Chapter 6: Intermolecular Forces and States of Matter

Summary

This chapter connects molecular structure to intermolecular forces (London dispersion, dipole-dipole, hydrogen bonding, and ion-dipole) and explores how these forces determine the properties of solids, liquids, and gases including the ideal gas law.

Concepts Covered

This chapter covers the following 35 concepts from the learning graph:

Intermolecular Forces
Intramolecular Forces
IMF vs Chemical Bonds
London Dispersion Forces
Polarizability
Dipole-Dipole Forces
Hydrogen Bonding
Hydrogen Bond Requirements
Ion-Dipole Forces
IMF Strength Comparison
States of Matter
Kinetic Molecular Theory
Solids
Crystalline Solids
Amorphous Solids
Liquids
Surface Tension
Viscosity
Capillary Action
Gases
Gas Pressure
Atmospheric Pressure
Ideal Gas Law
Ideal Gas Constant
Boyle's Law
Charles's Law
Avogadro's Law
Dalton's Law
Partial Pressures
Mole Fraction
Gas Stoichiometry
Molar Volume
Real Gases
Van der Waals Equation
Deviations from Ideality

Prerequisites

This chapter builds on concepts from:

Introduction

Why does water stay liquid at room temperature while methane, a molecule of similar mass, boils at $-162^\circ\text{C}$? Why does rubbing alcohol evaporate quickly off your skin while cooking oil barely evaporates at all? The answers lie in the forces that act between molecules — forces we call intermolecular forces (IMFs). These attractions are the hidden architects of the physical world, determining whether a substance is solid, liquid, or gas under ordinary conditions, and controlling properties as diverse as boiling point, surface tension, and viscosity.

In this chapter you will build a systematic understanding of intermolecular forces — what causes them, how to rank their strength, and how they shape macroscopic physical properties. You will then apply that understanding to the behavior of gases, learning the elegant mathematical laws that connect pressure, volume, temperature, and the amount of gas. The chapter closes by examining where those laws break down and how chemists correct for the real behavior of gas molecules.

Section 1: Intermolecular Forces vs. Intramolecular Forces

Defining the Two Types of Forces

The word intramolecular means "within a molecule." Intramolecular forces are the covalent and ionic bonds that hold atoms together inside a molecule or formula unit. Breaking these bonds requires enormous energy — on the order of hundreds of kilojoules per mole. When water boils, for example, you are not breaking the O–H covalent bonds inside each water molecule; those bonds stay intact. You are separating molecules from one another.

Intermolecular forces are the attractions and repulsions that act between separate molecules or between molecules and ions. These forces are generally much weaker than covalent bonds — typically 1 to 50 kJ/mol compared to 150–400 kJ/mol for most covalent bonds. Despite being weaker, IMFs are responsible for nearly every physical property we observe: boiling point, melting point, vapor pressure, surface tension, viscosity, solubility, and the three states of matter themselves.

IMF vs. Chemical Bonds — a critical distinction for the AP exam:

Chemical bonds (covalent, ionic, metallic) hold atoms together within a substance. They are intramolecular and are broken in chemical reactions.
Intermolecular forces act between separate particles (molecules, atoms, or ions). They determine physical properties and are overcome during phase changes such as melting and boiling.

When you read that water has a boiling point of $100^\circ\text{C}$, you are really reading a measurement of how strong the IMFs between water molecules are — specifically, how much kinetic energy water molecules must have to escape the liquid phase and become gas.

Section 2: Types of Intermolecular Forces

London Dispersion Forces

London dispersion forces (also called van der Waals dispersion forces or simply dispersion forces) are the most universal IMF — they exist between all atoms and molecules, whether polar or nonpolar. Even noble gases like helium and argon experience dispersion forces, which is why they can be liquefied at very low temperatures.

Dispersion forces arise from temporary dipoles. Even in a perfectly nonpolar molecule, the electron cloud is in constant, rapid motion. At any instant, the electrons may be distributed unevenly across the molecule, creating a temporary (or instantaneous) dipole — one end is momentarily slightly negative, the other slightly positive. This temporary dipole then induces a dipole in a neighboring molecule. The result is a fleeting attractive interaction. Millions of these fleeting interactions add up to a measurable force.

