p_{1} + p_{2}$ ? ::: Because $R$ , $T$ , and $V$ are shared by both gases (same box, same temperature), so the term is identical in each partial pressure. In $p_{i} = x_{i} P_{total}$ , why did temperature and volume vanish? ::: Dividing each gas's equation by the total cancels the common $R T / V$ , leaving only the headcount ratio $n_{i} / n_{total}$ . What physical assumption makes the pressures additive? ::: Ideal-gas molecules don't interact, so each gas's collisions are independent of the others.

Connections

Dalton's law of partial pressures — the parent topic this page derives visually.
Ideal Gas Equation — the Step 2 building block $p = n R T / V$ .
Kinetic Theory of Gases — why collisions (hence pressures) are independent and additive.
Mole Fraction and Concentration Terms — supplies $x_{i}$ for Step 6.
Vapour Pressure — the aqueous tension in Case C.
Real Gases and van der Waals Equation — where Step 3's independence assumption fails.