Dalton’s Law and Partial Pressures

Last updated: 17 August 2026

Diagnose → Learn → Master: Every study-note section includes 15 free MCQs — start with 5 Diagnostic MCQs to identify gaps in your understanding, then complete 10 Mastery MCQs near the end to check what you have learned.

Most gas-law calculations begin with a single gas, but real samples often contain several gases at the same time. Air is a mixture of gases, and a reaction vessel may contain reactants, products and unreactive gases together. Yet a pressure gauge gives only one reading: the total pressure.

Dalton’s Law allows us to look inside that total pressure mathematically. Instead of treating the mixture as one mysterious gas, we ask how much of the pressure is contributed by each component.

The key connection is between amount, mole fraction and partial pressure. Once you see why a gas containing a certain fraction of the particles contributes the same fraction of the total pressure, partial-pressure calculations become much more systematic.

Diagnostic Quiz

How Ready Are You for Partial Pressures?

Try these five questions before reading the chapter. They check whether your ideas about gas pressure, moles and mixtures are ready for Dalton’s Law.

5 quick prerequisite questions
Focus on pressure, moles and gas mixtures
Every answer option is explained

A container holds a mixture of N₂ and O₂. What does the partial pressure of N₂ represent?

The pressure exerted by N₂ only in the part of the container not occupied by O₂

Gases do not divide the container into separate regions. N₂ molecules spread throughout the whole available volume.

The pressure N₂ would exert if it alone occupied the same container at the same temperature

Correct. This is the meaning of partial pressure. The amount of N₂, container volume and temperature are kept the same while the other gases are imagined to be absent.

Half the total pressure because two gases are present

The pressure is shared equally only if the gases have equal mole fractions. Simply having two gases does not mean each contributes half.

The average pressure of N₂ and O₂

Partial pressure is an individual contribution to the total pressure, not an average of the pressures of the different gases.

Two gases have partial pressures of 35 kPa and 65 kPa. What is the total pressure?

30 kPa

This subtracts the two pressure contributions. Both gases contribute positively to the pressure on the container walls.

50 kPa

This is the average of 35 kPa and 65 kPa. Dalton’s Law requires the partial pressures to be added, not averaged.

65 kPa

This ignores the pressure contribution from the first gas.

100 kPa

Correct. P_total = 35 + 65 = 100 kPa.

A mixture contains 2.0 mol of gas A and 3.0 mol of gas B. What is the mole fraction of A?

0.20

The denominator must be the total number of moles, 2.0 + 3.0 = 5.0 mol. Dividing 2.0 by 10 would not represent the fraction of particles that are A.

0.60

0.60 is the mole fraction of B because 3.0/5.0 = 0.60.

0.40

Correct. n_total = 5.0 mol, so x_A = 2.0/5.0 = 0.40.

1.50

A mole fraction cannot exceed 1. It represents a fraction of the total amount.

Which quantity should be compared when determining the composition of a gas mixture for a partial-pressure calculation?

Number of moles

Correct. Mole fraction compares amounts in moles because moles are proportional to the number of gas particles.

Masses only

Equal masses of different gases can contain different numbers of molecules because their molar masses differ. Convert masses to moles first.

Relative molecular masses only

Mᵣ helps convert mass into moles but does not by itself tell us how much of each gas is present.

Densities only

Density alone does not give the fraction of the gas particles contributed by each component.

Gas A is in a rigid container at constant temperature. Gas B is added without reacting with A. What happens to the partial pressure of A?

It decreases because A is now a smaller fraction of the mixture

The mole fraction of A decreases, but its amount, temperature and container volume have not changed. Its partial pressure therefore does not decrease.

It remains unchanged

Correct. For gas A, p_A = n_ART/V. Since n_A, T and V are unchanged, p_A is unchanged.

It doubles

Adding B increases the total pressure but does not double the amount of A.

It becomes zero

Gas A is still present and continues colliding with the container walls, so it still contributes a partial pressure.

What is partial pressure?