The key concept that controls the strength of London dispersion forces is polarizability — the ease with which the electron cloud of an atom or molecule can be distorted to form a dipole. Larger atoms and molecules have more electrons and more diffuse, loosely held electron clouds, making them more polarizable. As a result:

Larger molecules have stronger London dispersion forces.
Molecules with more electrons have higher boiling points due to stronger dispersion.
Elongated ("rod-shaped") molecules have stronger dispersion forces than compact spherical molecules of the same formula, because they have more surface area for contact.

Examples illustrating the size trend for nonpolar molecules:

F$_2$ (bp $-188^\circ\text{C}$) < Cl$_2$ (bp $-34^\circ\text{C}$) < Br$_2$ (bp $59^\circ\text{C}$) < I$_2$ (bp $184^\circ\text{C}$)
CH$_4$ (bp $-161^\circ\text{C}$) < C$_2$H$_6$ (bp $-89^\circ\text{C}$) < C$_4$H$_{10}$ (bp $-1^\circ\text{C}$) < C$_8$H$_{18}$ (bp $126^\circ\text{C}$)

London dispersion forces are the only IMF present in nonpolar molecules, but they also contribute to the total intermolecular attraction in polar molecules.

Dipole-Dipole Forces

When a molecule has a permanent dipole moment — that is, when its electrons are distributed unevenly due to differences in electronegativity — it is a polar molecule. Dipole-dipole forces arise from the electrostatic attraction between the positive end of one polar molecule and the negative end of an adjacent polar molecule.

For dipole-dipole forces to be significant, two conditions must be met:

The molecule must contain polar bonds (bonds between atoms with different electronegativities).
The molecule must have an asymmetric geometry so that the bond dipoles do not cancel. (This is why CO$_2$ is nonpolar despite its polar bonds — the linear geometry causes the dipoles to cancel exactly.)

Polar molecules like HCl, SO$_2$, and acetone (CH$_3$COCH$_3$) experience dipole-dipole interactions in addition to London dispersion forces. Dipole-dipole forces are generally stronger than dispersion forces for molecules of similar size, leading to higher boiling points.

Hydrogen Bonding

Hydrogen bonding is a special, unusually strong type of dipole-dipole interaction. It occurs when a hydrogen atom bonded directly to a highly electronegative atom (nitrogen, oxygen, or fluorine) interacts with a lone pair of electrons on another electronegative atom (N, O, or F).

Hydrogen bond requirements — both criteria must be satisfied:

A hydrogen atom must be covalently bonded to N, O, or F (the hydrogen bond donor).
The hydrogen must be attracted to a lone pair on a nearby N, O, or F atom (the hydrogen bond acceptor).

The combination of a very electronegative atom and the tiny size of hydrogen creates an unusually large partial charge on the H atom, making these attractions much stronger than ordinary dipole-dipole forces. Hydrogen bonds have energies of roughly 10–40 kJ/mol, compared to 2–20 kJ/mol for typical dipole-dipole forces.

Hydrogen bonding explains several remarkable properties of water:

Water's boiling point ($100^\circ\text{C}$) is far higher than expected for a molecule of its small molar mass (18 g/mol). By comparison, H$_2$S (molar mass 34 g/mol) boils at $-60^\circ\text{C}$.
Ice is less dense than liquid water because hydrogen bonds force water molecules into a hexagonal open lattice structure.
Water has an unusually high specific heat capacity, making it an excellent temperature buffer for biological systems.
DNA's double helix is held together by hydrogen bonds between complementary base pairs.

Molecules that exhibit hydrogen bonding include: water (H$_2$O), ammonia (NH$_3$), hydrogen fluoride (HF), ethanol (C$_2$H$_5$OH), and all carboxylic acids, alcohols, and amines.

