Ideal Gases

SPEC 3.6.2.2 - THERMAL PHYSICS

Gas laws, the ideal gas equation, and work done by expanding gases

What is an Ideal Gas?

Ideal Gas Definition

Molecules are infinitely small (point masses)

Molecules exert no forces on each other

Obeys Boyle's law at all p and T

Internal energy depends only on T

Real Gases

Real gases approximate ideal behaviour at low pressure and high temperature (molecules far apart, moving fast).

The Gas Laws

The gas laws describe the experimental relationship between pressure (p), volume (V), and temperature (T) for a fixed mass of gas. They are empirical - derived from observation, not theory.

Boyle's Law

p ∝ 1/V (constant T)

Charles's Law

Pressure Law

p ∝ T (constant V)

Boyle's Law

The pressure of a fixed mass of gas is inversely proportional to its volume, as long as the temperature remains constant.

p V = constant

→

p_{1} V_{1} = p_{2} V_{2}

Conditions

Fixed mass of gas

Constant temperature (isothermal process)

When volume decreases, the same number of molecules are confined to a smaller space. They collide with the walls more frequently, causing higher pressure.

Particle Explanation

At constant T, molecules have the same average speed. Halving the volume means molecules travel half the distance between wall collisions, so they hit the walls twice as often → pressure doubles.

Worked Example - Boyle's Law01

A fixed mass of gas at a constant temperature and a pressure of 10⁵ Pa is compressed from 2 litres to 500 cm³. Calculate the new pressure of the gas.

1

Convert to SI units:

V_{1} = 2 L = 2 \times 1 0^{- 3} m^{3}

,

V_{2} = 500 cm^{3} = 500 \times 1 0^{- 6} m^{3}

2

Constant T → Boyle's law:

p_{1} V_{1} = p_{2} V_{2}

3

p_{2} = \frac{p _{1} V _{1}}{V _{2}} = \frac{1 0 ^{5} \times 2 \times 1 0 ^{- 3}}{500 \times 1 0 ^{- 6}} = 4 \times 1 0^{5} Pa

Answer: 4 × 10⁵ Pa (Volume ÷4 → Pressure ×4)

Charles's Law

The volume of a fixed mass of gas is directly proportional to its absolute temperature, as long as the pressure remains constant.

\frac{V}{T} = constant

→

\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}

Conditions

Fixed mass of gas

Constant pressure (isobaric process)

When temperature increases, molecules move faster and hit the walls harder. To keep pressure constant, the gas must expand so collisions become less frequent.

Particle Explanation

Higher T means greater average kinetic energy, so molecules move faster. The volume increases until the reduced collision frequency balances the harder impacts, maintaining constant pressure.

Worked Example - Charles's Law02

A fixed mass of gas at constant pressure with an initial volume of 5 litres is heated from 20°C to 80°C. Calculate the new volume.

1

Convert to Kelvin:

T_{1} = 20 + 273 = 293

K,

T_{2} = 80 + 273 = 353

K

2

Constant p → Charles's law:

\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}

3

V_{2} = \frac{V _{1} \times T _{2}}{T _{1}} = \frac{5 \times 1 0 ^{- 3} \times 353}{293} = 6.02 \times 1 0^{- 3} m^{3}

Answer: 6 × 10⁻³ m³ (or 6 litres)

The Pressure Law

Pressure Law

The pressure of a fixed mass of gas is directly proportional to its absolute temperature, as long as the volume remains constant.

\frac{p}{T} = constant

→

\frac{p _{1}}{T _{1}} = \frac{p _{2}}{T _{2}}

Conditions

Fixed mass of gas

Constant volume (isochoric process)

When temperature increases, molecules move faster. In a fixed volume, they hit the walls harder and more often, causing higher pressure.

Particle Explanation

Higher T means molecules have greater kinetic energy and higher speeds. Each collision transfers more momentum to the walls, and collisions happen more frequently → pressure increases proportionally with T.

Worked Example - Pressure Law03

During a spacewalk, an astronaut moves from shadow into full sun. The temperature of his oxygen tank rises from 200 K to 350 K. If the original pressure was 2 × 10⁵ Pa, what is the new pressure?

1

Tank has constant volume → use pressure law:

\frac{p _{1}}{T _{1}} = \frac{p _{2}}{T _{2}}

2

Rearrange:

p_{2} = \frac{p _{1} \times T _{2}}{T _{1}}

3

p_{2} = \frac{2 \times 1 0 ^{5} \times 350}{200} = 3.5 \times 1 0^{5} Pa

Answer: 3.5 × 10⁵ Pa

Absolute Zero

Absolute zero (0 K = -273°C) is the lowest possible temperature. At this point, particles have minimum possible energy (no kinetic energy) and the pressure and volume of an ideal gas would be zero.