Imagine a container holding gases A, B and C. The gases are completely mixed, so particles of all three gases move throughout the entire container and collide with its walls.

The pressure measured for the mixture is caused by all these collisions together. We can nevertheless separate the contributions mathematically.

The partial pressure of gas A, written p_A, is the pressure that gas A would exert if it alone occupied the same container at the same temperature.

Quantity	Meaning
p_A	Pressure contribution from gas A
p_B	Pressure contribution from gas B
p_C	Pressure contribution from gas C
P_total	Pressure exerted by all gases in the mixture together

Partial pressure is a contribution to the total pressure.

The gases do not occupy separate sections of the container. Each gas spreads throughout the available volume. Partial pressure simply tells us how much of the measured total pressure is attributable to one component.

Linking partial pressure to the ideal gas equation

If gas A behaves ideally, we can write the ideal gas equation for A alone:

p_AV = n_ART

Rearranging gives:

p_A = n_ART/V

This equation is important because it tells us directly what controls the partial pressure of A.

Increasing n_A increases p_A.
Increasing T at constant V increases p_A.
Increasing V at constant T decreases p_A.
The amount of another non-reacting gas does not appear in this equation.

At constant temperature and volume, the partial pressure is therefore directly proportional to the number of moles of that gas.

At constant T and V: p_A ∝ n_A

Dalton’s Law of Partial Pressures

Since each gas in an ideal mixture contributes independently to collisions with the walls of the container, the total pressure is obtained by adding all the individual partial pressures.

Dalton’s Law states that the total pressure of a mixture of ideal gases is the sum of the partial pressures of the individual gases.

P_total = p_A + p_B + p_C + …

For a mixture containing only two gases:

P_total = p_A + p_B

Why are the partial pressures added?

Pressure is produced when gas particles collide with the container walls. In the ideal-gas model, particles of A behave independently of particles of B. Collisions made by A and collisions made by B both contribute to the force exerted on the walls.

The total effect is therefore the sum of their separate contributions.

Incorrect idea: The total pressure is the average of the pressures of the different gases.

Correct reasoning: Every gas contributes to collisions with the container walls, so the individual partial pressures are added.

Simple example

Suppose a mixture contains three gases with partial pressures of 25 kPa, 40 kPa and 15 kPa.

P_total = 25 + 40 + 15 = 80 kPa

The total pressure is therefore 80 kPa.

Mole fraction describes the composition of a gas mixture

Dalton’s Law tells us how partial pressures combine, but we also need a way to calculate the pressure contribution of one gas from the composition of the mixture. This is where mole fraction becomes useful.

The mole fraction of gas A, written x_A, is the number of moles of A divided by the total number of moles of gas present.

x_A = n_A/n_total

where:

n_A is the amount of gas A;
n_total is the total amount of all gases in the mixture.

n_total = n_A + n_B + n_C + …

Mole fraction has no units

Both the numerator and denominator are measured in moles, so the units cancel. Mole fraction is therefore a dimensionless quantity.

Mole fraction	Interpretation
x_A = 0	No A is present
x_A = 0.20	A makes up 20% of the total amount of gas
x_A = 0.75	A makes up 75% of the total amount of gas
x_A = 1	The sample contains only A

Since every mole in the mixture must belong to one of its components, the mole fractions of all gases add to 1.

x_A + x_B + x_C + … = 1

Mole fraction tells us a gas’s share of the total number of moles.

If x_A = 0.30, then 30% of the amount of gas in the mixture is A. For an ideal gas mixture, A will also contribute 30% of the total pressure.

Connecting mole fraction to partial pressure

The most useful equation in Dalton’s Law calculations is:

p_A = x_AP_total

Rather than simply memorising this relationship, we can derive it from the ideal gas equation.

