Ideal Gas Equation and Mᵣ | A-Level Chemistry

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.

Gas calculations become much more powerful when pressure, volume, temperature and amount are considered together. Rather than learning a separate rule for every change in gas conditions, the ideal gas equation combines these variables into one relationship: pV = nRT.

At A-Level, you must do more than substitute numbers into a formula. You need to recognise which units are compatible with the gas constant, convert temperatures to kelvin, rearrange the equation accurately and explain how the measured mass of a gas can be used to determine its relative molecular mass.

Once the reasoning is organised, these questions become systematic. The chemistry lies in identifying what the gas sample represents; the calculation then follows from careful handling of moles, units and experimental data.

Diagnostic Quiz

How Ready Are You for pV = nRT?

Try these five questions before reading the chapter. They check the unit conversions, algebra and mole ideas that make ideal-gas calculations much easier.

5 quick prerequisite questions
Focus on units, kelvin and rearrangement
Every answer option is explained

A gas is at 27 °C. Which temperature should be substituted for T in pV = nRT?

27 K

27 is the Celsius reading, not the absolute temperature. The ideal gas equation requires temperature in kelvin.

246 K

Subtracting 273 is the wrong conversion direction. A Celsius temperature is converted to kelvin by adding 273.

300 K

Correct. T = 27 + 273 = 300 K, so 300 K is the value used in the ideal gas equation.

546 K

This adds 273 twice. Only one conversion from °C to K is required.

When R = 8.31 J mol⁻¹ K⁻¹ is used, which pressure–volume combination is directly compatible?

Pa and m³

Correct. 1 Pa m³ = 1 J, so pV is in joules and is compatible with R = 8.31 J mol⁻¹ K⁻¹.

Pa and dm³

Pa and dm³ introduce a factor-of-1000 mismatch. With R = 8.31, Pa should be paired with m³.

kPa and m³

kPa and m³ also introduce a factor-of-1000 mismatch. With R = 8.31, kPa should be paired with dm³.

atm and cm³

Atmospheres and cm³ require a differently expressed gas constant. They are not directly compatible with R = 8.31 J mol⁻¹ K⁻¹.

Which rearrangement of pV = nRT correctly gives the amount of gas, n?

n = pV/RT

Correct. Dividing both sides of pV = nRT by RT gives n = pV/RT.

n = RT/pV

This takes the reciprocal of the correct expression. nRT equals pV, so n is obtained by dividing pV by RT.

n = pRT/V

This moves the variables incorrectly. Both R and T multiply n and must therefore divide pV.

n = V/pRT

Pressure belongs in the numerator with volume. The correct rearrangement is n = pV/RT.

A gas has molar mass 44.0 g mol⁻¹. What is its relative molecular mass, Mᵣ?

44.0 g mol⁻¹

44.0 g mol⁻¹ is the molar mass. Mᵣ has the same numerical value but is a relative quantity and has no units.

44.0 g

Mᵣ is not a mass of one sample and does not have units of grams.

44.0

Correct. The molar mass is 44.0 g mol⁻¹, so the numerical value of Mᵣ is 44.0, with no units.

1/44.0

Relative molecular mass is not the reciprocal of molar mass. Their numerical values are equal when molar mass is expressed in g mol⁻¹.

A gas occupies 250 cm³. Which value is the same volume expressed in dm³?

0.0250 dm³

This divides by 10 000. Since 1000 cm³ = 1 dm³, divide by 1000.

0.250 dm³

Correct. 250 cm³ ÷ 1000 = 0.250 dm³.

2.50 dm³

This is ten times too large. A dm³ is larger than a cm³, so the numerical value should decrease on conversion.

250 000 dm³

This converts in the wrong direction by a very large factor. 1000 cm³, not 0.001 cm³, makes 1 dm³.

