Ideal Gas Law

Theory Exercises

The Ideal Gas Law

The ideal gas law is one of the most important equations in chemistry and physics. It describes the relationship between pressure, volume, temperature, and the amount of gas in a container. This law helps us understand and predict the behavior of gases under different conditions.

The Ideal Gas Law Equation

\[PV = nR_gT\]

Where:

P = Pressure (usually in atm, Pa, or mmHg)
V = Volume (usually in L or m³)
n = Amount of gas in moles
R_g = Universal gas constant (0.0821 L·atm/mol·K or 8.314 J/mol·K)
T = Temperature in Kelvin (K)

⚠️ Important: Temperature MUST always be in Kelvin for gas law calculations!

\[T(K) = T(°C) + 273.15\]

Prerequisites: Proportional Relationships

The gas laws are based on proportional relationships. If you need to review direct and inverse proportional relationships, check out the Science Equations and Proportional Relationships page first.

Quick reminder:

Direct proportion: When one variable increases, the other increases (y ∝ x)
Inverse proportion: When one variable increases, the other decreases (y ∝ 1/x)

Understanding the Gas Laws - Building Blocks

The ideal gas law combines three simpler gas laws. Now that we understand proportional relationships, let's explore how they apply to gas behavior:

1. Boyle's Law (Pressure vs. Volume) - Inverse Proportion

At constant temperature and amount:

\[P \propto \frac{1}{V} \quad \text{or} \quad PV = \text{constant} \quad \text{or} \quad P_1V_1 = P_2V_2\]

Relationship: Pressure and volume are inversely proportional
What it means: When you squeeze a gas (increase pressure), its volume decreases
Proportional thinking: If pressure doubles → volume halves
Example: Pressing down on a syringe plunger

2. Charles's Law (Volume vs. Temperature) - Direct Proportion

At constant pressure and amount:

\[V \propto T \quad \text{or} \quad \frac{V}{T} = \text{constant} \quad \text{or} \quad \frac{V_1}{T_1} = \frac{V_2}{T_2}\]

Relationship: Volume and temperature are directly proportional
What it means: When you heat a gas, it expands
Proportional thinking: If temperature doubles → volume doubles
Example: A balloon gets bigger when heated, smaller when cooled

3. Gay-Lussac's Law (Pressure vs. Temperature) - Direct Proportion

At constant volume and amount:

\[P \propto T \quad \text{or} \quad \frac{P}{T} = \text{constant} \quad \text{or} \quad \frac{P_1}{T_1} = \frac{P_2}{T_2}\]

Relationship: Pressure and temperature are directly proportional
What it means: When you heat a gas in a rigid container, pressure increases
Proportional thinking: If temperature doubles → pressure doubles
Example: A pressurized can gets more dangerous when heated

Quick Reference: Direct vs. Inverse Relationships in Gas Laws

🔗 Direct Proportional Relationships (Same Direction)

Volume ∝ Temperature (Charles's Law): Heat gas → volume increases
Pressure ∝ Temperature (Gay-Lussac's Law): Heat gas → pressure increases
Volume ∝ Amount of gas: More gas → larger volume

🔄 Inverse Proportional Relationships (Opposite Direction)

Pressure ∝ 1/Volume (Boyle's Law): Squeeze gas → pressure increases, volume decreases

Memory tip: Think of temperature as energy - more energy makes gases want to expand (direct relationships with V and P). Pressure and volume fight each other - squeeze one, the other fights back (inverse relationship).

How to Solve Gas Law Problems

Let's learn the systematic approach to solving gas problems:

Step-by-Step Problem Solving Method

Identify what you know (given values)
Identify what you need to find (unknown variable)
Convert all temperatures to Kelvin
Choose the appropriate equation
Rearrange to solve for the unknown
Substitute values and calculate
Check your answer (does it make sense?)

Which Equation to Use?

Variables Changing	Constants	Use This Equation
P and V	T and n	P₁V₁ = P₂V₂
V and T	P and n	V₁/T₁ = V₂/T₂
P and T	V and n	P₁/T₁ = P₂/T₂
Multiple variables	None or n only	PV = nRgT

Practical Applications and Examples

Example 1: Temperature Effect on Volume (Charles's Law)

Problem: A balloon contains 2.0 L of air at 20°C. What will be its volume if heated to 80°C at constant pressure? Step 1 - Given:

V₁ = 2.0 L
T₁ = 20°C = 20 + 273 = 293 K
T₂ = 80°C = 80 + 273 = 353 K
P and n are constant

Step 2 - Find: V₂ = ? Step 3 - Equation: Charles's Law

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2}\]

Step 4 - Solve for V₂:

\[V_2 = V_1 \times \frac{T_2}{T_1} = 2.0 \text{ L} \times \frac{353 \text{ K}}{293 \text{ K}} = 2.4 \text{ L}\]

Answer: The balloon will expand to 2.4 L when heated. Understanding: When you heat a gas at constant pressure, it expands proportionally to the temperature increase in Kelvin. Proportional check: Temperature ratio = 353/293 = 1.20, so volume should also increase by factor of 1.20: 2.0 × 1.20 = 2.4 L ✓

