Complete Gas Laws Cheat Sheet: Boyle’s, Charles’s, and Ideal Gas Laws

Introduction: Understanding Gas Behavior

Gas laws are fundamental principles that describe how gases behave under different conditions of temperature, pressure, and volume. These relationships form the cornerstone of numerous scientific and engineering applications, from weather forecasting to medical respiratory therapy. This comprehensive cheat sheet covers the major gas laws, their mathematical formulations, practical applications, and problem-solving strategies, serving as both a quick reference and a study aid for students, educators, and professionals working with gaseous systems.

Fundamental Gas Properties

Key Gas Variables

Common Pressure Unit Conversions

Temperature Conversions

Individual Gas Laws

Boyle’s Law

Statement: At constant temperature, the volume of a fixed amount of gas is inversely proportional to its pressure.

Mathematical Form: P₁V₁ = P₂V₂ (when T and n are constant)

Graphical Representation:

• P vs. V: Hyperbola
• P vs. 1/V: Straight line with positive slope
• PV vs. P: Horizontal line

Key Concepts:

• As pressure increases, volume decreases proportionally
• Doubling the pressure halves the volume
• The product PV remains constant

Example Problem: A gas occupies 2.5 L at 1.0 atm. What volume will it occupy at 2.5 atm if temperature remains constant?

Solution:

1. Given: P₁ = 1.0 atm, V₁ = 2.5 L, P₂ = 2.5 atm
2. Using Boyle’s Law: P₁V₁ = P₂V₂
3. V₂ = (P₁V₁)/P₂ = (1.0 atm × 2.5 L)/2.5 atm = 1.0 L

Charles’s Law

Statement: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature.

Mathematical Form: V₁/T₁ = V₂/T₂ (when P and n are constant)

Graphical Representation:

• V vs. T: Straight line with positive slope
• V vs. 1/T: Hyperbola
• V/T vs. T: Horizontal line

Key Concepts:

• As temperature increases, volume increases proportionally
• Doubling the absolute temperature doubles the volume
• The ratio V/T remains constant
• Volume would theoretically be zero at absolute zero (-273.15°C or 0 K)

Example Problem: A gas has a volume of 500 mL at 25°C. What will its volume be at 100°C if pressure remains constant?

Solution:

1. Given: V₁ = 500 mL, T₁ = 25°C = 298.15 K, T₂ = 100°C = 373.15 K
2. Using Charles’s Law: V₁/T₁ = V₂/T₂
3. V₂ = V₁(T₂/T₁) = 500 mL × (373.15 K/298.15 K) = 626 mL

Gay-Lussac’s Law

Statement: At constant volume, the pressure of a fixed amount of gas is directly proportional to its absolute temperature.

Mathematical Form: P₁/T₁ = P₂/T₂ (when V and n are constant)

Graphical Representation:

• P vs. T: Straight line with positive slope
• P vs. 1/T: Hyperbola
• P/T vs. T: Horizontal line

Key Concepts:

• As temperature increases, pressure increases proportionally
• Doubling the absolute temperature doubles the pressure
• The ratio P/T remains constant
• Pressure would theoretically be zero at absolute zero

Example Problem: A gas exerts a pressure of 1.5 atm at 30°C. What pressure will it exert at 80°C if volume remains constant?

Solution:

1. Given: P₁ = 1.5 atm, T₁ = 30°C = 303.15 K, T₂ = 80°C = 353.15 K
2. Using Gay-Lussac’s Law: P₁/T₁ = P₂/T₂
3. P₂ = P₁(T₂/T₁) = 1.5 atm × (353.15 K/303.15 K) = 1.75 atm

Avogadro’s Law

Statement: At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas.

Mathematical Form: V₁/n₁ = V₂/n₂ (when P and T are constant)

Key Concepts:

• As the amount of gas increases, volume increases proportionally
• Doubling the number of moles doubles the volume
• The ratio V/n remains constant

Example Problem: 2.0 moles of an ideal gas occupy 44.8 L at STP. What volume will 5.0 moles occupy under the same conditions?

Solution:

1. Given: n₁ = 2.0 mol, V₁ = 44.8 L, n₂ = 5.0 mol
2. Using Avogadro’s Law: V₁/n₁ = V₂/n₂
3. V₂ = V₁(n₂/n₁) = 44.8 L × (5.0 mol/2.0 mol) = 112 L

Combined Gas Law

Statement: Combines Boyle’s, Charles’s, and Gay-Lussac’s laws into a single expression.

