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

This page is a self-test. Read each problem, try it with pen and paper, then open the collapsible solution. Everything you need was built in the parent note. If a symbol feels new, that means: go re-read the parent, then come back.

Constants used throughout: gas constant $R = 0.0821 L\cdotatm\cdotmol^{- 1} K^{- 1}$ , and $1 atm = 760 mmHg$ . Temperature always in kelvin ( $T_{K} = T_{^{\circ} C} + 273$ ).

The three tools we lean on, and why each one:

Level 1 — Recognition

(Can you spot which tool applies and plug numbers in?)

Recall Solution L1·Q1

WHAT: we know moles and volume, want He's own pressure → T1. WHY T1: each gas fills the whole $5 L$ on its own (that is the definition of partial pressure — the gas alone in the shared volume). $p_{He} = \frac{n _{He} R T}{V} = \frac{1 \times 0.0821 \times 273}{5} = 4.48 atm$ Answer: $4.48 atm$ .

Recall Solution L1·Q2

WHAT: total pressure known, fraction known → T2. WHY: " $25%$ by moles" is the mole fraction: $x_{Ar} = 0.25$ . $p_{Ar} = x_{Ar} P_{total} = 0.25 \times 8 = 2 atm$ Answer: $2 atm$ .

Recall Solution L1·Q3

WHAT: gas over water → T3. WHY: the barometer sees everything pushing on it — dry O $_{2}$ plus water vapour. Subtract the vapour. $p_{O_{2}} = 745 - 25 = 720 mmHg$ Answer: $720 mmHg$ .

Level 2 — Application

Recall Solution L2·Q1

Route A (T1 — each alone): $p_{N_{2}} = \frac{2 \times 0.0821 \times 300}{20} = 2.463 atm$ $p_{O_{2}} = \frac{1 \times 0.0821 \times 300}{20} = 1.2315 atm$ $p_{CO_{2}} = \frac{1 \times 0.0821 \times 300}{20} = 1.2315 atm$ $P_{total} = 2.463 + 1.2315 + 1.2315 = 4.926 atm$ Route B (T2 — cross-check via total moles): $n_{total} = 4$ , so $P_{total} = \frac{4 \times 0.0821 \times 300}{20} = 4.926 atm ✓$ Then $x_{N_{2}} = 2/4 = 0.5$ , $p_{N_{2}} = 0.5 \times 4.926 = 2.463 atm$ ✓ — both routes agree. Answers: $p_{N_{2}} = 2.463$ , $p_{O_{2}} = p_{CO_{2}} = 1.232$ , $P_{total} = 4.926 atm$ .

Recall Solution L2·Q2

WHAT: we are given mass, but Dalton uses mole fraction → convert first ( $n = m / M$ ). $n_{O_{2}} = \frac{8}{32} = 0.25 mol, n_{N_{2}} = \frac{14}{28} = 0.5 mol$ $x_{O_{2}} = \frac{0.25}{0.25 + 0.5} = \frac{0.25}{0.75} = \frac{1}{3}$ $p_{O_{2}} = \frac{1}{3} \times 3 = 1 atm$ Answer: $1 atm$ . (Notice: equal masses would not give equal fractions — moles rule.)

Recall Solution L2·Q3

Step 1 (T3): $p_{H_{2}} = 760 - 27 = 733 mmHg = \frac{733}{760} = 0.9645 atm$ . Step 2 (T1 rearranged): $T = 27 + 273 = 300 K$ , $V = 0.250 L$ . $n = \frac{p V}{R T} = \frac{0.9645 \times 0.250}{0.0821 \times 300} = 0.00979 mol$ Answer: $\approx 9.79 \times 1 0^{- 3} mol$ .

Level 3 — Analysis

Figure — Dalton's law of partial pressures

Recall Solution L3·Q1

WHAT & WHY: partial pressure is defined at the final shared volume. When the valve opens, each gas now spreads through $V = 2 + 3 = 5 L$ . Each gas expands independently, so use Boyle's law ( $p_{1} V_{1} = p_{2} V_{2}$ at fixed $T$ , $n$ ) for each. N $_{2}$ : $p_{N_{2}} = \frac{3 \times 2}{5} = 1.2 atm$ . O $_{2}$ : $p_{O_{2}} = \frac{2 \times 3}{5} = 1.2 atm$ . Total (Dalton): $P_{total} = 1.2 + 1.2 = 2.4 atm$ . Answers: $p_{N_{2}} = p_{O_{2}} = 1.2 atm$ , $P = 2.4 atm$ .

Recall Solution L3·Q2

WHAT: invert T2 to get the fraction. $x_{CH_{4}} = \frac{p _{CH_{4}}}{P _{total}} = \frac{0.6}{2.5} = 0.24$ Answer: $24%$ by moles.