Kelvin Scale

Why Absolute Zero?

Both V-T and p-T graphs extrapolate to zero at -273°C

This defines the Kelvin scale where 0 K is absolute zero

On the Kelvin scale, the graph passes through the origin

Temperature Conversions

°C to K: T = θ + 273

K to °C: θ = T - 273

A change of 1 K = a change of 1°C

Worked Example - Finding Absolute Zero04

A gas has volume 64.0 cm³ at 0°C and 87.5 cm³ at 100°C. Using these results, calculate a value for absolute zero in °C.

1

V ∝ T (Charles's law) → straight line through absolute zero when V = 0

2

Gradient:

\frac{Δ V}{Δ T} = \frac{87.5 - 64.0}{100 - 0} = 0.235 cm^{3} °C^{- 1}

3

Line equation:

V = 0.235 T + 64.0

(using T = 0°C, V = 64.0)

4

At V = 0:

0 = 0.235 T + 64.0

→

T = - 64.0/0.235 = - 272° C

Answer: -272°C (close to accepted -273°C)

Combined Gas Law

When more than one variable changes, combine all three gas laws:

Combined Gas Law

\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}

Simplification

If one variable is constant, it cancels from both sides:ConstantEquationLawT constant

p_{1} V_{1} = p_{2} V_{2}

Boyle'sp constant

\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}

Charles'sV constant

\frac{p _{1}}{T _{1}} = \frac{p _{2}}{T _{2}}

Pressure

Worked Example - Combined Gas Law05

A balloon contains 0.50 m³ of helium at 1.0 × 10⁵ Pa and 20°C. The balloon rises to a height where the pressure is 0.50 × 10⁵ Pa and the temperature is -30°C. Calculate the new volume of the balloon.

1

Convert temperatures to Kelvin:

T_{1} = 20 + 273 = 293

K,

T_{2} = - 30 + 273 = 243

K

2

Both p and T change, use combined gas law:

\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}

3

Rearrange:

V_{2} = \frac{p _{1} V _{1} T _{2}}{T _{1} p _{2}}

4

Substitute:

V_{2} = \frac{1.0 \times 1 0 ^{5} \times 0.50 \times 243}{293 \times 0.50 \times 1 0 ^{5}}

5

Calculate:

V_{2} = \frac{1.215 \times 1 0 ^{7}}{1.465 \times 1 0 ^{7}} = 0.83 m^{3}

Answer: 0.83 m³

Ideal Gas Equation - Molar Form

Ideal Gas Equation

p V = n RT

SymbolQuantityUnit

p

PressurePa

V

Volumem³

n

Number of molesmol

R

Molar gas constant8.31 J mol⁻¹ K⁻¹

T

Absolute temperatureK

Worked Example - pV = nRT06

Calculate the number of moles of gas at pressure 10⁵ Pa, temperature 27°C, and volume 5 m³.

1

Convert temperature:

T = 27 + 273 = 300

K

2

Rearrange

p V = n RT

for n:

n = \frac{p V}{RT}

3

n = \frac{1 0 ^{5} \times 5}{8.31 \times 300} = \frac{5 \times 1 0 ^{5}}{2493} = 201

mol

Answer: 201 mol (or 2.0 × 10² mol)

Ideal Gas Equation - Molecular Form

Molecular Form

p V = N k T

Gas Constants

ConstantSymbolValueAvogadro constant

N_{A}

6.02 × 10²³ mol⁻¹Molar gas constant

R

8.31 J mol⁻¹ K⁻¹Boltzmann constant

k

1.38 × 10⁻²³ J K⁻¹Key relationships:

N = n N_{A}

and

k = \frac{R}{N _{A}}

Worked Example - pV = NkT07

A container holds 3.0 × 10²⁴ molecules at 400 K and 2.0 × 10⁵ Pa. Calculate the volume. (k = 1.38 × 10⁻²³ J K⁻¹)

1

Given N (molecules), use molecular form:

p V = N k T

2

Rearrange:

V = \frac{N k T}{p}

3

V = \frac{3.0 \times 1 0 ^{24} \times 1.38 \times 1 0 ^{- 23} \times 400}{2.0 \times 1 0 ^{5}} = \frac{16.56}{2.0 \times 1 0 ^{5}}

4

V = 8.28 \times 1 0^{- 5} m^{3}

Answer: 8.3 × 10⁻⁵ m³

Molar & Molecular Mass

Definitions

Molar mass (M): Mass of one mole (g mol⁻¹ or kg mol⁻¹)

Molecular mass (m): Mass of one molecule (kg)