Step 1: write the equation for gas A

p_AV = n_ART

Step 2: write the equation for the whole mixture

P_totalV = n_totalRT

Step 3: divide the two equations

Since both equations refer to the same temperature and container volume, V, R and T cancel.

p_A/P_total = n_A/n_total

But:

n_A/n_total = x_A

Therefore:

p_A/P_total = x_A

and hence:

p_A = x_AP_total

Amount fraction: x_A = n_A/n_total

Pressure fraction: p_A/P_total

For an ideal gas mixture, these two fractions are equal.

Why this relationship makes physical sense

Suppose 40% of the gas particles in a mixture are molecules of A. At the same temperature, those molecules have the same average kinetic-energy principle as the other ideal gases in the mixture.

Gas A therefore provides 40% of the particle contribution to pressure, so:

x_A = 0.40 ⟹ p_A = 0.40P_total

Calculating partial pressure from amounts of gas

A common examination question gives the number of moles of several gases and the total pressure. The calculation can be organised into three clear stages.

Find the total number of moles.
Find the mole fraction of the required gas.
Multiply its mole fraction by the total pressure.

Worked example 1: using amounts in moles

A container contains 2.00 mol of N₂ and 3.00 mol of O₂. The total pressure is 10.0 atm. Calculate the partial pressure of N₂.

First calculate the total amount of gas:

n_total = 2.00 + 3.00 = 5.00 mol

The mole fraction of N₂ is:

x_N₂ = 2.00/5.00 = 0.400

Therefore:

p_N₂ = 0.400 × 10.0 = 4.00 atm

Answer: p_N₂ = 4.00 atm.

Since the mole fraction of O₂ is 0.600, its partial pressure is 6.00 atm. As a check:

4.00 + 6.00 = 10.0 atm

When the amounts are given as masses

A common trap is to compare the masses directly. Partial pressure depends on the number of particles, so masses must first be converted to moles.

n = m/Mᵣ

Worked example 2: convert masses to moles first

A mixture contains 4.00 g of He and 16.0 g of O₂. The total pressure is 150 kPa. Calculate the partial pressure of each gas.

For helium:

n(He) = 4.00/4.00 = 1.00 mol

For oxygen:

n(O₂) = 16.0/32.0 = 0.500 mol

Therefore:

n_total = 1.00 + 0.500 = 1.50 mol

The mole fractions are:

x_He = 1.00/1.50 = 0.667

x_O₂ = 0.500/1.50 = 0.333

Hence:

p_He = 0.667 × 150 = 100 kPa

p_O₂ = 0.333 × 150 = 50.0 kPa

Answer: p_He = 100 kPa and p_O₂ = 50.0 kPa.

Do not use mass fraction in place of mole fraction.

4.00 g of He contains the same number of atoms as 32.0 g of O₂ contains O₂ molecules because both are 1.00 mol. Mass alone therefore does not measure the fraction of gas particles present.

Calculating a partial pressure directly from pV = nRT

Sometimes the amount of one component, the volume and the temperature are known. In that case, you do not need to calculate the total pressure first.

Apply the ideal gas equation directly to the required component:

p_AV = n_ART

so:

p_A = n_ART/V

Worked example 3: finding each pressure contribution directly

A 10.0 dm³ vessel contains 0.200 mol of Ne and 0.300 mol of Ar at 298 K. Calculate the partial pressures and total pressure. Use R = 8.31 kPa dm³ mol⁻¹ K⁻¹.

For Ne:

p_Ne = (0.200 × 8.31 × 298)/10.0 = 49.5 kPa

For Ar:

p_Ar = (0.300 × 8.31 × 298)/10.0 = 74.3 kPa

Dalton’s Law then gives:

P_total = 49.5 + 74.3 = 123.8 kPa

Answer: p_Ne = 49.5 kPa, p_Ar = 74.3 kPa and P_total = 123.8 kPa.

Working backwards from partial pressure

The relationship can also be rearranged to find the composition of a gas mixture.

x_A = p_A/P_total

Worked example 4: finding a mole fraction

The partial pressure of CO₂ in a gas mixture is 12.0 kPa. The total pressure is 80.0 kPa. Find the mole fraction of CO₂.

x_CO₂ = 12.0/80.0 = 0.150

Answer: x_CO₂ = 0.150.

