Ideal Gases

Contents

This chapter covers four key topics that build on each other, from the behaviour of ideal gas particles to the real-world deviations that occur under extreme conditions.

01

Ideal Gases

The assumptions and properties of an ideal gas model.

02

Molar Gas Volume

How one mole of any gas occupies the same volume under the same conditions.

03

The Ideal Gas Equation

Using PV = nRT to solve quantitative gas problems.

04

Real Gases

How and why real gases deviate from ideal behaviour.

What Is an Ideal Gas?

An ideal gas is a theoretical model used to simplify the behaviour of real gases. It is based on a set of assumptions about gas particles that make calculations much more manageable.

Assumptions of an Ideal Gas

These five assumptions form the foundation of the ideal gas model. Keeping them in mind helps you understand both when the model works — and when it breaks down.

⚡ Negligible Volume

Gas particles have no significant volume compared to the container they occupy.

🔄 Random Motion

Particles move in straight lines with constant, random motion in all directions.

💥 Elastic Collisions

All collisions between particles and walls are perfectly elastic — no kinetic energy is lost.

🚫 No Intermolecular Forces

There are no attractive or repulsive forces between particles except during collisions.

🌡️ KE ∝ Temperature

The average kinetic energy of particles is directly proportional to the absolute temperature in Kelvin.

Ideal Gases & Kinetic Theory

Kinetic Theory Assumptions

The kinetic theory describes how gas particles behave. It is built on 5 key assumptions:

🔀 Constant Random Motion

Gas molecules move in constant, rapid, random motion at all times.

📦 Negligible Volume

The volume of individual molecules is negligible compared to the total gas volume.

🚫 No Intermolecular Forces

Gas particles do not attract or repel each other between collisions.

⚡ Perfectly Elastic Collisions

No kinetic energy is lost during collisions between particles.

🌡️ KE ∝ Temperature

Average kinetic energy is directly proportional to absolute temperature (K).

Ideal vs. Real Gases

✅ Ideal Gas

Follows ALL assumptions of kinetic theory perfectly.

⚠️ Real Gas

Does not perfectly follow the model. Behaves most like an ideal gas at LOW PRESSURE and HIGH TEMPERATURE.

Factors Affecting Gas Volume

The volume a gas occupies depends on two variables:

🔵 Pressure (P)

Higher pressure → smaller volume

🔴 Temperature (T)

Higher temperature → larger volume

Section 4

Diagram: Elastic vs Inelastic Collisions

Elastic: particles bounce apart, kinetic energy conserved. Inelastic: particles stick together, kinetic energy lost.

Molar Gas Volume

At standard temperature and pressure (STP), defined as 0 °C (273 K) and 100 kPa, one mole of any ideal gas occupies 22.7 dm³. This is the molar gas volume.

This value is the same regardless of the identity of the gas — whether it is hydrogen, oxygen, or carbon dioxide. This follows directly from the ideal gas assumptions: particle identity does not affect volume because particle size and intermolecular forces are ignored.

Using Molar Gas Volume

You can use the molar gas volume to convert between moles and volume at STP using simple relationships.

Moles → Volume

Volume (dm³) = moles × 22.7

Example: 2 mol of O₂ at STP = 2 × 22.7 = 45.4 dm³

Volume → Moles

Moles = Volume (dm³) ÷ 22.7

Example: 11.35 dm³ of CO₂ at STP = 11.35 ÷ 22.7 = 0.5 mol

Molar Gas Volume & Gas Laws

💨 What is Gas Pressure?

Gases exert pressure by constantly colliding with the walls of their container.

  • Gas molecules are in constant motion
  • They collide with container walls repeatedly
  • Each collision exerts a tiny force
  • Millions of collisions per second = measurable pressure

Gas particles exert pressure by constantly colliding with the walls of the container

📦 Boyle's Law — Changing Volume (constant T)

What happens?

  • Decrease volume → molecules squashed together
  • More frequent collisions with container wall
  • Pressure increases

The Maths 🔢

P ∝ 1/V

PV = constant

P₁V₁ = P₂V₂

Decreasing volume → increased collision frequency → higher pressure

📊 Boyle's Law — Three Graph Forms

Three graphs that show Boyle's Law: P vs 1/V (straight line), P vs V (curve), PV vs P (straight line)

Charles' Law & Gay-Lussac's Law

🌡️ Charles' Law — Changing Temperature (constant P)

What happens?

