🌬️ Gas Properties & Kinetic Molecular Theory

Part 1 of 7 — Understanding Gas Behavior

Gases are everywhere — the air you breathe, the atmosphere above you, the propane in a grill tank. What makes gases unique among the states of matter is their ability to expand to fill any container, be easily compressed, and exert pressure on the walls of their container.

In this lesson, we explore the four key measurable properties of gases and the kinetic molecular theory (KMT) that explains gas behavior at the molecular level.

The Four Gas Variables

Every gas sample is described by four measurable quantities:

Pressure

Pressure is force per unit area: $P = F/A$. Gas molecules exert pressure by colliding with container walls.

Key conversions:

• $1 \text{ atm} = 760 \text{ mmHg} = 760 \text{ torr} = 101.325 \text{ kPa}$

Temperature

Temperature must always be in Kelvin for gas law calculations:

$T(K) = T(°C) + 273.15$

At $0 \text{ K}$ (absolute zero), molecular motion theoretically stops.

Volume

Volume is typically measured in liters (L) in chemistry. $1 \text{ L} = 1000 \text{ mL} = 0.001 \text{ m}^3$.

Amount

The amount of gas is measured in moles ($n$), which connects to the number of particles via Avogadro's number ($6.022 \times 10^{23}$).

Kinetic Molecular Theory (KMT)

The kinetic molecular theory describes an ideal gas with the following assumptions:

1. Gas particles are in constant, random motion — they travel in straight lines until they collide.
2. Collisions are perfectly elastic — no kinetic energy is lost during collisions.
3. Gas particles have negligible volume — the volume of individual molecules is tiny compared to the container.
4. No intermolecular forces — gas particles don't attract or repel each other.
5. Average kinetic energy is proportional to temperature (in Kelvin):

$KE_{\text{avg}} = \frac{3}{2}k_BT$

where $k_B = 1.38 \times 10^{-23}$ J/K is the Boltzmann constant.

Root Mean Square Speed

The average speed of gas molecules depends on temperature and molar mass:

$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$

where $R = 8.314$ J/(mol·K) and $M$ is the molar mass in kg/mol.

Key insight: Lighter molecules move faster at the same temperature.

Gas Properties Quiz 🎯

Unit Conversion Practice 🧮

Convert the following gas measurements:

1. Convert 2.50 atm to mmHg.

2. Convert 350 K to °C (to 3 significant figures).

3. Convert 0.500 L to mL.

KMT Concepts Check 🔍

Exit Quiz — Gas Properties & KMT ✅

📏 Boyle's, Charles's, and Avogadro's Laws

Part 2 of 7 — The Foundational Gas Laws

Before the ideal gas law was discovered, scientists identified three separate relationships between gas variables. Each law holds one or two variables constant while examining how the remaining variables relate.

These "simple" gas laws are the building blocks of $PV = nRT$.

Boyle's Law (Pressure–Volume)

At constant temperature and amount of gas:

$P_1V_1 = P_2V_2$

Pressure and volume are inversely proportional. When you compress a gas (decrease volume), the pressure increases — and vice versa.

Example

A gas occupies 4.00 L at 1.00 atm. What is the volume if the pressure increases to 2.00 atm?

$V_2 = \frac{P_1V_1}{P_2} = \frac{(1.00)(4.00)}{2.00} = 2.00 \text{ L}$

Conceptual Picture

When volume decreases, molecules hit the walls more often → more collisions per second → higher pressure.

Charles's Law (Volume–Temperature)

At constant pressure and amount of gas:

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

Volume and temperature are directly proportional (temperature in Kelvin!). Heat a gas and it expands; cool it and it contracts.

Example

A balloon has a volume of 2.50 L at 20°C. What volume will it have at 80°C? (Pressure constant)

$T_1 = 20 + 273.15 = 293.15 \text{ K}, \quad T_2 = 80 + 273.15 = 353.15 \text{ K}$

$V_2 = V_1 \times \frac{T_2}{T_1} = 2.50 \times \frac{353.15}{293.15} = 3.01 \text{ L}$

Why Kelvin?

If you use Celsius, 0°C would imply zero volume — which is nonsensical. Kelvin starts at absolute zero, where molecular motion stops.

Avogadro's Law (Volume–Amount)

At constant temperature and pressure:

$\frac{V_1}{n_1} = \frac{V_2}{n_2}$

Volume and amount of gas (in moles) are directly proportional. More molecules → more volume (at the same $T$ and $P$).

Example

3.00 mol of gas occupies 6.00 L. What volume will 5.00 mol occupy under the same conditions?

$V_2 = V_1 \times \frac{n_2}{n_1} = 6.00 \times \frac{5.00}{3.00} = 10.0 \text{ L}$

At STP

At standard temperature and pressure (STP: 0°C, 1 atm), 1 mole of any ideal gas occupies 22.4 L. This is the molar volume at STP.