CO₂ therefore makes up 15.0% of the total amount of gas in the mixture.

How changes to a gas mixture affect partial pressure

The equation:

p_A = n_ART/V

helps us distinguish between changes to the partial pressure and changes to the mole fraction.

Change	Effect on p_A	Reason
Add more A at constant T and V	Increases	n_A increases
Add B at constant T and V without reaction	Unchanged	n_A, T and V are unchanged
Compress the mixture at constant T	Increases	V decreases
Heat the mixture at constant V	Increases	T increases
Remove some A at constant T and V	Decreases	n_A decreases

Adding another gas: an important distinction

Suppose A is already present in a rigid container. Gas B is then added at constant temperature without reacting with A.

The total number of moles increases, so the mole fraction of A decreases. However, the amount of A itself has not changed.

p_A = n_ART/V

Since n_A, T and V are unchanged, p_A remains unchanged. Gas B simply adds another pressure contribution, causing the total pressure to increase.

Mole fraction of A: decreases when B is added because n_total increases.

Partial pressure of A: remains unchanged if n_A, T and V remain unchanged.

Changing the pressure without changing composition

If a gas mixture is compressed without adding, removing or reacting any gas, the numbers of moles of all components remain in the same ratio. Their mole fractions therefore remain constant.

If the total pressure doubles, every partial pressure also doubles.

p_A = x_AP_total

What if gases in the mixture react?

Dalton’s Law is most straightforward for gases that do not react with one another. If a reaction occurs, the initial amounts no longer represent the gas mixture whose final pressure is being measured.

For example:

NH₃(g) + HCl(g) → NH₄Cl(s)

Mixing NH₃ and HCl removes gaseous particles as solid NH₄Cl forms. It would therefore be wrong to use the initial mole fractions of NH₃ and HCl to calculate the final partial pressures.

The correct sequence

Write the balanced chemical equation.
Use stoichiometry to determine the amounts that react.
Calculate the amounts of gaseous substances remaining or formed.
Add these amounts to obtain the final n_total.
Calculate the final mole fractions.
Calculate the partial pressures.

Reaction first, Dalton’s Law second.

Partial pressures describe the gases actually present at the time the pressure is considered. If a reaction changes the gaseous composition, complete the stoichiometric calculation before using mole fractions.

When does Dalton’s Law work best?

The simple relationship between mole fraction and partial pressure assumes ideal-gas behaviour. In an ideal gas mixture, the particles have negligible volume and intermolecular attractions have no significant effect on the pressure.

Real gases behave most like ideal gases when:

the pressure is relatively low; and
the temperature is relatively high.

Under these conditions, the particles are relatively far apart and intermolecular attractions are less important.

Situation	Approach
Ideal, non-reacting gas mixture	Apply Dalton’s Law directly
Gases undergo a reaction	Find the post-reaction composition first
Real gases at low pressure and high temperature	Ideal treatment is usually a good approximation
Real gases at high pressure or low temperature	Greater deviation from ideal behaviour may occur

Why partial pressure matters in real situations

The behaviour of a gas in a mixture often depends on its partial pressure rather than on the total pressure alone. For example, increasing the total pressure of a breathing-gas mixture increases the partial pressures of its components if their mole fractions remain constant.

This is why the composition of breathing gases becomes important at high pressures: the same mole fraction can correspond to a much larger partial pressure when the total pressure increases.