  • Gas is heated → particles gain kinetic energy
  • More frequent collisions with container walls
  • To keep pressure constant → gas must expand
  • Volume increases

The Maths 🔢

V ∝ T

V/T = constant

V₁/T₁ = V₂/T₂

Graph of V vs T (in Kelvin) = straight line through origin

Increasing temperature → increased collision frequency → gas expands to maintain constant pressure (a); V is directly proportional to T in Kelvin (b)

🔥 Gay-Lussac's Law — Changing Temperature (constant V)

What happens?

  • Temperature increases at constant volume
  • Molecules gain more kinetic energy
  • Particles move faster → collide more frequently
  • Pressure increases

The Maths 🔢

P ∝ T

P/T = constant

P₁/T₁ = P₂/T₂

Graph of P vs T (in Kelvin) = straight line through origin

Increasing temperature at constant volume → increased collision frequency → higher pressure (a); T is directly proportional to P (b)

Combined Gas Law & Worked Examples

⚗️ Combining All Three Gas Laws → PV = nRT

Each gas law holds one variable constant. Combining all three gives the ideal gas equation.

Boyle's Law

PV = constant (constant T)

Charles' Law

V/T = constant (constant P)

Gay-Lussac's Law

P/T = constant (constant V)

Step 1: Combine → PV/T = constant

Step 2: PV = constant × T

Step 3: The constant = n × R (moles × gas constant)

Step 4: ∴ PV = nRT ✅

📝 Worked Example 1 — Finding New Temperature

At 25°C and 100 kPa, a gas occupies 20 dm³. The volume is decreased to 10 dm³ at constant pressure. Calculate the new temperature in °C.

Given info:

  • V₁ = 20 dm³
  • V₂ = 10 dm³
  • T₁ = 25 + 273 = 298 K
  • P is constant → use V₁/T₁ = V₂/T₂

Step 1 — Rearrange for T₂: T₂ = V₂ × T₁ / V₁

Step 2 — Substitute values: T₂ = (10 × 298) / 20

Step 3 — Calculate: T₂ = 2980 / 20 = 149 K

Step 4 — Convert to °C: 149 − 273 = −124°C

📝 Worked Example 2 — Finding New Pressure

A 2.00 dm³ container of oxygen at 80 kPa is heated from 20°C to 70°C. The volume expands to 2.25 dm³. What is the final pressure?

Given info:

  • P₁ = 80 kPa
  • V₁ = 2.00 dm³, V₂ = 2.25 dm³
  • T₁ = 20 + 273 = 293 K
  • T₂ = 70 + 273 = 343 K

Step 1 — Rearrange for P₂: P₂ = P₁V₁T₂ / (V₂T₁)

Step 2 — Substitute values: P₂ = (80 × 2.00 × 343) / (2.25 × 293)

Step 3 — Numerator: 80 × 2.00 × 343 = 54,880

Step 4 — Denominator: 2.25 × 293 = 659.25

Step 5 — Calculate: P₂ = 54,880 / 659.25 = 83 kPa

Boyle's Law — P and V

As pressure increases, volume decreases — if temperature and moles are kept constant. This is Boyle's Law.

P₁V₁ = P₂V₂

If P doubles → V halves

Example: P = 100 kPa, V = 2 dm³ → if P becomes 200 kPa, V = 1 dm³

Boyle's Law: P vs V (at constant T and n)

The curve is a hyperbola — P × V = constant at fixed T and n.

Charles's Law — V and T

As temperature increases, volume increases — if pressure and moles are kept constant. This is Charles's Law. Temperature must always be in Kelvin.

V₁/T₁ = V₂/T₂

If T doubles → V doubles

Example: V = 2 dm³ at 300 K → if T becomes 600 K, V = 4 dm³

Charles's Law: V vs T (at constant P and n)

The graph is a straight line through the origin — V is directly proportional to T (in Kelvin).

Gay-Lussac's Law — P and T

As temperature increases, pressure increases — if volume and moles are kept constant. This is Gay-Lussac's Law.

P₁/T₁ = P₂/T₂

If T doubles → P doubles

Example: P = 100 kPa at 300 K → if T becomes 450 K, P = 150 kPa

Gay-Lussac's Law: P vs T (at constant V and n)

Pressure is directly proportional to temperature in Kelvin — a straight line through the origin.

The Ideal Gas Equation

The ideal gas equation combines Boyle's Law, Charles's Law, and Avogadro's Law into a single, powerful relationship that works under any conditions — not just STP.

PV = nRT

P — Pressure

Measured in Pascals (Pa) or kPa. Convert kPa → Pa by × 1000.

