Dalton's law of partial pressures: Worked examples — Chemistry deep dive

Before we start, one reminder of the two tools we lean on. Everything below is either:

the ideal gas law $p V = n R T$ — a single gas's pressure from its moles, volume, temperature (see Ideal Gas Equation), or
Dalton's law $P_{total} = \sum_{i} p_{i}$ and its child $p_{i} = x_{i} P_{total}$ (the parent note).

Nothing new is invented here — we just aim these two tools at every awkward angle.

Cell	Case class	The twist it tests	Example
C1	Moles given, same volume	Plain additive pressures	Ex 1
C2	Percentages / mole fraction	$p_{i} = x_{i} P$ backwards & forwards	Ex 2
C3	Masses given (not moles)	Convert mass → moles first	Ex 3
C4	Gases from different containers merged	Re-compute each $p_{i}$ at the new shared $V$	Ex 4
C5	Gas collected over water	Subtract aqueous tension	Ex 5
C6	Degenerate: single gas, or $x_{i} \to 0$ , $x_{i} \to 1$	Limiting behaviour, sanity of the formula	Ex 6
C7	Reacting gases — Dalton breaks mid-problem	Recompute moles after reaction, then apply	Ex 7
C8	Real-world word problem (a diver)	Strip the story down to $p_{i} = x_{i} P$	Ex 8

C1 — Moles given, same volume

Worked example Example 1 (cell C1)

$4 mol$ of He and $1 mol$ of Ar share a rigid $8.0 L$ flask at $273 K$ . Find each partial pressure and the total.

Forecast: More moles of He, so its partial pressure should be bigger than Ar's — by a factor of $4$ , since they share the same $V, T$ . Guess before reading on.

Step 1 — Partial pressure of He. $p_{He} = \frac{n _{He} R T}{V} = \frac{4 \times 0.0821 \times 273}{8.0} = 11.21 atm$ Why this step? By the definition of partial pressure, we imagine He alone in the same flask and use $p V = n R T$ . The other gas is invisible to it.

Step 2 — Partial pressure of Ar. $p_{Ar} = \frac{1 \times 0.0821 \times 273}{8.0} = 2.80 atm$ Why this step? Same reasoning, now imagining Ar alone.

Step 3 — Add them (Dalton). $P_{total} = 11.21 + 2.80 = 14.01 atm$ Why this step? Independent molecules ⇒ independent, additive collisions ⇒ pressures sum.

Verify: Shortcut — total moles $= 5$ , so $P = \frac{5 \times 0.0821 \times 273}{8.0} = 14.01 atm$ . ✓ Matches. Ratio $p_{He} / p_{Ar} = 11.21/2.80 = 4$ , exactly the mole ratio — our forecast holds. Units: $\frac{mol \cdot L atm mol ^{- 1} K ^{- 1} \cdot K}{L} = atm$ ✓

C2 — Percentages / mole fraction

Worked example Example 2 (cell C2)

Dry air is (by moles) about $78%$ N $_{2}$ , $21%$ O $_{2}$ , $1%$ Ar. Atmospheric pressure is $1.00 atm$ . Find the three partial pressures, and confirm they rebuild the total.

Forecast: Each partial pressure is just its percentage of $1 atm$ — so roughly $0.78$ , $0.21$ , $0.01 atm$ . This is almost too easy; the point is to check they sum back to $1$ .

Step 1 — Mole fractions. $x_{N_{2}} = 0.78, x_{O_{2}} = 0.21, x_{Ar} = 0.01$ . Why this step? "Percentage by moles" is the mole fraction — no conversion needed (contrast Ex 3, where it's a mass percent).

Step 2 — Apply $p_{i} = x_{i} P$ . $p_{N_{2}} = 0.78 \times 1.00 = 0.78, p_{O_{2}} = 0.21, p_{Ar} = 0.01 atm$ Why this step? $p_{i} = x_{i} P$ is Dalton's law rearranged — each gas gets its share of the total (see Mole Fraction and Concentration Terms).

