The behaviour of gases can be described by simple mathematical laws that relate pressure, volume, temperature, and amount of gas. These laws combine into the ideal gas equation pV = nRT, which is one of the most powerful tools in physical chemistry.
12.1 Gas Laws: Boyle's, Charles', and Gay-Lussac's
Definition
Pressure
The pressure of a gas is a measure of the number of collisions of gas particles with each other and with the sides of the container.
Definition
Boyle's Law
The pressure of a fixed quantity of gas is inversely proportional to the volume so long as the temperature remains constant.
Definition
Charles' Law
The volume of an enclosed sample of gas is directly proportional to its Kelvin temperature provided pressure and amount remain constant.
Formula
Boyle's Law
p = pressure (Pa or kPa), V = volume (dm³ or m³); temperature constant
SI unit: Pa·dm³ (or Pa·m³)
Formula
Charles' Law
V = volume (dm³), T = absolute temperature (K); pressure constant
SI unit: dm³·K⁻¹
Formula
Gay-Lussac's Law
p = pressure (kPa), T = absolute temperature (K); volume constant
SI unit: kPa·K⁻¹
Formula
Celsius to Kelvin conversion
T(K) = absolute temperature; T(°C) = Celsius temperature
SI unit: K
Watch Out
ALWAYS convert temperatures to Kelvin before substituting into gas law equations. Using °C gives the wrong answer every time. T(K) = T(°C) + 273.
Worked Example
A gas has a volume of 4,0 dm³ at a pressure of 200 kPa. What is its volume at 50 kPa if temperature is constant?
Given
- V₁ = 4,0 dm³
- p₁ = 200 kPa
- p₂ = 50 kPa
- T = constant
Find
V₂ = ?
Solution
- 1p₁V₁ = p₂V₂
- 2200 × 4,0 = 50 × V₂
- 3V₂ = 800/50 = 16 dm³
Worked Example
A gas occupies 3,0 dm³ at 27 °C. What volume does it occupy at 127 °C at constant pressure?
Given
- V₁ = 3,0 dm³
- T₁ = 27 °C = 300 K
- T₂ = 127 °C = 400 K
Find
V₂ = ?
Solution
- 1T₁ = 27 + 273 = 300 K; T₂ = 127 + 273 = 400 K
- 2V₁/T₁ = V₂/T₂
- 33,0/300 = V₂/400
- 4V₂ = (3,0 × 400)/300 = 4,0 dm³
Practice Question
A gas at 150 kPa and 300 K is heated at constant volume to 450 K. Calculate the new pressure.
(4 marks)
12.2 The General Gas Equation
When pressure, volume, AND temperature all change simultaneously, we combine the three gas laws into one equation called the general (combined) gas equation. The amount of gas (moles) must remain constant.
Formula
General gas equation
p = pressure (kPa), V = volume (dm³), T = temperature (K); n constant
SI unit: kPa·dm³·K⁻¹
Three Gas Laws vs General Equation
| Property | Individual Law | Condition |
|---|---|---|
| Boyle's Law (p₁V₁ = p₂V₂) | Temperature constant | T fixed; p and V change |
| Charles' Law (V₁/T₁ = V₂/T₂) | Pressure constant | p fixed; V and T change |
| Gay-Lussac's (p₁/T₁ = p₂/T₂) | Volume constant | V fixed; p and T change |
| General Equation | None (p, V, T all vary) | All three change; n constant |
Worked Example
A gas occupies 2,0 dm³ at 100 kPa and 300 K. If the pressure changes to 200 kPa and the temperature changes to 600 K, what is the new volume?
Given
- p₁ = 100 kPa, V₁ = 2,0 dm³, T₁ = 300 K
- p₂ = 200 kPa, T₂ = 600 K
Find
V₂ = ?
Solution
- 1p₁V₁/T₁ = p₂V₂/T₂
- 2(100 × 2,0)/300 = (200 × V₂)/600
- 30,6667 = 200V₂/600
- 4V₂ = 0,6667 × 600/200 = 2,0 dm³
Exam Tip
A quick sanity check: if you increase both pressure and temperature, the pressure increase reduces volume but the temperature increase raises volume — they can cancel out (as in the example above). Always check whether your answer makes physical sense.
Worked Example
A gas at STP (0 °C, 101,3 kPa) has a volume of 5,60 dm³. Calculate its volume at 25 °C and 80 kPa.