V — Volume

Measured in m³. Convert dm³ → m³ by ÷ 1000.

n — Moles

The amount of gas in mol.

R — Gas Constant

8.314 J mol⁻¹ K⁻¹ — given in the IB Data Booklet.

T — Temperature

Must always be in Kelvin (K). Convert °C → K by + 273.

Worked Example: Ideal Gas Equation

The Problem

Calculate the volume occupied by 0.25 mol of an ideal gas at a temperature of 27 °C and a pressure of 100 kPa.

Given Information

  • n = 0.25 mol
  • T = 27 + 273 = 300 K
  • P = 100 × 1000 = 100,000 Pa
  • R = 8.314 J mol⁻¹ K⁻¹

Step-by-Step Solution

Rearrange PV = nRT for V:

V = nRT ÷ P

V = (0.25 × 8.314 × 300) ÷ 100,000

V = 623.55 ÷ 100,000

V = 6.24 × 10⁻³ m³

The Ideal Gas Equation

⚗️ PV = nRT

The ideal gas equation links pressure, volume, temperature and moles for any ideal gas under any conditions.

P — Pressure

Measured in Pascals (Pa)

Convert kPa → Pa: × 1000

V — Volume

Measured in m³

dm³ → m³: ÷ 1000

cm³ → m³: ÷ 1,000,000

n — Moles

Amount of gas in mol

R — Gas Constant

8.31 J K⁻¹ mol⁻¹

(given in IB Data Booklet)

T — Temperature

Must be in Kelvin (K)

Convert °C → K: + 273

⚠️ Unit Conversions — Don't Lose Marks!

Volume

  • dm³ → m³: ÷ 1000
  • cm³ → m³: ÷ 1,000,000
  • m³ → cm³: × 1,000,000
  • 1 m³ = 10⁶ cm³

Pressure

  • kPa → Pa: × 1000
  • Pa → kPa: ÷ 1000
  • Always use Pa in PV = nRT

Temperature

  • °C → K: + 273
  • K → °C: − 273
  • NEVER use °C in the equation

Worked Example 1 — Finding Volume

The Problem

Calculate the volume, in dm³, occupied by 0.781 mol of oxygen at a pressure of 220 kPa and a temperature of 21°C.

Given Information

  • n = 0.781 mol
  • P = 220 kPa → 220,000 Pa (× 1000)
  • T = 21°C → 294 K (+ 273)
  • R = 8.31 J K⁻¹ mol⁻¹
  • Find: V = ?

Step-by-Step Solution

Step 1 — Rearrange PV = nRT for V:

V = nRT / P

Step 2 — Substitute values:

V = (0.781 × 8.31 × 294) / 220,000

Step 3 — Calculate numerator:

0.781 × 8.31 = 6.490

6.490 × 294 = 1,908.1

Step 4 — Divide by pressure:

V = 1,908.1 / 220,000 = 0.00867 m³

Step 5 — Convert m³ → dm³ (× 1000):

V = 0.00867 × 1000 = 8.67 dm³

Worked Example 2 — Finding Pressure

The Problem

Calculate the pressure of a gas, in kPa, given that 0.20 moles of the gas occupy 10.1 dm³ at a temperature of 25°C.

Given Information

  • n = 0.20 mol
  • V = 10.1 dm³ → 0.0101 m³ (÷ 1000)
  • T = 25°C → 298 K (+ 273)
  • R = 8.31 J K⁻¹ mol⁻¹
  • Find: P = ?

Step-by-Step Solution

Step 1 — Rearrange PV = nRT for P:

P = nRT / V

Step 2 — Substitute values:

P = (0.20 × 8.31 × 298) / 0.0101

Step 3 — Calculate numerator:

0.20 × 8.31 = 1.662

1.662 × 298 = 495.3

Step 4 — Divide by volume:

P = 495.3 / 0.0101 = 49,037 Pa

Step 5 — Convert Pa → kPa (÷ 1000):

P = 49,037 / 1000 = 49 kPa (2 sig figs)

Worked Example 3 — Finding Temperature

The Problem

Calculate the temperature of a gas, in °C, if 0.047 moles of the gas occupy 1.2 dm³ at a pressure of 100 kPa.

Given Information

  • n = 0.047 mol
  • V = 1.2 dm³ → 0.0012 m³ (÷ 1000)
  • P = 100 kPa → 100,000 Pa (× 1000)
  • R = 8.31 J K⁻¹ mol⁻¹
  • Find: T = ?