Key Relationships

n = \frac{m _{t o t a l}}{M}

N = n \times N_{A}

M = m \times N_{A}

Common Molar Masses

GasM (g mol⁻¹)Hydrogen (H₂)2Nitrogen (N₂)28Oxygen (O₂)32Carbon dioxide (CO₂)44

Worked Example - Molar Mass08

A container holds 64 g of oxygen gas (O₂) at a pressure of 1.5 × 10⁵ Pa and temperature 300 K. Calculate the volume of the container. (Molar mass of O₂ = 32 g mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹)

1

Calculate number of moles:

n = \frac{m}{M} = \frac{64}{32} = 2

mol

2

Use ideal gas equation:

p V = n RT

3

Rearrange:

V = \frac{n RT}{p}

4

Substitute:

V = \frac{2 \times 8.31 \times 300}{1.5 \times 1 0 ^{5}}

5

Calculate:

V = \frac{4986}{1.5 \times 1 0 ^{5}} = 0.033 m^{3}

Answer: 0.033 m³ (or 33 litres)

Work Done by an Expanding Gas

Work Done Formula

When a gas expands at constant pressure, it does work on its surroundings. From

W = F d

and

F = p A

:

W = p Δ V

where W (J), p (Pa), ΔV (m³)

Key Points

Work done = area under p-V graph

Expansion (ΔV > 0): gas does work on surroundings

Compression (ΔV < 0): work done on gas

Unit Conversions

1 litre = 10⁻³ m³ | 1 cm³ = 10⁻⁶ m³

Worked Example - Work Done09

An ideal gas at 2 × 10⁵ Pa is in a 6 litre cylinder. How much work is done by the gas if it expands slowly to 6.05 litres?

1

Change in volume:

Δ V = 6.05 - 6.00 = 0.05 L = 0.05 \times 1 0^{- 3} m^{3}

2

Use:

W = p Δ V

3

W = 2 \times 1 0^{5} \times 0.05 \times 1 0^{- 3} = 2 \times 1 0^{5} \times 5 \times 1 0^{- 5} = 10 J

Answer: 10 J

Standard Temperature & Pressure

STP Values

Standard Temperature: 0°C (273 K)

Standard Pressure: 1.01 × 10⁵ Pa

Molar Volume at STP

At STP, 1 mole of any ideal gas occupies approximately 22.4 litres (0.0224 m³).

Verify:

V = \frac{n RT}{p} = \frac{1 \times 8.31 \times 273}{1.01 \times 1 0 ^{5}} = 0.0224 m^{3}

Exam Tip

STP is sometimes referred to in questions. Remember these standard values!

10

Worked Example - Using STP

At STP, a sample of gas occupies 44.8 litres. How many moles of gas are present? If the gas is carbon dioxide (CO₂, molar mass = 44 g mol⁻¹), what is the mass of the sample?

1

At STP, 1 mole of gas occupies 22.4 litres

2

Number of moles:

n = \frac{V}{22.4} = \frac{44.8}{22.4} = 2

mol

3

Mass = moles × molar mass:

m = n \times M = 2 \times 44 = 88

g

Answer: 2 moles, mass = 88 g

Practice

Question 1: Gas is stored at 8 atmospheres in a 3000 cm³ cylinder. What volume of gas is released when pressure is reduced to 1 atmosphere?

1

Boyle's law:

p_{1} V_{1} = p_{2} V_{2}

2

V_{2} = \frac{8 \times 3000}{1} = 24000 cm^{3}

3

Volume released = 24000 - 3000 = 21000 cm³

Practice

Question 2: Air in a closed cylinder at 50 000 Pa and 27°C is heated to 227°C. What is the new pressure?

1

Convert:

T_{1} = 300

K,

T_{2} = 500

K

2

Pressure law:

p_{2} = \frac{p _{1} T _{2}}{T _{1}} = \frac{50000 \times 500}{300}

3

p_{2} =

83 000 Pa

Practice

Question 3: 4 litres of gas at 20°C is heated to 100°C at constant pressure. Find the new volume in m³.

1

Convert:

T_{1} = 293

K,

T_{2} = 373

K

2

Charles's law:

V_{2} = \frac{V _{1} T _{2}}{T _{1}} = \frac{4 \times 1 0 ^{- 3} \times 373}{293}

3

V_{2} =

5.1 × 10⁻³ m³

Summary

Key Equations

LawFormulaCondition

Boyle's

p_{1} V_{1} = p_{2} V_{2}

Constant T

Charles's

\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}

Constant p

Pressure

\frac{p _{1}}{T _{1}} = \frac{p _{2}}{T _{2}}

Constant V

Combined

\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}

Fixed mass

Ideal (molar)

p V = n RT

R = 8.31

Ideal (molecular)

p V = N k T

k = 1.38×10⁻²³

Work done

W = p Δ V

Constant p

Exam Checklist

Always convert T to Kelvin

Check units (Pa, m³, K)

Work = area under p-V graph