Common mistakes with Dalton’s Law

Common mistake	Why it is wrong	Correct approach
Using mass fraction instead of mole fraction	Pressure depends on the number of gas particles, represented by moles	Convert each mass to moles before calculating x_i
Using only two gases in n_total when more gases are present	Mole fraction must be based on the complete gaseous mixture	Add the moles of every gas present
Giving mole fraction a unit	It is a ratio of two amounts measured in mol	Write mole fraction without units
Allowing mole fractions to add to more or less than 1	Every gas particle must belong to one component of the mixture	Check that Σx_i = 1
Averaging the partial pressures	Dalton’s Law requires pressure contributions to be added	Use P_total = Σp_i
Using p_A = n_AP_total	n_A is an amount, not a fraction of the mixture	Use p_A = (n_A/n_total)P_total
Mixing pressure units	Values in Pa, kPa and atm cannot be added directly	Convert all pressures to the same unit first
Using initial moles after the gases have reacted	The reaction changes the number of gaseous particles present	Carry out the stoichiometry first
Thinking adding another gas must reduce p_A	At fixed T and V, p_A depends on n_A, not directly on the amount of B	Use p_A = n_ART/V to analyse the change
Obtaining p_A greater than P_total	A single component cannot contribute more pressure than the complete mixture	Check that 0 ≤ x_A ≤ 1 and p_A ≤ P_total

Partial-pressure calculation checklist

Identify the gases

List every gas present in the final mixture.
Check whether any gases react.
Use post-reaction amounts where necessary.
Distinguish partial pressure from total pressure.

Find the composition

Convert masses to moles if required.
Calculate n_total.
Use x_i = n_i/n_total.
Check that all mole fractions add to 1.

Choose the relationship

Use P_total = Σp_i to add pressure contributions.
Use p_i = x_iP_total when composition is known.
Use p_iV = n_iRT when V and T are known.
Use x_i = p_i/P_total when working backwards.

Check the answer

Use consistent pressure units.
Ensure 0 ≤ x_i ≤ 1.
Ensure p_i ≤ P_total.
Confirm that all partial pressures add to P_total.

Mastery Quiz

Dalton’s Law and Partial Pressures Mastery Quiz

Now apply the chapter to gas-mixture calculations, mole fractions, changing conditions and reacting gases. Several questions require more than one step.

10 application-focused questions
Includes calculations and misconception testing
Every answer option is explained

A mixture contains 0.200 mol of N₂ and 0.300 mol of O₂ at a total pressure of 250 kPa. What is the partial pressure of N₂?

50.0 kPa

This would correspond to a mole fraction of 0.200 rather than the actual fraction 0.200/0.500 = 0.400.

100 kPa

Correct. x_N₂ = 0.200/(0.200 + 0.300) = 0.400, so p_N₂ = 0.400 × 250 = 100 kPa.

150 kPa

150 kPa is the partial pressure of O₂ because its mole fraction is 0.600.

200 kPa

This would require a mole fraction of 0.800, which is not consistent with the given amounts.

A mixture contains 4.00 g of He and 16.0 g of O₂. The total pressure is 180 kPa. What is the partial pressure of O₂?

36.0 kPa

This comes from using the mass fraction 16/(4 + 16) incorrectly. Partial pressure depends on mole fraction, not mass fraction.

90.0 kPa

The two gases do not contain equal numbers of moles. 4.00 g He is 1.00 mol, whereas 16.0 g O₂ is only 0.500 mol.

60.0 kPa

Correct. n(He) = 1.00 mol and n(O₂) = 0.500 mol, so x_O₂ = 0.500/1.50 = 1/3. Therefore p_O₂ = 60.0 kPa.

144 kPa

This corresponds to using the oxygen mass fraction of 0.80, which is not the correct particle fraction.

Three gases have partial pressures of 35 kPa, 42 kPa and 18 kPa. What is the total pressure?

95 kPa

Correct. Dalton’s Law gives P_total = 35 + 42 + 18 = 95 kPa.

31.7 kPa

This is the average of the three values. Partial pressures must be added, not averaged.

77 kPa

This adds only the first two gases and ignores the 18 kPa contribution.

59 kPa

Subtracting one partial pressure is not consistent with Dalton’s Law. All gases present contribute positively to the total pressure.

In a mixture, the partial pressure of gas X is 24.0 kPa and the total pressure is 80.0 kPa. What is the mole fraction of X?

0.20

A mole fraction of 0.20 would give a partial pressure of only 0.20 × 80.0 = 16.0 kPa.