Step-by-Step Solution

Step 1 — Rearrange PV = nRT for T:

T = PV / nR

Step 2 — Substitute values:

T = (100,000 × 0.0012) / (0.047 × 8.31)

Step 3 — Calculate numerator:

100,000 × 0.0012 = 120

Step 4 — Calculate denominator:

0.047 × 8.31 = 0.3906

Step 5 — Divide:

T = 120 / 0.3906 = 307.24 K

Step 6 — Convert K → °C (− 273):

T = 307.24 − 273 = 34.24°C = 34°C (2 sig figs)

Worked Example 4 — Finding Molar Mass

The Problem

A flask of volume 1000 cm³ contains 6.39 g of a gas. The pressure in the flask was 300 kPa and the temperature was 23°C. Calculate the molar mass of the gas.

Given Information

  • mass = 6.39 g
  • V = 1000 cm³ → 0.001 m³ (÷ 1,000,000)
  • P = 300 kPa → 300,000 Pa (× 1000)
  • T = 23°C → 296 K (+ 273)
  • R = 8.31 J K⁻¹ mol⁻¹
  • Find: M (molar mass) = ?

Step-by-Step Solution

Step 1 — Rearrange PV = nRT for n:

n = PV / RT

Step 2 — Substitute values:

n = (300,000 × 0.001) / (8.31 × 296)

Step 3 — Calculate numerator:

300,000 × 0.001 = 300

Step 4 — Calculate denominator:

8.31 × 296 = 2,459.8

Step 5 — Calculate moles:

n = 300 / 2,459.8 = 0.12 mol

Step 6 — Calculate molar mass using M = mass / n:

M = 6.39 / 0.12 = 53 g mol⁻¹ (2 sig figs)

Real Gases

No gas is truly ideal. Under conditions of high pressure and low temperature, real gases deviate significantly from ideal behaviour — because the ideal assumptions no longer hold.

How Real Gases Deviate

The two key assumptions that break down for real gases are the absence of intermolecular forces and the negligible volume of particles.

High Pressure

At high pressures, gas particles are forced close together. The actual volume of the particles becomes significant, so the gas takes up more volume than an ideal gas would predict.

Low Temperature

At low temperatures, particles move slowly and intermolecular attractive forces become significant. Particles are pulled toward each other, reducing the pressure below ideal predictions.

Polar Molecules

Gases with strong intermolecular forces (e.g. HCl, NH₃, H₂O) deviate more than non-polar gases like He or H₂, which behave most ideally.

Real vs Ideal Gas: PV/nRT vs Pressure

Z = 1 means ideal behaviour. Z < 1 means attractive forces dominate (low T, moderate P). Z > 1 means particle volume dominates (high P).

Real Gases — Deviations from Ideal Behaviour

🔬 What Does the Ideal Gas Model Assume?

PV = nRT is built on two key assumptions that break down in the real world.

📦 Negligible Volume

Gas particles are assumed to take up NO space — their own volume is ignored.

🚫 No Intermolecular Forces

Gas particles are assumed to have NO attractive or repulsive forces between them.

⚠️ When Do Real Gases Deviate?

Real gases deviate most from ideal behaviour under two conditions:

🌡️ Low Temperature

  • Particles move slowly
  • Intermolecular attractions become significant
  • Pressure is lower than ideal predictions
  • Assumption of no forces BREAKS DOWN

🔴 High Pressure

  • Particles are forced close together
  • Particle volume becomes significant
  • Less space available for movement
  • Assumption of negligible volume BREAKS DOWN

At low temperatures and high pressures, real gases deviate significantly from the ideal gas equation. The higher the pressure and the lower the temperature, the greater the deviation.

📦 Deviation 1 — Particle Volume at High Pressure

  • Ideal gases assume particles take up NO space
  • At high pressure → particles are forced close together
  • Particle volume becomes a significant fraction of total volume
  • Less space available for movement
  • Gas takes up MORE volume than ideal predictions

At high pressures, the fraction of space taken up by molecules becomes substantial.

🔗 Deviation 2 — Intermolecular Forces at Low Temperature

What happens at low T?

  • Particles move more slowly
  • Intermolecular attractive forces become significant
  • Particles are pulled toward each other
  • Fewer and weaker collisions with container walls
  • Pressure is LOWER than ideal predictions

Ideal vs Real

  • Ideal: no forces → P = nRT/V
  • Real (low T): attractive forces → P < nRT/V
  • Real (high P): particle volume → P > nRT/V