Mathematical Form: P₁V₁/T₁ = P₂V₂/T₂ (when n is constant)

Key Concepts:

• Allows calculation of unknown gas properties when amount of gas is constant
• Simplifies to individual gas laws when appropriate variables are held constant
• The ratio PV/T remains constant

Example Problem: A gas has a volume of 3.0 L at 2.0 atm and 27°C. What will its volume be at 1.0 atm and 127°C?

Solution:

1. Given: P₁ = 2.0 atm, V₁ = 3.0 L, T₁ = 27°C = 300 K, P₂ = 1.0 atm, T₂ = 127°C = 400 K
2. Using the Combined Gas Law: P₁V₁/T₁ = P₂V₂/T₂
3. V₂ = V₁(P₁/P₂)(T₂/T₁) = 3.0 L × (2.0 atm/1.0 atm) × (400 K/300 K) = 8.0 L

Ideal Gas Law

Fundamental Equation and Variations

Ideal Gas Law Equation: PV = nRT

Where:

• P = pressure
• V = volume
• n = number of moles
• R = universal gas constant
• T = absolute temperature (Kelvin)

Universal Gas Constant (R) Values in Different Units:

Variations of the Ideal Gas Law:

Standard Conditions

Key Calculations Using the Ideal Gas Law

Finding Moles: n = PV/RT

Finding Volume: V = nRT/P

Finding Pressure: P = nRT/V

Finding Temperature: T = PV/nR

Example Problem 1: Calculate the number of moles in 2.5 L of gas at 1.5 atm and 25°C.

Solution:

1. Given: V = 2.5 L, P = 1.5 atm, T = 25°C = 298.15 K, R = 0.08206 L·atm/(mol·K)
2. Using n = PV/RT: n = (1.5 atm × 2.5 L)/(0.08206 L·atm/(mol·K) × 298.15 K) = 0.153 mol

Example Problem 2: A gas sample contains 0.25 mol and exerts a pressure of 2.0 atm at 20°C. What is its volume?

Solution:

1. Given: n = 0.25 mol, P = 2.0 atm, T = 20°C = 293.15 K, R = 0.08206 L·atm/(mol·K)
2. Using V = nRT/P: V = (0.25 mol × 0.08206 L·atm/(mol·K) × 293.15 K)/2.0 atm = 3.01 L

Gas Mixtures

Dalton’s Law of Partial Pressures

Statement: The total pressure of a gas mixture equals the sum of the partial pressures of each component.

Mathematical Form: Ptotal = P₁ + P₂ + P₃ + … + Pₙ

Partial Pressure: Pᵢ = xᵢ × Ptotal, where xᵢ is the mole fraction of gas i

Mole Fraction: xᵢ = nᵢ/ntotal, where nᵢ is the number of moles of gas i

Key Concepts:

• Each gas in a mixture behaves independently
• Partial pressure is the pressure that each gas would exert if it occupied the container alone
• Mole fractions sum to 1: x₁ + x₂ + … + xₙ = 1

Example Problem: A gas mixture contains 2.0 mol O₂, 5.0 mol N₂, and 3.0 mol CO₂. If the total pressure is 5.0 atm, what is the partial pressure of each gas?

Solution:

1. Calculate total moles: ntotal = 2.0 + 5.0 + 3.0 = 10.0 mol
2. Calculate mole fractions:
   • xO₂ = 2.0/10.0 = 0.20
   • xN₂ = 5.0/10.0 = 0.50
   • xCO₂ = 3.0/10.0 = 0.30
3. Calculate partial pressures:
   • PO₂ = 0.20 × 5.0 atm = 1.0 atm
   • PN₂ = 0.50 × 5.0 atm = 2.5 atm
   • PCO₂ = 0.30 × 5.0 atm = 1.5 atm
4. Verify: 1.0 + 2.5 + 1.5 = 5.0 atm ✓

Graham’s Law of Diffusion and Effusion

Statement: The rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass.

