The Ideal Gas Equation and Gas Calculations

This lesson focuses on advanced applications of the ideal gas equation pV = nRT, including calculating molar mass from gas density, collecting gases over water, and combined gas law problems. These calculations are frequently examined at A-Level.

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Review of the Ideal Gas Equation

pV = nRT

Symbol	Quantity	SI Unit
p	Pressure	Pa (pascals)
V	Volume	m³
n	Amount of substance	mol
R	Gas constant	8.314 J K⁻¹ mol⁻¹
T	Temperature	K (kelvin)

Key Assumptions of Ideal Gas Behaviour

An ideal gas is one that:

1. Has negligible intermolecular forces between particles.
2. Has particles that occupy negligible volume compared to the volume of the container.
3. Undergoes perfectly elastic collisions (no energy is lost).
4. Obeys pV = nRT exactly at all temperatures and pressures.

Real gases deviate from ideal behaviour at high pressures (particles are close together, so their volume is significant and intermolecular forces become important) and low temperatures (particles move slowly, so intermolecular forces have a greater effect).

Key Point: Ideal gas behaviour is best approximated at high temperatures and low pressures, where particles are far apart and moving fast.

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Molar Volume

At standard conditions, one mole of an ideal gas occupies a characteristic volume called the molar volume.

At RTP (room temperature and pressure, approximately 298 K and 101.3 kPa): Molar volume ≈ 24.0 dm³ mol⁻¹

We can verify this with pV = nRT:

V = nRT / p = (1 × 8.314 × 298) / 101 300 = 2477.6 / 101 300 = 0.02446 m³ = 24.5 dm³

(The slight difference from 24.0 is because 24.0 dm³ is a rounded value often used in calculations at approximately 293 K.)

At STP (standard temperature and pressure, 273 K and 100 kPa): V = (1 × 8.314 × 273) / 100 000 = 2269.7 / 100 000 = 0.02270 m³ = 22.7 dm³

Exam Tip: Unless told otherwise, use 24.0 dm³ mol⁻¹ as the molar volume at RTP in your calculations. The AQA data sheet gives this value.

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Calculating Molar Mass from Gas Data

The ideal gas equation can be combined with n = m/M to find the molar mass of a gas.

Since n = m/M, substituting into pV = nRT:

pV = (m/M)RT

Rearranging: M = mRT / (pV)

Worked Example 1: Molar Mass from pV = nRT

0.250 g of a gaseous compound occupies 98.0 cm³ at 100 kPa and 373 K. Calculate its molar mass.

p = 100 000 Pa V = 98.0 × 10⁻⁶ m³ = 9.80 × 10⁻⁵ m³ m = 0.250 g T = 373 K R = 8.314 J K⁻¹ mol⁻¹

M = mRT / (pV)

M = (0.250 × 8.314 × 373) / (100 000 × 9.80 × 10⁻⁵)

M = 775.3 / 9.80

M = 79.1 g mol⁻¹

Worked Example 2: Identifying a Gas

A gas has a density of 1.78 g dm⁻³ at 298 K and 100 kPa. Calculate its molar mass and suggest its identity.

Density = mass / volume, so in 1 dm³ (= 1 × 10⁻³ m³), mass = 1.78 g.

M = mRT / (pV)

M = (1.78 × 8.314 × 298) / (100 000 × 1.00 × 10⁻³)

M = 4408.5 / 100.0

M = 44.1 g mol⁻¹

A molar mass of 44.1 g mol⁻¹ corresponds to carbon dioxide (CO₂): M = 12.0 + 2(16.0) = 44.0 g mol⁻¹. ✓

It could also be propane (C₃H₈): M = 3(12.0) + 8(1.0) = 44.0 g mol⁻¹. Additional tests would be needed to distinguish between these.

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Gas Density

The density of a gas can be calculated from the ideal gas equation.

From pV = nRT and n = m/M:

ρ = m/V = pM / (RT)

where ρ = density (g m⁻³ if M is in g mol⁻¹, p in Pa, T in K).

Worked Example 3: Gas Density

Calculate the density of oxygen gas (O₂) at 298 K and 101.3 kPa.

M(O₂) = 32.0 g mol⁻¹

ρ = pM / (RT) = (101 300 × 32.0) / (8.314 × 298)

ρ = 3 241 600 / 2477.6

ρ = 1308 g m⁻³ = 1.31 g dm⁻³

Worked Example 4: Comparing Gas Densities

Which is denser at the same temperature and pressure: nitrogen (N₂) or carbon dioxide (CO₂)?

Since ρ = pM / (RT), at the same T and p, the denser gas has the larger molar mass.

M(N₂) = 28.0 g mol⁻¹ M(CO₂) = 44.0 g mol⁻¹

CO₂ is denser than N₂. This is why CO₂ sinks and can be collected by downward delivery (upward displacement of air).

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Collecting Gases

Over Water