Gas Law Identification Quiz 🎯

Gas Law Calculations 🧮

1. A gas at 3.00 atm occupies 12.0 L. What is the volume at 1.50 atm? (constant T and n, in L)

2. A gas occupies 500 mL at 27°C. What volume will it occupy at 127°C at constant pressure? (in mL, round to nearest whole number)

3. 4.00 mol of gas occupies 10.0 L at a given T and P. How many moles would occupy 25.0 L? (in mol, to 3 significant figures)

Match the Relationship 🔍

Exit Quiz — Gas Laws ✅

⚗️ The Ideal Gas Law

Part 3 of 7 — PV = nRT

The three simple gas laws (Boyle's, Charles's, Avogadro's) each describe a two-variable relationship while holding others constant. The ideal gas law combines them all into one powerful equation:

$PV = nRT$

This single equation lets you calculate any one variable if you know the other three.

The Equation and R

$PV = nRT$

The Gas Constant R

The value of $R$ depends on your pressure unit:

Most AP Chemistry problems use $R = 0.0821$ L·atm/(mol·K).

Rearranged Forms

• Solve for $V$: $V = \frac{nRT}{P}$
• Solve for $P$: $P = \frac{nRT}{V}$
• Solve for $n$: $n = \frac{PV}{RT}$
• Solve for $T$: $T = \frac{PV}{nR}$

Worked Examples

Example 1: Find Volume

What volume does 0.500 mol of gas occupy at 1.20 atm and 25°C?

$T = 25 + 273.15 = 298.15 \text{ K}$

$V = \frac{nRT}{P} = \frac{(0.500)(0.0821)(298.15)}{1.20} = 10.2 \text{ L}$

Example 2: Find Pressure

2.00 mol of gas is in a 15.0 L container at 300 K. What is the pressure?

$P = \frac{nRT}{V} = \frac{(2.00)(0.0821)(300)}{15.0} = 3.28 \text{ atm}$

Example 3: Find Moles

A gas at 2.50 atm and 350 K occupies 5.00 L. How many moles?

$n = \frac{PV}{RT} = \frac{(2.50)(5.00)}{(0.0821)(350)} = 0.435 \text{ mol}$

Unit Conversion Reminder

Always check:

• $P$ in atm (if using $R = 0.0821$)
• $V$ in liters
• $T$ in Kelvin
• $n$ in moles

Ideal Gas Law Concept Check 🎯

PV = nRT Calculations 🧮

Use $R = 0.0821$ L·atm/(mol·K). Round to appropriate significant figures.

1. What volume (in L) does 1.00 mol of ideal gas occupy at STP (0°C, 1.00 atm)? (to 3 significant figures)

2. What pressure (in atm) is exerted by 3.50 mol of gas in a 20.0 L container at 400 K? (to 3 significant figures)

3. How many moles of gas are in a 10.0 L container at 2.00 atm and 27°C? (to 3 significant figures)

Quick Concept Sort 🔍

Exit Quiz — The Ideal Gas Law ✅

⚖️ Molar Mass & Gas Density

Part 4 of 7 — Connecting Gases to Molar Mass

The ideal gas law can be rearranged to find the molar mass of an unknown gas, or to calculate the density of a gas at any temperature and pressure. These are powerful lab techniques used frequently in AP Chemistry.

Molar Mass from Gas Data

Starting with $PV = nRT$ and substituting $n = m/M$ (where $m$ = mass, $M$ = molar mass):

$PV = \frac{m}{M}RT$

Solving for molar mass:

$M = \frac{mRT}{PV}$

Example

A 0.325 g sample of gas occupies 225 mL at 100°C and 0.960 atm. Find the molar mass.

$T = 100 + 273.15 = 373.15 \text{ K}, \quad V = 0.225 \text{ L}$

$M = \frac{(0.325)(0.0821)(373.15)}{(0.960)(0.225)} = \frac{9.957}{0.216} = 46.1 \text{ g/mol}$

This matches ethanol (C₂H₅OH), which has $M = 46.07$ g/mol.

Lab Application: Dumas Method

In the Dumas method for finding molar mass:

1. Vaporize a liquid in a flask of known volume
2. Measure the temperature and atmospheric pressure
3. Condense the vapor and weigh it
4. Use $M = mRT/(PV)$

Gas Density

Density is mass per volume: $d = m/V$. From $PV = (m/M)RT$:

$\frac{m}{V} = \frac{PM}{RT}$

$\boxed{d = \frac{PM}{RT}}$

Key Observations

• Gas density increases with pressure (more molecules per volume)
• Gas density decreases with temperature (gas expands)
• Gas density increases with molar mass (heavier molecules)

Example

What is the density of O₂ ($M = 32.00$ g/mol) at STP?