The ideal gas equation

The behaviour of an ideal gas is described by the equation:

Symbol	Quantity	Common compatible unit	Meaning
p	Pressure	Pa or kPa	Force exerted per unit area by collisions of gas particles
V	Volume	m³ or dm³	Space occupied by the gas sample
n	Amount of gas	mol	Number of moles of gas particles
R	Molar gas constant	8.31 J mol⁻¹ K⁻¹	Constant linking the four gas variables
T	Thermodynamic temperature	K	Temperature measured from absolute zero

The equation is written as:

pV = nRT

The left-hand side represents pressure multiplied by volume. The right-hand side represents the amount of gas multiplied by the gas constant and absolute temperature. The units used for p and V must match the units represented by R.

What does the equation tell us?

The equation shows that the four gas variables are not independent. If the amount of gas is fixed, a change in one variable must be accompanied by a change in one or more of the others.

For example, increasing the temperature of a gas while keeping its volume fixed increases its pressure. The particles have greater average kinetic energy, so their collisions with the container walls are more frequent and more forceful.

At constant n and T, increasing V decreases p.
At constant n and V, increasing T increases p.
At constant p and T, increasing n increases V.

For one particular sample of gas, the equation may be rearranged according to the unknown quantity. The numerical value of R remains constant.

pV = nRT

Choosing compatible units

Most errors in ideal-gas calculations come from substituting values in incompatible units. The numerical value R = 8.31 can be used with more than one convenient pressure–volume pair because the product pV must have units of energy.

Since 1 Pa m³ = 1 J and 1 kPa dm³ = 1 J, both of the following combinations are valid when R = 8.31 J mol⁻¹ K⁻¹.

Pressure unit	Volume unit	Temperature unit	Suitable value of R
Pa	m³	K	8.31 J mol⁻¹ K⁻¹
kPa	dm³	K	8.31 J mol⁻¹ K⁻¹
Pa	dm³	K	Not compatible without changing R
kPa	m³	K	Not compatible without changing R

Converting pressure

Pressure is often given in kPa but may need to be expressed in Pa if the volume is converted to m³.

1 kPa = 1000 Pa.
101 kPa = 101 000 Pa.
Divide by 1000 to convert Pa to kPa.

Converting volume and temperature

Volumes may be given in cm³ or dm³. Temperature may be given in °C. These values must be converted before substitution into the ideal gas equation.

T/K = temperature/°C + 273

The more precise conversion is +273.15, but +273 is normally suitable unless the question requires greater precision. A temperature of 27 °C is therefore 300 K.

The volume conversions follow from the relationships between centimetres, decimetres and metres.

1 dm³ = 1000 cm³ = 1.00 × 10⁻³ m³

Therefore, 250 cm³ is 0.250 dm³ or 2.50 × 10⁻⁴ m³.

1 cm³ = 1.00 × 10⁻⁶ m³

A reliable check is to confirm that the pressure–volume product has units of joules.

Use paired units rather than converting mechanically.

With R = 8.31, use Pa with m³ or kPa with dm³. Temperature must always be in K, and amount must be in mol. Writing the units beside each value before substitution prevents most factor-of-1000 errors.

Why must temperature be in kelvin?

The ideal gas equation requires an absolute temperature scale. On the kelvin scale, zero represents absolute zero, the theoretical temperature at which an ideal gas would have minimum thermal motion.

0 °C = 273 K, not 0 K.
25 °C = 298 K.

Celsius temperature cannot be used directly because its zero point is arbitrary and does not represent zero thermal energy.

Doubling a temperature from 100 K to 200 K has a physical meaning.
Doubling a temperature from 100 °C to 200 °C does not double the absolute temperature.

Substituting a Celsius value therefore gives an incorrect proportional relationship and usually produces a seriously wrong answer.

Incorrect substitution: using T = 25 for a gas at 25 °C.

Correct substitution: T = 25 + 273 = 298 K.

Rearranging pV = nRT

Before inserting numbers, rearrange the equation so that the required quantity is alone. This makes the calculation easier to follow and reduces calculator errors.