Mathematical Form: Rate₁/Rate₂ = √(M₂/M₁)

Key Concepts:

• Lighter gases diffuse and effuse faster than heavier gases
• The ratio of diffusion rates equals the inverse ratio of the square roots of their molar masses
• Applies to both diffusion (movement through another gas) and effusion (movement through a tiny hole)

Example Problem: Hydrogen gas (H₂) effuses through a porous barrier 4.0 times faster than an unknown gas. What is the molar mass of the unknown gas?

Solution:

1. Given: RateH₂/RateX = 4.0, MH₂ = 2.0 g/mol
2. Using Graham’s Law: RateH₂/RateX = √(MX/MH₂)
3. 4.0 = √(MX/2.0 g/mol)
4. Square both sides: 16.0 = MX/2.0 g/mol
5. MX = 16.0 × 2.0 g/mol = 32.0 g/mol (possibly O₂)

Kinetic Molecular Theory of Gases

Key Postulates

1. Gases consist of tiny particles (molecules or atoms) in constant, random motion
2. Gas particles are separated by large distances; actual volume of particles is negligible
3. Gas particles have negligible intermolecular forces
4. Collisions between particles and with container walls are perfectly elastic
5. Average kinetic energy of gas particles is directly proportional to absolute temperature

Important Equations

Where:

• k = Boltzmann constant (1.38 × 10⁻²³ J/K)
• T = absolute temperature (K)
• R = gas constant (8.314 J/(mol·K))
• NA = Avogadro’s number (6.022 × 10²³ particles/mol)
• M = molar mass (kg/mol)

Example Problem: Calculate the root mean square velocity of nitrogen molecules (N₂) at 25°C.

Solution:

1. Given: T = 25°C = 298.15 K, M(N₂) = 28.0 g/mol = 0.0280 kg/mol
2. Using vrms = √(3RT/M): vrms = √(3 × 8.314 J/(mol·K) × 298.15 K/0.0280 kg/mol) vrms = √(265,553 J/kg) = √(265,553 m²/s²) = 515 m/s

Non-Ideal Gas Behavior

Van der Waals Equation

Equation: (P + a(n/V)²)(V – nb) = nRT

Where:

• a accounts for attractive forces between molecules
• b accounts for the volume occupied by gas molecules themselves

Compressibility Factor: Z = PV/nRT

• Z = 1 for ideal gases
• Z < 1 indicates attractive forces dominate
• Z > 1 indicates repulsive forces dominate

Conditions When Gases Behave Non-Ideally

Example Problem: Calculate the pressure of 1.00 mol of CO₂ in a 1.00 L container at 300 K using: a) Ideal gas law b) Van der Waals equation (For CO₂: a = 3.64 L²·atm/mol² and b = 0.0427 L/mol)

Solution: a) Using ideal gas law: P = nRT/V = (1.00 mol × 0.08206 L·atm/(mol·K) × 300 K)/1.00 L = 24.6 atm

b) Using Van der Waals equation: P = nRT/(V – nb) – a(n/V)² P = (1.00 mol × 0.08206 L·atm/(mol·K) × 300 K)/(1.00 L – 1.00 mol × 0.0427 L/mol) – 3.64 L²·atm/mol² × (1.00 mol/1.00 L)² P = 24.6 atm/(0.9573) – 3.64 atm = 25.7 atm – 3.64 atm = 22.1 atm

Problem-Solving Strategies

1. General Problem-Solving Approach

1. Identify the known and unknown variables

   • Write down all given information
   • Convert all units to a consistent system (preferably SI)
   • Identify what you need to find

2. Select the appropriate gas law

   • Determine which variables are changing and which remain constant
   • Choose the law that relates the variables you have with the ones you need

3. Set up the equation and solve

   • Substitute the known values
   • Rearrange to isolate the unknown variable
   • Calculate and double-check your work

4. Verify your answer

   • Check that units are correct
   • Ensure the answer makes physical sense
   • Confirm magnitude with estimation

2. Typical Problem Types

3. Common Conversion Factors and Relationships

Practical Applications of Gas Laws

Laboratory Applications

Real-World Applications

Common Mistakes and How to Avoid Them

This cheat sheet provides a comprehensive overview of gas laws and their applications. Remember that while these laws accurately describe gas behavior under many conditions, real gases may deviate from ideal behavior, particularly at high pressures and low temperatures. Always consider the specific conditions of your system when applying these principles.