Given
- p₁ = 101,3 kPa, V₁ = 5,60 dm³, T₁ = 273 K
- p₂ = 80 kPa, T₂ = 298 K
Find
V₂
Solution
- 1p₁V₁/T₁ = p₂V₂/T₂
- 2(101,3 × 5,60)/273 = (80 × V₂)/298
- 32,079 = 80V₂/298
- 4V₂ = 2,079 × 298/80 = 7,74 dm³
Practice Question
A gas has p₁ = 120 kPa, V₁ = 3,0 dm³, T₁ = 400 K. It is compressed to V₂ = 1,5 dm³ and cooled to T₂ = 200 K. Find p₂.
(5 marks)
12.3 The Ideal Gas Law: pV = nRT
Definition
Ideal gas
An ideal gas is a hypothetical gas whose behaviour perfectly obeys all the ideal gas laws under all conditions of temperature and pressure. Ideal gas particles are assumed to have negligible volume, to exert no intermolecular forces on one another, and to undergo perfectly elastic collisions. Real gases approach ideal behaviour at low pressures and high temperatures.
Definition
Real gas
A real gas does not perfectly obey the ideal gas laws because its particles have a finite volume and do exert intermolecular forces on one another. Real gases deviate most from ideal behaviour at high pressures (particles are close together, volume matters) and at low temperatures (particles move slowly, attractions are significant).
Formula
Ideal gas equation
p = pressure (Pa), V = volume (m³), n = moles (mol), R = 8,314 J·K⁻¹·mol⁻¹, T = temperature (K)
SI unit: Pa·m³ = J
Watch Out
Unit consistency is critical in pV = nRT. If p is in Pa, use V in m³ (1 dm³ = 0,001 m³). If p is in kPa, convert to Pa first (multiply by 1000). R = 8,314 J·K⁻¹·mol⁻¹ = 8,314 Pa·m³·K⁻¹·mol⁻¹.
The ideal gas equation combines ALL the gas laws into a single equation that also includes the amount of gas (n). It allows you to calculate any one of p, V, n, or T if the other three are known. Real gases deviate from ideal behaviour at very high pressures (molecules get close, volume matters) and very low temperatures (intermolecular forces become significant).
Ideal Gas vs Real Gas
| Property | Ideal Gas | Real Gas |
|---|---|---|
| Particle volume | Zero (point particles) | Has finite volume |
| Intermolecular forces | None | Exist (van der Waals, etc.) |
| Obeys pV = nRT | Always | At moderate T and p only |
| Deviates at | Never | High p, low T |
Worked Example
Calculate the pressure exerted by 2 mol of an ideal gas in a 10 dm³ container at 27 °C. (R = 8,314 J·K⁻¹·mol⁻¹)
Given
- n = 2 mol
- V = 10 dm³ = 0,010 m³
- T = 27 °C = 300 K
- R = 8,314 J·K⁻¹·mol⁻¹
Find
p = ?
Solution
- 1pV = nRT
- 2p = nRT/V
- 3p = (2 × 8,314 × 300) / 0,010
- 4p = 4988,4 / 0,010
- 5p = 498 840 Pa ≈ 499 kPa
Worked Example
What volume does 0,5 mol of ideal gas occupy at 200 kPa and 127 °C?
Given
- n = 0,5 mol
- p = 200 kPa = 200 000 Pa
- T = 127 °C = 400 K
- R = 8,314 J·K⁻¹·mol⁻¹
Find
V = ?
Solution
- 1V = nRT/p
- 2V = (0,5 × 8,314 × 400) / 200 000
- 3V = 1662,8 / 200 000
- 4V = 0,008314 m³ = 8,314 dm³
Exam Tip
Memory aid for the variables: 'Please Visit New Regions Twice' → p, V, n, R, T. The ideal gas equation rearranges to p = nRT/V, V = nRT/p, n = pV/RT, or T = pV/nR depending on what you need to find.
Practice Question
A sample of gas has a pressure of 150 kPa, a volume of 4,0 dm³, and a temperature of 57 °C. Calculate the number of moles of gas present. (R = 8,314 J·K⁻¹·mol⁻¹)
(5 marks)
Practice Question
Under what conditions does a real gas deviate most from ideal gas behaviour? Give a reason for each condition.
(4 marks)
Real World
The ideal gas law is used by engineers to design storage tanks for compressed gases like oxygen (hospitals) and LPG (cooking gas). The equation predicts how much gas is stored at a given pressure and temperature.