Finding the amount of gas

Divide both sides of pV = nRT by RT:

n = pV ÷ RT

This form is used when pressure, volume and temperature are known and the number of moles is required.

The value obtained for n can then be used in stoichiometric calculations or linked to mass using n = m/Mᵣ.

Finding pressure

Divide both sides by V:

The resulting expression is:

p = nRT ÷ V

Similarly, volume is found by dividing both sides by p:

V = nRT ÷ p

Finding temperature

Divide both sides by nR. The answer is obtained directly in kelvin.

T = pV ÷ nR
Convert to °C only if the question specifically asks for a Celsius temperature.
Temperature/°C = T/K − 273

Rearrange first, substitute second.

A clearly rearranged equation shows the examiner which relationship is being used. It also makes unit checks easier because the required quantity is visible on the left-hand side.

Using pV = nRT in gas calculations

Consider 250 cm³ of a gas at 101 kPa and 27 °C. To determine the amount of gas, first choose a compatible set of units.

p = 101 kPa, V = 0.250 dm³, T = 300 K

Using kPa with dm³ avoids unnecessary powers of ten:

n = pV ÷ RT = (101 × 0.250) ÷ (8.31 × 300) = 1.01 × 10⁻² mol

The same answer would be obtained using 101 000 Pa and 2.50 × 10⁻⁴ m³. The two methods are equivalent because both pressure–volume products are measured in joules.

State the converted values before substitution.
Keep extra calculator digits during intermediate steps.
Round the final answer to an appropriate number of significant figures.

Determining a value for Mᵣ

If the mass, pressure, volume and temperature of a gas sample are known, the ideal gas equation can be combined with the mole–mass relationship to determine the relative molecular mass of the substance.

Begin with the mole–mass relationship

For a pure molecular substance of mass m:

n = m ÷ Mᵣ

Substitute into the ideal gas equation

Replace n in pV = nRT with m/Mᵣ.

This gives:

pV = (m ÷ Mᵣ)RT

The equation now links the measurable gas data directly to the unknown relative molecular mass.

m is the measured mass of the gas or vapour sample.
p, V and T describe the conditions of that same sample.
R must be used with compatible pressure and volume units.

Rearrange for Mᵣ

Multiply both sides by Mᵣ and divide by pV:

The working formula is therefore:

Use mass in grams when the required numerical value is Mᵣ.
Use kelvin for temperature.
Use Pa with m³ or kPa with dm³.

Mᵣ = mRT ÷ pV

Strictly, this calculation first gives the molar mass in g mol⁻¹ when m is measured in grams. The numerical value of that molar mass is equal to the relative molecular mass, Mᵣ.

Molar mass = 44.0 g mol⁻¹ corresponds to Mᵣ = 44.0

Relative molecular mass is a ratio and therefore has no units.

Do not attach g mol⁻¹ to Mᵣ.

The molar mass has units of g mol⁻¹. Relative molecular mass, Mᵣ, is dimensionless. Their numerical values are the same, which is why gas calculations often move directly from a calculated molar mass to a value for Mᵣ.

A reliable four-step method

Identify the data: write down m, p, V and T with their units.
Convert the units: make p and V compatible with R, and convert T to K.
Substitute: use Mᵣ = mRT/pV.
Interpret: report Mᵣ without units and check whether the value is chemically reasonable.

Measured quantity	Example reading	Converted value used	Reason
Mass, m	0.176 g	0.176 g	Gives molar mass numerically in g mol⁻¹
Pressure, p	100 kPa	100 kPa	Compatible with volume in dm³
Volume, V	100 cm³	0.100 dm³	1000 cm³ = 1 dm³
Temperature, T	27 °C	300 K	Absolute temperature is required

Incorrect idea: Mᵣ = pV/mRT because the letters are simply moved across the equals sign.

Correct reasoning: starting from pV = (m/Mᵣ)RT and rearranging algebraically gives Mᵣ = mRT/pV.