$d = \frac{PM}{RT} = \frac{(1.00)(32.00)}{(0.0821)(273.15)} = \frac{32.00}{22.43} = 1.43 \text{ g/L}$

Comparing Gases

At the same $T$ and $P$, the ratio of gas densities equals the ratio of molar masses:

$\frac{d_1}{d_2} = \frac{M_1}{M_2}$

Molar Mass & Density Concepts 🎯

Molar Mass & Density Calculations 🧮

Use $R = 0.0821$ L·atm/(mol·K).

1. A 1.56 g sample of gas occupies 1.00 L at 27°C and 1.00 atm. What is the molar mass? (in g/mol, to 3 significant figures)

2. What is the density of N₂ ($M = 28.02$ g/mol) at 25°C and 1.00 atm? (in g/L, to 3 significant figures)

3. A gas has a density of 3.17 g/L at STP. What is its molar mass? (in g/mol, to 3 significant figures)

Density Relationships 🔍

Exit Quiz — Molar Mass & Density ✅

🎈 Dalton's Law of Partial Pressures

Part 5 of 7 — Gas Mixtures

Most gases we encounter are actually mixtures — air itself is about 78% N₂, 21% O₂, and 1% Ar plus trace gases. Dalton's Law tells us how to handle the pressure contributions from each gas in a mixture.

Dalton's Law

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

$P_{\text{total}} = P_1 + P_2 + P_3 + \cdots$

Each partial pressure is the pressure that gas would exert if it alone occupied the entire container.

Using the Ideal Gas Law

Since $PV = nRT$, each gas independently obeys:

$P_i = \frac{n_iRT}{V}$

And the total:

$P_{\text{total}} = \frac{n_{\text{total}}RT}{V}$

Example

A 10.0 L container at 300 K holds 0.200 mol N₂ and 0.300 mol O₂.

$P_{N_2} = \frac{(0.200)(0.0821)(300)}{10.0} = 0.493 \text{ atm}$

$P_{O_2} = \frac{(0.300)(0.0821)(300)}{10.0} = 0.739 \text{ atm}$

$P_{\text{total}} = 0.493 + 0.739 = 1.232 \text{ atm}$

Mole Fraction

The mole fraction ($\chi$) of a component is the fraction of total moles that it contributes:

$\chi_i = \frac{n_i}{n_{\text{total}}}$

The partial pressure is related to mole fraction by:

$P_i = \chi_i \times P_{\text{total}}$

Example

A mixture has 2.0 mol He and 3.0 mol Ne at a total pressure of 5.0 atm.

$\chi_{He} = \frac{2.0}{2.0 + 3.0} = 0.40$

$P_{He} = 0.40 \times 5.0 = 2.0 \text{ atm}$

$\chi_{Ne} = \frac{3.0}{5.0} = 0.60, \quad P_{Ne} = 0.60 \times 5.0 = 3.0 \text{ atm}$

Note: All mole fractions must add up to 1.0: $\chi_{He} + \chi_{Ne} = 0.40 + 0.60 = 1.00$ ✓

Gas Collection Over Water

When a gas is collected by displacement of water, the collected gas is mixed with water vapor. You must subtract the vapor pressure of water:

$P_{\text{gas}} = P_{\text{total}} - P_{\text{H}_2\text{O}}$

The vapor pressure of water depends on temperature (values are given in data tables).

Example

Oxygen is collected over water at 25°C. The total pressure is 752 mmHg. The vapor pressure of water at 25°C is 23.8 mmHg.

$P_{O_2} = 752 - 23.8 = 728 \text{ mmHg} = 0.958 \text{ atm}$

This corrected pressure is then used in $PV = nRT$ to find the moles of dry gas collected.

Partial Pressure Concepts 🎯

Partial Pressure Calculations 🧮

1. A flask contains 0.50 mol Ar and 1.50 mol Ne. The total pressure is 4.00 atm. What is the partial pressure of Ar? (in atm)

2. Hydrogen gas is collected over water at 22°C ($P_{H_2O}$ = 19.8 mmHg). The total pressure is 745 mmHg. What is the pressure of the dry hydrogen? (in mmHg, to 3 significant figures)

3. A mixture has $\chi_{CO_2} = 0.30$ and the total pressure is 2.50 atm. What is $P_{CO_2}$? (in atm)

Dalton's Law Concepts 🔍

Exit Quiz — Dalton's Law ✅

🧪 Problem-Solving Workshop

Part 6 of 7 — Mixed Gas Law Calculations

Now it's time to put all the gas laws together. In this workshop, you'll practice selecting the right equation, converting units, and solving multi-step problems — exactly the way they appear on the AP Chemistry exam.