Example 2: Pressure Effect on Volume (Boyle's Law)

Problem: A gas occupies 5.0 L at 1.0 atm. What pressure is needed to compress it to 2.0 L at constant temperature? Step 1 - Given:

V₁ = 5.0 L
P₁ = 1.0 atm
V₂ = 2.0 L
T and n are constant

Step 2 - Find: P₂ = ? Step 3 - Equation: Boyle's Law

\[P_1V_1 = P_2V_2\]

Step 4 - Solve for P₂:

\[P_2 = \frac{P_1V_1}{V_2} = \frac{1.0 \text{ atm} \times 5.0 \text{ L}}{2.0 \text{ L}} = 2.5 \text{ atm}\]

Answer: A pressure of 2.5 atm is needed to compress the gas to 2.0 L. Understanding: When you compress a gas to a smaller volume, the pressure increases inversely. Proportional check: Volume decreased by factor of 2.0/5.0 = 0.4, so pressure should increase by factor of 1/0.4 = 2.5: 1.0 × 2.5 = 2.5 atm ✓

Example 3: Fixed Volume, Changing Temperature (Gay-Lussac's Law)

Problem: A sealed container holds gas at 2.0 atm and 25°C. What will be the pressure if heated to 100°C? Step 1 - Given:

P₁ = 2.0 atm
T₁ = 25°C = 25 + 273 = 298 K
T₂ = 100°C = 100 + 273 = 373 K
V and n are constant (sealed container)

Step 2 - Find: P₂ = ? Step 3 - Equation: Gay-Lussac's Law

\[ \frac{P_1}{T_1} = \frac{P_2}{T_2}\]

Step 4 - Solve for P₂:

\[P_2 = P_1 \times \frac{T_2}{T_1} = 2.0 \text{ atm} \times \frac{373 \text{ K}}{298 \text{ K}} = 2.5 \text{ atm}\]

Answer: The pressure increases to 2.5 atm when heated. Proportional check: Temperature increased by factor of 373/298 = 1.25, so pressure should also increase by factor of 1.25: 2.0 × 1.25 = 2.5 atm ✓ ⚠️ Safety Note: This is why pressurized containers can be dangerous when heated!

Example 4: Using the Full Ideal Gas Law

Problem: How many moles of gas are in a 10.0 L container at 2.0 atm and 27°C? Step 1 - Given:

V = 10.0 L
P = 2.0 atm
T = 27°C = 27 + 273 = 300 K
Rg = 0.0821 L·atm/mol·K

Step 2 - Find: n = ? Step 3 - Equation: Ideal Gas Law

\[PV = nR_gT\]

Step 4 - Solve for n:

\[n = \frac{PV}{R_gT} = \frac{2.0 \text{ atm} \times 10.0 \text{ L}}{0.0821 \text{ L·atm/mol·K} \times 300 \text{ K}} = 0.81 \text{ mol}\]

Answer: The container holds 0.81 moles of gas.

Key Relationships to Remember

When Amount and Volume Are Fixed:

Scenario: A sealed rigid container (like a steel tank)

If temperature increases → pressure increases
If temperature decreases → pressure decreases
Relationship: P ∝ T (directly proportional)

When Amount and Pressure Are Fixed:

Scenario: A flexible container at constant atmospheric pressure (like a balloon)

If temperature increases → volume increases
If temperature decreases → volume decreases
Relationship: V ∝ T (directly proportional)

When Amount and Temperature Are Fixed:

Scenario: Compressing or expanding gas at room temperature

If pressure increases → volume decreases
If pressure decreases → volume increases
Relationship: P ∝ 1/V (inversely proportional)

Common Mistakes to Avoid

❌ Don't Forget These:

Always convert temperature to Kelvin! (Add 273 to Celsius)
Check units are consistent (atm with L, Pa with m³)
Identify which variables are constant before choosing an equation
Use the correct value of Rg (0.0821 for atm·L, 8.314 for J)
Double-check your final answer - does it make physical sense?

Real-World Applications

Weather and Atmosphere

Hot air balloons: Heating air makes it less dense, causing the balloon to rise
Weather patterns: Temperature differences create pressure differences that drive wind
Altitude effects: Lower pressure at high altitude affects breathing and boiling points

Technology and Industry

Car tires: Temperature changes affect tire pressure
Scuba diving: Understanding pressure changes with depth
Refrigeration: Gas compression and expansion for cooling
Aerosol cans: Pressure relationships for spray mechanisms

Safety Considerations

Pressure vessels: Never heat sealed containers
Gas cylinders: Store in cool, well-ventilated areas
Car safety: Check tire pressure when temperature changes significantly

Practice Strategy

To master gas law problems:

Learn to identify which variables change and which stay constant
Practice unit conversions, especially temperature to Kelvin
Start with simple problems using individual gas laws
Work up to complex problems using the full ideal gas law
Always check if your answer makes physical sense

Remember: The ideal gas law is a powerful tool that helps us understand and predict how gases behave. Master the relationships between P, V, n, and T, and you'll be able to solve any gas problem!