Using gas density to determine Mᵣ

Sometimes the gas data are given as a density rather than as a separate mass and volume. The same equation can be rewritten in terms of density.

Density, ρ, is defined as mass divided by volume.

Therefore, m/V can replace the mass and volume terms in the Mᵣ expression.

Deriving the density form

Begin with Mᵣ = mRT/pV and group m/V together:

Step	Relationship	Result
1	Mᵣ = mRT/pV	Place m and V together
2	Mᵣ = (m/V)RT/p	Recognise that m/V = ρ
3	Mᵣ = ρRT/p	Density form of the equation

The density form is therefore:

Mᵣ = ρRT ÷ p

Why is the density form useful?

It allows Mᵣ to be found without choosing a particular sample volume. A gas density measured at a known pressure and temperature contains the same mass-to-volume information required by the original equation.

Conditions must refer to the same gas sample

The density, pressure and temperature must all describe the gas under the same conditions. Gas density changes when pressure or temperature changes.

A density measured at one temperature cannot be combined with a pressure and temperature from a different experiment.

Matching density units

If pressure is in kPa and R = 8.31 is used with dm³, density should be in g dm⁻³. If pressure is in Pa and volume is treated in m³, density should be in g m⁻³.

The numerical value changes with the volume unit, so density units must never be omitted during working.

Mᵣ = ρRT/p

Use ρ in g dm⁻³ with p in kPa, or ρ in g m⁻³ with p in Pa. In either case, temperature must be in K. The resulting molar mass in g mol⁻¹ has the same numerical value as Mᵣ.

Worked example using density

A gas has density 1.77 g dm⁻³ at 100 kPa and 298 K. Its relative molecular mass can be determined directly from the density form.

Quantity	Value used	Substitution
ρ	1.77 g dm⁻³	Mᵣ = (1.77 × 8.31 × 298) ÷ 100 = 43.8
R	8.31 J mol⁻¹ K⁻¹
T	298 K
p	100 kPa

The value is approximately 44, which could be consistent with a gas such as CO₂ under conditions where its behaviour is close to ideal.

Ideal behaviour and experimental limitations

The equation pV = nRT describes an ideal gas exactly. Real gases approximate ideal behaviour most closely at low pressure and high temperature, where particles are far apart and intermolecular attractions are relatively unimportant.

Condition or issue	Effect on the gas or measurement	Possible consequence for calculated Mᵣ
High pressure	Particles are closer and their own volume becomes more significant	Greater deviation from ideal behaviour
Low temperature	Intermolecular attractions become more important	Greater deviation from ideal behaviour
Gas leak	The measured mass may not correspond to the measured gas volume	Calculated Mᵣ becomes unreliable
Incomplete vaporisation	Not all the sample is present as gas	Gas data do not represent the full measured mass
Temperature not equilibrated	The gas is not actually at the recorded temperature	Incorrect value is substituted for T

Ideal-gas assumption: particles have negligible volume and no intermolecular attractions except during elastic collisions.

Real experiment: particles occupy space, attractions may matter, and measured values carry uncertainty. The calculated Mᵣ is therefore an experimental estimate rather than a perfectly exact value.

Worked examples using pV = nRT

Example 1: finding the amount of gas

Calculate the amount of gas in 250 cm³ at 101 kPa and 27 °C.

Use kPa with dm³ and convert the temperature to kelvin.

V = 250 ÷ 1000 = 0.250 dm³; T = 27 + 273 = 300 K

Substitute into n = pV/RT.

n = (101 × 0.250) ÷ (8.31 × 300) = 1.01 × 10⁻² mol

Answer: n = 1.01 × 10⁻² mol.

Example 2: determining Mᵣ from mass and gas data

A 0.176 g sample of a gas occupies 100 cm³ at 100 kPa and 27 °C. Determine Mᵣ.

Convert 100 cm³ to 0.100 dm³ and 27 °C to 300 K.