0.24

The numerical value of the partial pressure cannot simply be converted into a fraction. It must be divided by the total pressure.

3.33

This uses the reciprocal, 80/24. A mole fraction must lie between 0 and 1.

0.300

Correct. x_X = p_X/P_total = 24.0/80.0 = 0.300.

Gas A is in a rigid vessel at constant temperature. An unreactive gas B is added. Which statement is correct?

Both the mole fraction and partial pressure of A decrease

The mole fraction decreases, but p_A does not. The amount of A, temperature and volume remain unchanged.

The mole fraction of A decreases, but its partial pressure remains unchanged

Correct. n_total increases, so x_A decreases. However, p_A = n_ART/V remains unchanged.

The mole fraction of A remains unchanged, but its partial pressure increases

Adding B changes the composition, so the mole fraction of A decreases.

Both the mole fraction and partial pressure of A increase

Adding B increases neither the amount of A nor its fraction of the mixture.

A mixture initially contains 0.30 mol NH₃, 0.20 mol HCl and 0.50 mol Ar. NH₃ and HCl react completely according to NH₃(g) + HCl(g) → NH₄Cl(s). The final total pressure is 120 kPa. What is the partial pressure of the remaining NH₃?

36 kPa

36 kPa comes from using the initial NH₃ fraction 0.30/1.00. The reaction changes the gaseous composition before the final pressure is considered.

12 kPa

After reaction, 0.10 mol NH₃ remains, but it is not 0.10 of the final gas mixture. The final gaseous total is 0.10 + 0.50 = 0.60 mol.

20 kPa

Correct. HCl is limiting, so 0.20 mol NH₃ reacts and 0.10 mol remains. Ar remains at 0.50 mol. Thus x_NH₃ = 0.10/0.60 = 1/6 and p_NH₃ = (1/6) × 120 = 20 kPa.

60 kPa

This would require NH₃ to make up half of the final gaseous mixture. It actually contributes only 0.10 of the 0.60 mol remaining.

A 1.00 dm³ vessel contains 0.0400 mol of gas A at 300 K. What is the partial pressure of A? Use R = 8.31 kPa dm³ mol⁻¹ K⁻¹.

9.97 kPa

This is too small by a factor of 10. Apply p_A = n_ART/V using the given values directly.

24.9 kPa

This does not follow from nRT/V. The product 0.0400 × 8.31 × 300 is close to 100.

997 kPa

This is too large by a factor of 10. No additional dm³-to-cm³ conversion is required.

99.7 kPa

Correct. p_A = (0.0400 × 8.31 × 300)/1.00 = 99.7 kPa.

A gas mixture has x_O₂ = 0.210. The mixture is compressed at constant temperature without changing its composition, causing the total pressure to rise from 200 kPa to 500 kPa. What is the new partial pressure of O₂?

105 kPa

Correct. The composition is unchanged, so x_O₂ remains 0.210. Hence p_O₂ = 0.210 × 500 = 105 kPa.

42.0 kPa

42.0 kPa was the original partial pressure at 200 kPa. Compression increases both the total pressure and every partial pressure.

210 kPa

This treats 0.210 as though it were 42% of the new pressure. The correct fraction remains 21.0%.

500 kPa

500 kPa is the pressure of the complete mixture, not the oxygen contribution.

A student calculates x_A = 0.25, x_B = 0.35 and x_C = 0.50 for a mixture containing only A, B and C. What can be concluded?

The calculation must be correct because every value is between 0 and 1

Individual mole fractions must lie between 0 and 1, but they must also add to exactly 1 for the complete mixture.

At least one mole fraction is wrong because the three values add to 1.10

Correct. For a mixture containing only A, B and C, x_A + x_B + x_C must equal 1.

Gas C must have a partial pressure of 0.50 kPa

Mole fraction has no units and cannot be converted into a partial pressure without knowing the total pressure.

The mole fraction of A should be 25

A mole fraction is expressed as a decimal or fraction between 0 and 1. A value of 25 would be impossible.