Ion-Dipole Forces

Ion-dipole forces are the attractive forces between an ion (a charged particle, either cation or anion) and the oppositely charged end of a polar molecule. These are the strongest of the intermolecular forces and are critical for understanding how ionic compounds dissolve in polar solvents like water.

When table salt (NaCl) dissolves in water, the positive Na$^+$ ions attract the oxygen end (partial negative charge) of water molecules, while the negative Cl$^-$ ions attract the hydrogen end (partial positive charge) of water molecules. This process is called hydration. The energy released when ions are hydrated partially compensates for the energy required to break the ionic lattice, allowing the salt to dissolve.

Ion-dipole forces are why "like dissolves like" works so well: polar solvents like water dissolve ionic and polar substances because the ion-dipole and dipole-dipole forces between solute and solvent are strong enough to pull the solute apart and surround individual particles.

IMF Strength Comparison

Understanding the relative strength of intermolecular forces is essential for predicting boiling points, solubility, and other physical properties. The general ranking from weakest to strongest is:

London Dispersion < Dipole-Dipole < Hydrogen Bonding < Ion-Dipole

However, this ranking has important nuances. A large nonpolar molecule with many electrons can have stronger dispersion forces than a small polar molecule has dipole-dipole forces. Always compare forces in context.

IMF Type	What Molecules	Relative Strength	Effect on Boiling Point	Example
London Dispersion	All molecules and atoms	Weakest (but grows with size)	Increases with molar mass	He, CH$_4$, I$_2$
Dipole-Dipole	Polar molecules	Moderate	Significantly higher than nonpolar analogs	HCl, SO$_2$, acetone
Hydrogen Bonding	Molecules with N–H, O–H, or F–H bonds	Strong	Very high for molecule size	H$_2$O, NH$_3$, HF
Ion-Dipole	Ion + polar molecule	Strongest	Governs dissolution in polar solvents	Na$^+$ in H$_2$O

Diagram: IMF Strength Explorer — Boiling Points vs. IMF Type

IMF Strength Explorer — Boiling Points vs. IMF Type

Type: MicroSim sim-id: imf-strength-explorer
Library: p5.js
Status: Specified

Learning objective: Students will compare the boiling points of molecules with different intermolecular forces and identify the relationship between IMF type and boiling point (Bloom L4: Analyze — compare, differentiate, draw conclusions).

Canvas size: 800 × 450 px, responsive to window resize.

Layout: - Left panel (60% width): A vertical bar chart showing boiling points (in °C, y-axis from −200 to 200) for 8 molecules arranged side by side in four color-coded groups. - Group 1 (gray, London Dispersion only): CH₄ (−161°C), C₄H₁₀ (−1°C), I₂ (184°C) - Group 2 (blue, Dipole-Dipole): HCl (−85°C), SO₂ (−10°C) - Group 3 (red, Hydrogen Bonding): NH₃ (−33°C), HF (19°C), H₂O (100°C) - A dashed horizontal line at 0°C labeled "Room Temperature Reference" - Right panel (40% width): An info box that updates when the user clicks a bar, showing: - Molecule name and chemical formula - IMF type(s) present - Molar mass - Boiling point - Brief explanation of why the boiling point is high or low

Interactions: - Clicking a bar highlights it and updates the info panel. - A dropdown menu lets students filter by IMF type ("Show All", "London Dispersion Only", "Dipole-Dipole", "Hydrogen Bonding"). - Hovering over a bar shows a tooltip with the boiling point value.

Colors: London Dispersion bars = medium gray; Dipole-Dipole bars = cornflower blue; Hydrogen Bonding bars = tomato red. Selected bar gets a gold border highlight.

Labels: Y-axis labeled "Boiling Point (°C)", X-axis shows molecule names at 45° angle. A color legend in the top right of the chart area.

Pedagogical note: The chart makes visible the pattern that hydrogen-bonded molecules punch above their weight — H₂O boils far higher than molecules of similar mass without hydrogen bonding.