Problem-Solving Strategy

1. Identify what you know and what you need to find.
2. Choose the right law:
   • One gas, two sets of conditions → Combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
   • One set of conditions → Ideal gas law: $PV = nRT$
   • Gas mixtures → Dalton's law: $P_{\text{total}} = \sum P_i$
   • Unknown molar mass → $M = mRT/(PV)$
   • Gas density → $d = PM/(RT)$
3. Convert all units — Kelvin for $T$, liters for $V$, atm for $P$ (if using $R = 0.0821$).
4. Solve and check — does the answer make physical sense?

The Combined Gas Law

When the amount of gas is constant but P, V, and T all change:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

This reduces to Boyle's or Charles's law when one variable is held constant.

STP Problems

At STP (0°C = 273.15 K, 1.00 atm):

• 1 mol of ideal gas = 22.4 L
• This gives a quick shortcut for many calculations

Example 1: Volume at STP

What volume does 0.750 mol of CO₂ occupy at STP?

$V = 0.750 \times 22.4 = 16.8 \text{ L}$

Example 2: Moles from Volume at STP

A sample of gas occupies 5.60 L at STP. How many moles?

$n = \frac{5.60}{22.4} = 0.250 \text{ mol}$

Example 3: Combined Gas Law

A gas at 2.00 atm, 10.0 L, and 300 K is changed to 1.00 atm and 600 K. New volume?

$V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{(2.00)(10.0)(600)}{(300)(1.00)} = 40.0 \text{ L}$

Gas Stoichiometry

When gases appear in chemical reactions, you can use the ideal gas law with stoichiometry:

$\text{grams} \rightarrow \text{moles} \rightarrow \text{mole ratio} \rightarrow \text{moles of gas} \rightarrow PV = nRT$

Example

How many liters of O₂ at 25°C and 1.00 atm are produced from the decomposition of 49.0 g of KClO₃?

$2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$

Step 1: Moles of KClO₃ ($M = 122.55$ g/mol): $n = 49.0/122.55 = 0.400 \text{ mol}$

Step 2: Moles of O₂: $0.400 \times \frac{3}{2} = 0.600 \text{ mol O}_2$

Step 3: Volume: $V = \frac{nRT}{P} = \frac{(0.600)(0.0821)(298.15)}{1.00} = 14.7 \text{ L}$

Problem Type Identification 🎯

Calculation Workshop 🧮

1. A 5.00 L gas sample at 2.00 atm and 400 K is cooled to 200 K and compressed to 1.00 L. What is the new pressure? (in atm)

2. At STP, how many grams of CO₂ ($M = 44.01$ g/mol) occupy 11.2 L? (in g, to 3 significant figures)

3. A mixture of 0.30 mol He and 0.70 mol Ar has a total pressure of 5.00 atm. What is the partial pressure of Ar? (in atm)

Quick Decision Guide 🔍

Exit Quiz — Problem Solving ✅

🏆 Synthesis & AP Review

Part 7 of 7 — Ideal vs. Real Gases & AP Exam Preparation

You've mastered the ideal gas law and all its variations. Now we examine when the ideal gas model breaks down, introduce the van der Waals equation for real gases, and tackle AP-style free-response questions.

Ideal vs. Real Gases

The ideal gas law works well under many conditions, but real gases deviate from ideal behavior when:

When Do Gases Deviate Most?

• High pressure → molecules are close together → volume of molecules matters, attractions are significant
• Low temperature → molecules move slowly → attractions have more effect
• Near the boiling point → gas is close to condensing → strong intermolecular forces

When Is Ideal Behavior Best?

• Low pressure → molecules far apart → negligible volume and attractions
• High temperature → fast-moving molecules → overcome attractions easily
• Noble gases and small nonpolar molecules → weakest intermolecular forces

The van der Waals Equation

To correct for real gas behavior:

$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$

• $a$ = attraction parameter (larger for polar molecules with strong IMFs)
• $b$ = size parameter (larger for bigger molecules)

Example Values

Notice: H₂O has a large $a$ (strong H-bonds) but small $b$ (small molecule). He has tiny values for both (noble gas, very small).

Ideal vs. Real Gas Quiz 🎯

AP-Style Calculation Practice 🧮

1. 0.500 mol of an ideal gas at 1.00 atm and 273 K occupies what volume? (in L, to 3 significant figures)

2. A real gas has $a = 3.59$ L²·atm/mol² and $b = 0.043$ L/mol. For 1.00 mol in a 0.500 L container at 500 K, calculate the ideal gas pressure first: $P_{\text{ideal}} = nRT/V$. (in atm, to 3 significant figures)

3. What is the corrected van der Waals pressure for the same gas? Use $P = nRT/(V-nb) - an^2/V^2$. (in atm, to 3 significant figures)

Real Gas Behavior Trends 🔍

AP Free-Response Style Questions ✅