Mᵣ = mRT ÷ pV

Mᵣ = (0.176 × 8.31 × 300) ÷ (100 × 0.100) = 43.9

To three significant figures, the calculated value is:

Mᵣ = 43.9

Example 3: identifying a volatile substance

A 0.293 g sample of a volatile liquid is completely vaporised. The vapour occupies 150 cm³ at 98.0 kPa and 77 °C. Determine Mᵣ and suggest a possible molecular formula if the empirical formula is C₂H₅.

Convert V = 0.150 dm³ and T = 350 K, then apply Mᵣ = mRT/pV.

Mᵣ = (0.293 × 8.31 × 350) ÷ (98.0 × 0.150) = 58.0.
Empirical formula mass of C₂H₅ = 29.0.

The multiplier is 58.0/29.0 = 2, so a possible molecular formula is C₄H₁₀.

Common mistakes with the ideal gas equation

Common mistake	Why it is wrong	Correct approach
Using temperature in °C	The equation requires absolute temperature	Convert using T/K = temperature/°C + 273
Using kPa with m³ while keeping R = 8.31	The pressure–volume units are incompatible	Use Pa with m³ or kPa with dm³
Treating 1 cm³ as 10⁻³ m³	The length conversion must be cubed	Use 1 cm³ = 10⁻⁶ m³
Using Mᵣ = pV/mRT	The equation has been rearranged incorrectly	Derive Mᵣ = mRT/pV from pV = (m/Mᵣ)RT
Writing g mol⁻¹ after Mᵣ	Mᵣ is a relative quantity and has no units	Use units only for molar mass, not Mᵣ
Using the mass of the container as m	m must be the mass of gas or vapour only	Subtract the empty-container mass where necessary
Mixing data from different conditions	p, V, T and density must describe the same gas state	Use measurements taken for the same sample under the same conditions
Rounding intermediate values too early	Premature rounding increases the final error	Keep extra digits and round only the final answer
Assuming every real gas behaves ideally	Real gases may deviate, especially at high pressure and low temperature	Treat the result as an experimental estimate and consider the conditions

Ideal-gas calculation checklist

Identify the variables

Write down p, V, n, T and any mass given.
Identify the quantity being asked for.
Use values for the same gas sample.
Note whether density is given.

Convert the units

Convert °C to K.
Use Pa with m³ or kPa with dm³.
Convert cm³ carefully.
Keep mass in g when finding Mᵣ numerically.

Choose the relationship

Use pV = nRT for gas variables.
Use n = m/Mᵣ for mass and moles.
Use Mᵣ = mRT/pV for a gas sample.
Use Mᵣ = ρRT/p when density is given.

Check the final answer

Give n in mol where required.
Report Mᵣ without units.
Use appropriate significant figures.
Check that the value is chemically reasonable.

Mastery Quiz

Ideal Gas Equation and Mᵣ Mastery Quiz

Now apply the chapter to multi-step gas calculations, experimental Mᵣ determination and unfamiliar situations where unit choices and assumptions matter.

10 application-focused questions
Includes calculations and experimental reasoning
Every answer option is explained

A gas occupies 500 cm³ at 100 kPa and 298 K. What amount of gas is present? Use R = 8.31 J mol⁻¹ K⁻¹.

2.02 × 10⁻² mol

Correct. Convert 500 cm³ to 0.500 dm³, then n = pV/RT = (100 × 0.500)/(8.31 × 298) = 2.02 × 10⁻² mol.

2.02 × 10⁻⁵ mol

This is smaller by a factor of 1000, consistent with mixing an unnecessary m³ conversion with pressure still in kPa.

20.2 mol

This is larger by a factor of 1000. The 500 cm³ volume must be converted to 0.500 dm³ when pressure is in kPa.

0.168 mol

This does not follow n = pV/RT with compatible units. Both the unit conversion and division by RT must be handled together.

A 0.0400 mol sample of an ideal gas occupies 1.00 dm³ at 300 K. What is its pressure?

9.97 kPa

This is too small by a factor of 10. Using p = nRT/V gives approximately 99.7 kPa.