Section 3: States of Matter and Kinetic Molecular Theory

Kinetic Molecular Theory

Kinetic Molecular Theory (KMT) is the conceptual framework that explains the behavior of matter in terms of the motion of its particles. The core postulates of KMT are:

All matter is made of tiny particles (atoms, molecules, or ions) in constant, random motion.
The average kinetic energy of the particles in a sample is directly proportional to the absolute temperature (in Kelvin): $ KE_{avg} \propto T $.
Particles collide with each other and with the walls of their container. Collisions between particles are perfectly elastic (no net energy loss).
At higher temperatures, particles move faster on average.

KMT explains the states of matter — solid, liquid, and gas — as arising from the competition between IMFs (which pull particles together) and kinetic energy (which drives particles apart). When kinetic energy dominates, particles escape into the gas phase. When IMFs dominate, particles are locked into a solid. The liquid state is an intermediate: particles are close together but have enough energy to flow past one another.

Solids

In the solid state, particles are held in fixed positions by strong intermolecular forces. Particles vibrate in place but do not translate (move from location to location). Solids have:

Definite shape and definite volume
High density compared to the gas phase
Very low compressibility
Slow diffusion

There are two structural categories of solids:

Crystalline solids have particles arranged in a highly ordered, repeating three-dimensional pattern called a crystal lattice. This long-range order gives crystalline solids sharp, well-defined melting points. Examples:

Ionic crystals: NaCl, CaCO$_3$ — ions arranged in regular lattice, high melting points
Metallic crystals: copper, iron — metal cations in a "sea of electrons"
Molecular crystals: ice (H$_2$O), naphthalene (C$_{10}$H$_8$) — molecules held by IMFs
Covalent network solids: diamond (C), quartz (SiO$_2$) — covalently bonded 3D networks, extremely high melting points

Amorphous solids lack long-range order. Their particles are arranged randomly, like a snapshot of a liquid that has been frozen in place. Amorphous solids do not have a sharp melting point; instead, they soften gradually over a range of temperatures. Glass, rubber, and many plastics are amorphous solids.

Liquids

In the liquid state, particles are close together (density near that of the solid) but have enough kinetic energy to slide past one another, giving liquids their characteristic ability to flow and take the shape of their container. Three important physical properties of liquids arise directly from intermolecular forces:

Surface tension is a measure of the energy required to increase the surface area of a liquid. Molecules at the interior of a liquid are surrounded by neighbors on all sides and experience attractive forces in every direction. Molecules at the surface have neighbors only below and to the sides, so they experience a net inward pull. This inward pull creates the tendency of the liquid surface to contract to the smallest possible area — the phenomenon responsible for water drops forming spheres and water striders walking on water. Liquids with stronger IMFs have higher surface tension.

Viscosity is a measure of a liquid's resistance to flow. A highly viscous liquid (like honey or motor oil) flows slowly because the molecules experience strong attractive forces and/or are large and tangled, making it difficult for them to slide past one another. Water is much less viscous than honey at room temperature. Viscosity decreases as temperature increases because rising kinetic energy helps molecules overcome IMFs and flow more freely.

Capillary action is the ability of a liquid to flow through narrow tubes against gravity, driven by the interplay of two forces:

Adhesive forces: attractions between the liquid and the walls of the tube (e.g., water and glass)
Cohesive forces: attractions between liquid molecules themselves (IMFs)

When adhesion is stronger than cohesion, the liquid climbs the tube walls and a concave meniscus forms (water in glass). When cohesion is stronger than adhesion, the liquid pulls away from the walls and a convex meniscus forms (mercury in glass). Capillary action is critical in biology: plants use it to draw water from roots to leaves, and blood moves through narrow capillaries partly through this mechanism.