Recall Solution L3·Q3

WHAT: average molar mass is the mole-fraction-weighted average — the same $x_{i}$ Dalton uses. Moles: N $_{2}$ from $\frac{3 \times 2}{R T}$ , O $_{2}$ from $\frac{2 \times 3}{R T}$ — both give $p V = 6$ , so equal moles, $x = 0.5$ each. $\overset{ˉ}{M} = x_{N_{2}} M_{N_{2}} + x_{O_{2}} M_{O_{2}} = 0.5 (28) + 0.5 (32) = 30 g\cdotmol^{- 1}$ Answer: $30 g\cdotmol^{- 1}$ .

Level 4 — Synthesis

Recall Solution L4·Q1

WHY Dalton alone is not enough: Dalton's law needs the current moles. A reaction changes moles, so we must first do the stoichiometry, then apply Dalton. Reaction: $2 H_{2} + O_{2} \to 2 H_{2} O$ . $1 mol$ O $_{2}$ consumes $2 mol$ H $_{2}$ and makes $2 mol$ H $_{2}$ O. Left over: H $_{2}$ : $4 - 2 = 2 mol$ ; O $_{2}$ : $0$ ; H $_{2}$ O: $2 mol$ . New total moles: $n_{f} = 2 + 0 + 2 = 4 mol$ (was $n_{i} = 5$ ). At fixed $V, T$ , pressure $\propto$ moles: $P_{f} = P_{i} \cdot \frac{n _{f}}{n _{i}} = 10 \times \frac{4}{5} = 8 atm$ Answer: $8 atm$ . (Dalton applies to the equilibrium/final mixture — see Kinetic Theory of Gases for why pressure tracks moles.)

Recall Solution L4·Q2

Subtle point: the volume of the tube is shared by both the gas and the water vapour — they occupy the same $V$ . So we can use the total pressure with the total moles, OR the gas's partial pressure with the gas's moles. Use the gas we know moles of. $p_{gas} = 760 - 14 = 746 mmHg = \frac{746}{760} = 0.9816 atm$ . $T = 17 + 273 = 290 K$ . $V = \frac{n _{gas} R T}{p _{gas}} = \frac{0.02 \times 0.0821 \times 290}{0.9816} = 0.485 L$ Answer: $\approx 0.485 L$ ( $485 mL$ ).

Level 5 — Mastery

Recall Solution L5·Q1

Let the shared mass be $m$ grams each. $n_{He} = \frac{m}{4}, n_{CH_{4}} = \frac{m}{16}$ $x_{He} = \frac{m /4}{m /4 + m /16} = \frac{1/4}{1/4 + 1/16} = \frac{4/16}{4/16 + 1/16} = \frac{4}{5} = 0.8$ $x_{CH_{4}} = 1 - 0.8 = 0.2$ $p_{He} = 0.8 \times 4 = 3.2 atm, p_{CH_{4}} = 0.2 \times 4 = 0.8 atm$ Answers: $p_{He} = 3.2 atm$ , $p_{CH_{4}} = 0.8 atm$ . The light gas dominates the pressure because equal mass = many more light molecules.

Recall Solution L5·Q2

Ideal part: total moles $= 2$ in $V = 2 L$ : $P_{ideal} = \frac{n _{total} R T}{V} = \frac{2 \times 0.0821 \times 300}{2} = 24.63 atm$ Real part (qualitative): CO $_{2}$ has significant attractive intermolecular forces (large van der Waals $a$ ). Attractions pull molecules inward, softening wall collisions → the real total pressure is lower than $24.63 atm$ . He, being nearly ideal, barely deviates. So Dalton's additivity overestimates here. See Real Gases and van der Waals Equation. Answer: $P_{ideal} = 24.63 atm$ ; real total is somewhat lower due to CO $_{2}$ attractions.

Recall Solution L5·Q3

Step 1 (T3): dry mixture pressure $= 760 - 27 = 733 mmHg$ . This is $P_{dry total}$ — the water vapour is a separate component and is not part of the "dry" fractions. Step 2 (T2): among the dry gases, $x_{H_{2}} = 0.30$ , so $p_{H_{2}} = 0.30 \times 733 = 219.9 mmHg$ Answer: $\approx 220 mmHg$ .

Recall One-line self-audit before you leave

For every problem you either (a) used T1 because moles+volume were given, (b) used T2 because a total pressure and a fraction were given, or (c) used T3 because the gas was over water. If a reaction appeared, you did stoichiometry first. If real gases appeared, you flagged that Dalton over/under-estimates. Did you? Then you have mastered Dalton's law of partial pressures.

Connections

Ideal Gas Equation — every T1 and T3 step is $P V = n R T$ .
Mole Fraction and Concentration Terms — the $x_{i}$ behind T2 and average molar mass.
Kinetic Theory of Gases — why $P \propto n$ at fixed $V, T$ .
Real Gases and van der Waals Equation — L5·Q2's deviation.
Vapour Pressure — aqueous tension in every over-water problem.