Verify: $0.78 + 0.21 + 0.01 = 1.00 atm = P_{total}$ ✓ Because $\sum x_{i} = 1$ , the parts always rebuild the whole. The figure below shows this "slicing" of the total.

Figure — Dalton's law of partial pressures

C3 — Masses given, not moles

Worked example Example 3 (cell C3)

A vessel holds $28 g$ of N $_{2}$ and $8 g$ of He. Total pressure is $6.0 atm$ . Find each partial pressure.

Forecast: He is very light ( $M = 4$ ), so $8 g$ is a lot of moles — I bet He, despite weighing less, dominates the pressure. Guess the ordering first.

Step 1 — Convert masses to moles. Molar masses: N $_{2} = 28 g mol^{- 1}$ , He $= 4 g mol^{- 1}$ . $n_{N_{2}} = \frac{28}{28} = 1 mol, n_{He} = \frac{8}{4} = 2 mol$ Why this step? Pressure depends on the number of molecules; $n = m / M$ turns weight into count.

Step 2 — Mole fractions. Total $= 3 mol$ . $x_{N_{2}} = \frac{1}{3} = 0.333, x_{He} = \frac{2}{3} = 0.667$ Why this step? Now we're allowed to use fractions — because they're mole fractions.

Step 3 — Partial pressures. $p_{N_{2}} = 0.333 \times 6.0 = 2.0 atm, p_{He} = 0.667 \times 6.0 = 4.0 atm$ Why this step? $p_{i} = x_{i} P$ .

Verify: $2.0 + 4.0 = 6.0 atm$ ✓ And yes — the lighter gas has the higher partial pressure (forecast confirmed), because it packed more molecules into the same $8 g$ . Had we (wrongly) used mass fractions, we'd have said He is $8/36 = 22%$ — completely wrong.

C4 — Gases from different containers, then merged

Worked example Example 4 (cell C4)

A $2.0 L$ bulb of O $_{2}$ at $3.0 atm$ is connected to a $3.0 L$ bulb of N $_{2}$ at $1.0 atm$ ; both at constant $T$ . The valve opens and gases mix through the total $5.0 L$ . Find each partial pressure and the total.

Forecast: Both gases expand into more volume, so each partial pressure should drop below its original value. Total should sit somewhere between $1$ and $3 atm$ .

Step 1 — Use Boyle's law (constant $n, T$ ) for each gas separately. For a fixed amount of gas at fixed $T$ , $p V$ is constant, so $p_{new} = p_{old} \frac{V _{old}}{V _{new}}$ . $p_{O_{2}} = 3.0 \times \frac{2.0}{5.0} = 1.2 atm$ Why this step? The O $_{2}$ molecules — same count, same $T$ — now roam $5.0 L$ instead of $2.0 L$ . Its partial pressure is what it exerts alone in that full $5.0 L$ .

Step 2 — Same for N $_{2}$ . $p_{N_{2}} = 1.0 \times \frac{3.0}{5.0} = 0.60 atm$ Why this step? N $_{2}$ 's molecules now spread across $5.0 L$ instead of $3.0 L$ .

Step 3 — Add (Dalton). $P_{total} = 1.2 + 0.60 = 1.80 atm$ Why this step? Now that both are expressed at the same $V$ and $T$ , Dalton lets us add.

Verify: Both partial pressures dropped ( $3.0 \to 1.2$ , $1.0 \to 0.60$ ) as forecast, and $1.80$ lies between $1$ and $3$ ✓ Sanity: total moles-worth $= p_{O_{2}} V + p_{N_{2}} V$ scaling is consistent because $p V$ conserved for each species.

C5 — Gas collected over water

Worked example Example 5 (cell C5)

$O_{2}$ is collected over water at $2 3^{\circ} C$ ; the total pressure inside the tube reads $760 mmHg$ ; the aqueous tension at $2 3^{\circ} C$ is $21 mmHg$ . The volume is $250 mL$ . Find the moles of dry O $_{2}$ .