99.7 kPa

Correct. p = nRT/V = (0.0400 × 8.31 × 300)/1.00 = 99.7 kPa when V is in dm³.

997 kPa

This is too large by a factor of 10. The correct substitution gives 99.7 kPa.

2.77 × 10⁻⁵ kPa

This results from an incorrect rearrangement. Pressure is obtained by multiplying nRT and dividing by V.

A 0.880 g gas sample occupies 500 cm³ at 100 kPa and 300 K. What is its Mᵣ?

22.0

This is about half the correct value. Using the complete expression Mᵣ = mRT/pV gives approximately 43.9.

43.9

Correct. V = 0.500 dm³, so Mᵣ = (0.880 × 8.31 × 300)/(100 × 0.500) = 43.9.

87.8

This is about twice the correct value. Check the 500 cm³ = 0.500 dm³ conversion and the placement of pV in the denominator.

43.9 g mol⁻¹

43.9 is the correct numerical value, but the question asks for Mᵣ. Relative molecular mass has no units.

A gaseous compound has empirical formula CH₂ and an experimentally determined Mᵣ of 56. What is its molecular formula?

CH₂

The empirical formula mass of CH₂ is 14. Since 56/14 = 4, the molecular formula contains four empirical-formula units.

C₂H₄

C₂H₄ has Mᵣ = 28, only half the experimental value.

C₃H₆

C₃H₆ has Mᵣ = 42, which does not match the experimental value of 56.

C₄H₈

Correct. Empirical formula mass = 14 and 56/14 = 4, so (CH₂)₄ = C₄H₈.

A gas has density 1.78 g dm⁻³ at 100 kPa and 298 K. Which value is closest to its Mᵣ?

22.0

This is about half the correct value. Using Mᵣ = ρRT/p gives approximately 44.1.

44.1

Correct. Mᵣ = (1.78 × 8.31 × 298)/100 = 44.1 when ρ is in g dm⁻³ and p is in kPa.

74.0

This does not result from the density form of the ideal gas equation with the stated conditions.

4410

This is larger by a factor of 100. The kPa–dm³ unit pairing already matches R = 8.31, so no extra pressure conversion is needed.

A student correctly measures m, p and V for a gas at 25 °C but substitutes T = 25 instead of 298 in Mᵣ = mRT/pV. How will the calculated Mᵣ compare with the correct value?

It will be too small

Correct. Mᵣ is directly proportional to T in this expression. Using 25 instead of 298 makes the calculated value much too small.

It will be too large

Using a smaller temperature reduces the numerator mRT, so the calculated Mᵣ decreases rather than increases.

It will be unchanged

Temperature appears explicitly in the numerator, so changing its numerical value changes the calculated Mᵣ.

It cannot be predicted

The direction is predictable because all other measured quantities are unchanged and Mᵣ is proportional to T.

Which set of values can be substituted directly into pV = nRT with R = 8.31 without further pressure or volume conversion?

p = 100 kPa and V = 2.00 dm³

Correct. kPa and dm³ are a compatible pair because 1 kPa dm³ = 1 J.

p = 100 kPa and V = 2.00 m³

kPa should be paired with dm³ when R = 8.31. Using m³ here introduces a factor-of-1000 error.

p = 100 000 Pa and V = 2.00 dm³

Pa should be paired with m³, not dm³, when R = 8.31.

p = 1.00 atm and V = 2000 cm³

These units require a differently expressed gas constant or conversion to one of the compatible SI-derived pairs.

In an experiment to determine Mᵣ, some of the measured liquid sample fails to vaporise, but its full mass is still used as m. The measured p, V and T describe only the vapour. What is the most likely effect?

The calculated Mᵣ is too high

Correct. The mass in the numerator is too large for the actual amount of vapour represented by pV/RT, so Mᵣ = mRT/pV is overestimated.

The calculated Mᵣ is too low

Using too large a value of m increases, rather than decreases, the calculated Mᵣ.