A mixture contains 8.0 g of He and 32.0 g of O₂ at a total pressure of 120 kPa. A student claims that He contributes only 24 kPa because it accounts for 20% of the total mass. Which statement is correct?

The student is correct because partial pressure is proportional to mass

Partial pressure is proportional to the number of moles at fixed temperature and volume, not directly to mass.

The He partial pressure is 40 kPa

8.0 g He is 2.0 mol, while 32.0 g O₂ is 1.0 mol. Helium therefore makes up two-thirds, not one-third, of the moles.

The He partial pressure is 80 kPa because He contributes two-thirds of the total moles

Correct. n(He) = 8.0/4.0 = 2.0 mol and n(O₂) = 32.0/32.0 = 1.0 mol. Thus x_He = 2/3 and p_He = (2/3) × 120 = 80 kPa.

The He partial pressure is 96 kPa because helium is the lighter gas

Being lighter does not automatically determine a gas’s pressure fraction. The actual number of moles must be calculated.

Frequently asked questions

Questions students often ask about Dalton’s Law

These questions connect the equations to the particle model and clarify several ideas that can otherwise seem contradictory.

Does each gas occupy only its own fraction of the container?

No. Each gas spreads throughout the entire available volume. When we say gas A has a particular mole fraction, we are describing its fraction of the gas particles, not its fraction of the physical container.

Why does mole fraction determine partial pressure?

For ideal gases at the same temperature and volume, p_i = n_iRT/V. Pressure is therefore proportional to the number of moles. A gas containing 30% of the total moles contributes 30% of the total pressure.

Why do we use mole fraction instead of mass fraction?

Gas pressure arises from particles colliding with the walls. The number of particles is proportional to the number of moles. Equal masses of gases with different Mᵣ values do not generally contain equal numbers of molecules.

Can a partial pressure be greater than the total pressure?

No. Since p_A = x_AP_total and 0 ≤ x_A ≤ 1, the partial pressure of one component cannot exceed the pressure of the complete mixture.

If the total pressure doubles, does every partial pressure double?

Yes, provided the composition of the mixture remains unchanged. Each mole fraction remains constant, so p_i = x_iP_total means that every partial pressure changes in the same proportion as the total pressure.

Why can adding another gas leave the original gas’s partial pressure unchanged?

At constant temperature and volume, the original gas still has the same number of moles. From p_A = n_ART/V, its partial pressure therefore remains unchanged. The added gas contributes an additional partial pressure and raises the total pressure.

Does adding another gas always leave the original partial pressure unchanged?

No. That conclusion applies when the temperature and volume remain constant and the gases do not react. If adding the second gas changes the volume, temperature or amount of the original gas through reaction, its partial pressure may also change.

Can Dalton’s Law still be used if the gases initially react?

You should first use stoichiometry to determine which gases remain and how many moles of each are present after reaction. Partial pressures can then be calculated for the final gaseous mixture.

Is mole fraction the same as percentage composition?

A mole fraction can be converted into a mole percentage by multiplying by 100. For example, x_A = 0.25 corresponds to 25 mol% A. This should not be confused with percentage by mass.

Is partial pressure affected by the identity or Mᵣ of an ideal gas?

For an ideal gas at fixed temperature and volume, its partial pressure is given by p = nRT/V. The equation contains the number of moles but not Mᵣ. Different ideal gases containing the same number of moles at the same T and V therefore exert the same pressure.

Why is Dalton’s Law associated with ideal gases?

The simple additive treatment assumes that each gas behaves independently. Real gases can deviate when intermolecular attractions and the finite volume of particles become important, especially at high pressure and low temperature.

Dalton’s Law turns the pressure of a gas mixture into a simple set of individual contributions. The total pressure is the sum of the partial pressures, while the mole fraction of each ideal gas tells us what fraction of that total pressure it contributes. By converting amounts to moles, calculating the correct mole fractions and using the composition of the gases actually present, partial-pressure questions become a logical extension of the ideal gas equation rather than a separate set of rules to memorise.