Forecast: The barometer reading includes invisible water vapour, so dry O $_{2}$ pressure must be slightly less than $760 mmHg$ — by exactly $21$ .

Step 1 — Subtract aqueous tension. $p_{O_{2}} = 760 - 21 = 739 mmHg = \frac{739}{760} = 0.9724 atm$ Why this step? Over water the gas is saturated with vapour; the tube's pressure is $p_{dry} + p_{water}$ , so peel off the water part (see Vapour Pressure).

Step 2 — Convert temperature and volume. $T = 23 + 273 = 296 K$ , $V = 0.250 L$ . Why this step? The ideal gas law needs Kelvin and litres (with $R = 0.0821$ ).

Step 3 — Ideal gas law for the dry gas. $n = \frac{p _{O_{2}} V}{R T} = \frac{0.9724 \times 0.250}{0.0821 \times 296} = 0.01001 mol$ Why this step? Only the dry gas obeys $p V = n R T$ with the number of O $_{2}$ molecules we care about.

Verify: $\approx 0.0100 mol$ — a reasonable amount for $250 mL$ near room conditions ( $22.4 L$ per mole would give $\sim 0.011$ at STP; we're close, slightly less because $T > 273$ ) ✓

C6 — Degenerate cases (single gas, $x \to 0$ , $x \to 1$ )

Worked example Example 6 (cell C6)

(a) A flask contains only CO $_{2}$ at $2.5 atm$ . What is its partial pressure and mole fraction? (b) Into that flask we inject a trace of Ne: $x_{Ne} = 0.001$ . What is $p_{Ne}$ , and what happens to it as $x_{Ne} \to 0$ ?

Forecast: (a) With one gas, "partial" and "total" must be the same thing. (b) A trace gas should push almost nothing.

Step 1 — Single-gas limit. With one component, $x_{CO_{2}} = \frac{n _{CO_{2}}}{n _{CO_{2}}} = 1$ , so $p_{CO_{2}} = x_{CO_{2}} P = 1 \times 2.5 = 2.5 atm = P_{total}$ Why this step? When the sum has one term, $\sum_{i} p_{i} = p_{1} = P_{total}$ trivially. The law degenerates into "the gas's own pressure."

Step 2 — Trace-gas limit. Suppose total is held at $2.5 atm$ with $x_{Ne} = 0.001$ : $p_{Ne} = 0.001 \times 2.5 = 0.0025 atm$ Why this step? $p_{i} = x_{i} P$ scales linearly with $x_{i}$ .

Step 3 — Take the limit. As $x_{Ne} \to 0$ , $p_{Ne} = x_{Ne} \cdot 2.5 \to 0$ . As $x_{Ne} \to 1$ , the other gases' fractions $\to 0$ and $p_{Ne} \to P_{total}$ . Why this step? This confirms the formula smoothly connects "gas absent" ( $p = 0$ ) to "gas is everything" ( $p = P_{total}$ ) — no blow-ups, no gaps.

Verify: (a) $p_{CO_{2}} = 2.5 = P$ ✓ (b) $p_{Ne} = 0.0025 atm$ , and $lim_{x \to 0} x P = 0$ ✓ Both edges behave; the formula is trustworthy everywhere in between.

C7 — Reacting gases: where Dalton breaks (and how to save it)

Worked example Example 7 (cell C7)

$2 mol$ H $_{2}$ and $1 mol$ O $_{2}$ are sealed in a rigid $10 L$ vessel at $400 K$ and sparked: $2 H_{2} + O_{2} \to 2 H_{2} O (g)$ Find the final total pressure (all species gaseous). Compare with the naive "just add initial partials" answer.

Forecast: The reaction turns $3 mol$ of reactant gas into $2 mol$ of water vapour — fewer molecules, so the pressure should drop versus naively summing initial partials.