The calculated Mᵣ is unaffected

The calculation assumes that the measured mass and gas data refer to the same gaseous sample. Here they do not.

Only the calculated temperature is affected

Temperature is independently measured. The error arises because the mass used does not correspond to the amount of vapour present.

Which conditions generally make a real gas behave most like an ideal gas?

High pressure and low temperature

These conditions bring particles closer together and make intermolecular attractions and particle volume more significant.

High pressure and high temperature

High temperature helps reduce the relative importance of attractions, but high pressure still forces particles close together.

Low pressure and low temperature

Low pressure increases separation, but low temperature makes intermolecular attractions more important.

Low pressure and high temperature

Correct. Particles are far apart at low pressure and have high kinetic energy at high temperature, so attractions and particle volume matter less.

A fixed amount of an ideal gas is kept at constant temperature. Its volume is doubled. What happens to its pressure?

It doubles

At constant n and T, pV is constant. If V doubles, p must decrease, not increase.

It halves

Correct. From pV = nRT, nRT is constant. Doubling V therefore requires p to become half its original value.

It remains unchanged

Pressure cannot remain unchanged if volume changes while n and T are fixed, because pV must remain constant.

It becomes zero

Doubling the volume does not remove the gas or its particle collisions. The pressure is reduced to half, not zero.

Frequently asked questions

Questions students often ask about pV = nRT and Mᵣ

These questions connect the calculation method to the assumptions and unit choices that commonly cause confusion in H2 Chemistry.

Can I always use kPa and dm³ instead of Pa and m³?

Yes, when R = 8.31 is used, kPa and dm³ form a compatible pair because 1 kPa dm³ = 1 J. Do not mix kPa with m³ or Pa with dm³ unless you change the numerical value and units of R accordingly.

Why is R sometimes written as 8.314 rather than 8.31?

The values differ only in the number of significant figures quoted. Use the value supplied in the question or data booklet and keep sufficient calculator digits until the final answer.

Is Mᵣ the same as molar mass?

They have the same numerical value when molar mass is expressed in g mol⁻¹, but they are not the same type of quantity. Molar mass has units of g mol⁻¹, whereas Mᵣ is a dimensionless relative value.

Can pV = nRT be used for a mixture of gases?

Yes. The equation can describe the whole mixture using the total number of moles and total pressure, provided the mixture behaves approximately ideally. For one component, use its partial pressure and number of moles.

Why does a gas-density calculation need a stated temperature and pressure?

Gas density is not fixed independently of conditions. Heating generally increases volume and lowers density, while increasing pressure generally compresses the gas and raises density.

What happens if some vapour condenses during the experiment?

The measured gas volume and pressure would no longer represent all of the sample mass as vapour. The assumptions used in Mᵣ = mRT/pV would fail, so the calculated value would be unreliable.

Should I round the temperature after converting to kelvin?

Normally, use the precision appropriate to the data. For routine A-Level work, adding 273 and using the resulting whole-number kelvin value is often sufficient, but avoid unnecessary rounding in intermediate steps.

Does a calculated Mᵣ identify the gas uniquely?

Not necessarily. Different substances can have the same or very similar Mᵣ values. Other chemical information, such as an empirical formula, composition data or characteristic reactions, may be needed to identify the substance.

Why might the experimental Mᵣ differ from the value calculated from a formula?

Possible causes include measurement uncertainty, gas leakage, incomplete vaporisation, condensation, an incorrect temperature or pressure reading, impurities and non-ideal gas behaviour.

The ideal gas equation turns pressure, volume and temperature measurements into chemical information about the amount of gas present. By combining pV = nRT with n = m/Mᵣ, the same measurements can reveal a value for relative molecular mass. The algebra is straightforward once the units are compatible: use kelvin, match Pa with m³ or kPa with dm³, and remember that Mᵣ is reported without units. With this method secure, gas data can be linked confidently to stoichiometry, molecular formulae and experimental analysis.