Step 1 — Naive (wrong) approach, to see the trap. Initial total moles $= 3$ : $P_{naive} = \frac{3 \times 0.0821 \times 400}{10} = 9.85 atm$ Why show this? This is the answer a careless student gives — it ignores that gas got consumed.

Step 2 — Do the stoichiometry. $2 mol$ H $_{2}$ reacts with exactly $1 mol$ O $_{2}$ (they're in the perfect $2 : 1$ ratio, nothing left over) to give $2 mol$ H $_{2}$ O vapour. Final moles: H $_{2} = 0$ , O $_{2} = 0$ , H $_{2}$ O $= 2 mol$ . Total $n_{final} = 2 mol$ . Why this step? Dalton only applies to the actual gas present after the change.

Step 3 — Apply the gas law to the final mixture. $P_{final} = \frac{2 \times 0.0821 \times 400}{10} = 6.57 atm$ Why this step? Same $V, T$ , but now the true final mole count.

Verify: $6.57 < 9.85$ — pressure dropped as forecast, because $3 mol \to 2 mol$ ✓ Ratio $6.57/9.85 = 0.667 = 2/3$ , exactly the mole ratio ✓ Lesson: apply Dalton after the reaction, never before. (Underlying molecular reason: Kinetic Theory of Gases — fewer molecules ⇒ fewer wall collisions ⇒ less pressure.)

C8 — Real-world word problem (a diver)

Worked example Example 8 (cell C8)

A scuba diver breathes a mix that is $x_{O_{2}} = 0.16$ and $x_{He} = 0.84$ . At $30 m$ depth the total pressure is $4.0 atm$ . Divers care about the partial pressure of oxygen (too low = blackout, too high = toxic). Find $p_{O_{2}}$ at depth, and compare with its value at the surface ( $1.0 atm$ ).

Forecast: The oxygen fraction stays $0.16$ , but total pressure is $4 \times$ bigger underwater — so $p_{O_{2}}$ should be about $4 \times$ its surface value.

Step 1 — Strip the story to the formula. All the "diving" language is decoration; the physics is one line: $p_{O_{2}} = x_{O_{2}} P_{total}$ . Why this step? Recognising which formula the word problem hides is the real skill.

Step 2 — At depth. $p_{O_{2}} = 0.16 \times 4.0 = 0.64 atm$ Why this step? Direct application of $p_{i} = x_{i} P$ .

Step 3 — At the surface. $p_{O_{2}}^{surface} = 0.16 \times 1.0 = 0.16 atm$ Why this step? Same mix, lower total pressure.

Verify: $0.64/0.16 = 4$ — exactly the pressure ratio, as forecast ✓ This is why divers reduce O $_{2}$ fraction at depth: at $4 atm$ a normal-air $x_{O_{2}} = 0.21$ would give $0.84 atm$ of O $_{2}$ — heading toward toxic. Units: (dimensionless fraction) $\times$ atm $=$ atm ✓

Recall Quick self-test across the matrix

Which cell needs mass→mole conversion first? ::: C3 — Dalton uses mole fractions, never mass fractions. Two bulbs at different pressures are joined — what do you do before adding? ::: Recompute each $p_{i}$ at the new shared volume (C4) using $p V =$ const per gas. A gas is collected over water; what must you subtract? ::: The aqueous tension (saturated water vapour pressure) — cell C5. Gases react inside the flask — when do you apply Dalton? ::: Only after stoichiometry gives the final moles — cell C7. As a gas's mole fraction $\to 0$ , what happens to its partial pressure? ::: $p_{i} = x_{i} P \to 0$ — cell C6.

Connections

Dalton's law of partial pressures — the parent this page drills.
Ideal Gas Equation — the $p V = n R T$ used in C1, C4, C5, C7.
Mole Fraction and Concentration Terms — the $x_{i}$ used throughout.
Kinetic Theory of Gases — why fewer molecules mean less pressure (C7).
Real Gases and van der Waals Equation — where these clean answers start to fail.
Vapour Pressure — the aqueous tension of C5.

Worked examples — Dalton's law of partial pressures

The scenario matrix