Gases

In the gas phase, particles are far apart — typically separated by distances roughly 10 times their own diameter — and move rapidly in random directions. IMFs are essentially negligible at these distances (or are assumed to be negligible in the ideal gas model). Gases:

Have no definite shape or volume — they expand to fill their container
Have very low densities compared to liquids and solids
Are highly compressible
Mix completely and uniformly with other gases
Exert pressure on the walls of their container through collisions

Section 4: Gas Pressure

Understanding Gas Pressure

Gas pressure is the force exerted per unit area by gas molecules colliding with a surface. By KMT, every collision of a gas molecule with the container wall transfers a small amount of momentum to the wall, and the cumulative effect of billions of such collisions per second is a measurable force — pressure.

\[ P = \frac{F}{A} \]

where $ P $ is pressure, $ F $ is force, and $ A $ is area.

Pressure units encountered in AP Chemistry:

Pascal (Pa): the SI unit; 1 Pa = 1 N/m²
Atmosphere (atm): 1 atm = 101,325 Pa ≈ 101.3 kPa
Millimeters of mercury (mmHg): 1 atm = 760 mmHg
Torr: 1 torr = 1 mmHg (virtually identical for our purposes)

Atmospheric pressure is the pressure exerted by the weight of the atmosphere above us. At sea level, atmospheric pressure is 1 atm = 760 mmHg = 101.325 kPa. Atmospheric pressure decreases with altitude because there is less air above. A mercury barometer measures atmospheric pressure: the atmosphere pushes down on a mercury reservoir, forcing mercury up a sealed tube. At 1 atm, the mercury column rises to exactly 760 mm.

Section 5: The Gas Laws

Chemists discovered through careful experimentation that the pressure, volume, temperature, and amount of a gas are mathematically related in simple ways. These relationships, known as the gas laws, apply to ideal gases — gases that follow KMT assumptions perfectly.

Boyle's Law: Pressure and Volume

Boyle's Law states that at constant temperature and constant amount of gas, the pressure and volume of a gas are inversely proportional:

\[ P_1 V_1 = P_2 V_2 \quad (\text{constant } T, n) \]

As you squeeze a gas into a smaller volume, the molecules collide with the container walls more frequently, increasing pressure. A balloon compressed to half its volume will have twice the pressure. Boyle's Law explains how your lungs work: when your diaphragm contracts and lung volume increases, the pressure inside your lungs falls below atmospheric pressure, forcing air to flow in.

Charles's Law: Volume and Temperature

Charles's Law states that at constant pressure and constant amount of gas, the volume of a gas is directly proportional to its absolute temperature (in Kelvin):

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \quad (\text{constant } P, n) \]

As temperature increases, molecules move faster and collide with the walls more forcefully. For the pressure to remain constant, the container must expand. This is why a balloon deflates when brought into a cold room and expands when warmed.

Critical note: Temperature must always be in Kelvin when using gas laws. $ T(K) = T(°C) + 273.15 $. Using Celsius will give wrong answers every time.

Avogadro's Law: Volume and Moles

Avogadro's Law states that at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas:

\[ \frac{V_1}{n_1} = \frac{V_2}{n_2} \quad (\text{constant } T, P) \]

Equal volumes of different gases at the same temperature and pressure contain equal numbers of molecules — a profound result. Adding more molecules to a container at constant T and P requires the container to expand. Avogadro's Law is the foundation for the concept of molar volume.

Gas Law Summary Table

Law	Constant Variables	Relationship	Equation	Physical Example
Boyle's Law	$ T, n $	$ P $ and $ V $ inversely proportional	$ P_1V_1 = P_2V_2 $	Compressing a syringe
Charles's Law	$ P, n $	$ V $ and $ T $ directly proportional	$ \frac{V_1}{T_1} = \frac{V_2}{T_2} $	Balloon in cold air
Avogadro's Law	$ T, P $	$ V $ and $ n $ directly proportional	$ \frac{V_1}{n_1} = \frac{V_2}{n_2} $	Inflating a tire
Ideal Gas Law	—	All four variables	$ PV = nRT $	Any ideal gas calculation

Section 6: The Ideal Gas Law

Combining the Gas Laws

Combining Boyle's Law, Charles's Law, and Avogadro's Law yields the Ideal Gas Law, a single equation that relates all four gas variables simultaneously:

\[ PV = nRT \]

where:

$ P $ = pressure (in atm, or consistent with R)
$ V $ = volume (in liters)
$ n $ = number of moles of gas
$ R $ = ideal gas constant = $ 0.08206 \text{ L·atm·mol}^{-1}\text{·K}^{-1} $ (most common value for chemistry)
$ T $ = temperature in Kelvin

The ideal gas constant $ R $ has different numerical values depending on the pressure units you use:

$ R = 0.08206 \text{ L·atm·mol}^{-1}\text{·K}^{-1} $ (use with P in atm)
$ R = 8.314 \text{ J·mol}^{-1}\text{·K}^{-1} $ (use with P in Pa, V in m³; also used in thermodynamics)
$ R = 62.36 \text{ L·mmHg·mol}^{-1}\text{·K}^{-1} $ (use with P in mmHg)

The choice of $ R $ value must match your pressure units — this is one of the most common sources of error on the AP exam.

Molar Volume

A powerful consequence of Avogadro's Law and the ideal gas law is that one mole of any ideal gas occupies the same volume under the same conditions. At standard temperature and pressure (STP) — defined as $ 0^\circ\text{C}$ (273.15 K) and 1 atm — the molar volume of an ideal gas is:

\[ V_{molar} = \frac{nRT}{P} = \frac{(1 \text{ mol})(0.08206 \text{ L·atm·mol}^{-1}\text{·K}^{-1})(273.15 \text{ K})}{1 \text{ atm}} = 22.4 \text{ L/mol} \]

This is a number worth memorizing: 22.4 L/mol at STP. It means that 22.4 liters of oxygen at STP contains exactly 1 mole (6.022 × 10²³ molecules), and so does 22.4 liters of helium, nitrogen, or any other ideal gas at the same conditions.

Note: The College Board now often uses a "new STP" of 0°C and 100 kPa, which gives a molar volume of approximately 22.7 L/mol. Always check which definition your problem uses.

Gas Stoichiometry

The ideal gas law makes it possible to perform stoichiometry calculations when one or more reactants or products is a gas. The standard approach:

Balance the chemical equation.
Use stoichiometric mole ratios to convert between moles of different species.
Apply $ PV = nRT $ to convert between moles of gas and a measured quantity (volume, pressure).

Example: How many liters of H$_2$ gas at $25^\circ\text{C}$ and 1.00 atm are produced when 2.00 g of Zn reacts with excess hydrochloric acid?

\[ \text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g) \]

Moles of Zn: $ \frac{2.00 \text{ g}}{65.38 \text{ g/mol}} = 0.03059 \text{ mol Zn} $

Moles of H$_2$: $ 0.03059 \text{ mol Zn} \times \frac{1 \text{ mol H}_2}{1 \text{ mol Zn}} = 0.03059 \text{ mol H}_2 $

Volume of H$_2$: $ V = \frac{nRT}{P} = \frac{(0.03059)(0.08206)(298)}{1.00} = 0.748 \text{ L} $

Gas stoichiometry simply weaves $ PV = nRT $ into the familiar mole-map framework you already know.

Diagram: Ideal Gas Law Interactive Simulator

Ideal Gas Law Interactive Simulator

Type: MicroSim sim-id: ideal-gas-law-simulator
Library: p5.js
Status: Specified

Learning objective: Students will manipulate pressure, volume, temperature, and moles of an ideal gas and observe how changes to one variable affect the others, applying the ideal gas law quantitatively (Bloom L3: Apply — use the equation; Bloom L4: Analyze — predict the direction of change).

Canvas size: 800 × 450 px, responsive to window resize.

Layout: - Left panel (50% width): Animated visualization of a cylinder with a moveable piston. The interior of the cylinder shows ~30–80 animated dots (gas molecules) bouncing around in straight lines with elastic wall collisions. Faster dots (higher T) are colored warm orange; slower dots (lower T) are colored cool blue. The piston position visually reflects the current volume (higher volume = piston higher up). The pressure is shown by how frequently dots hit the walls. - Right panel (50% width): Four labeled sliders and a readout display. - Pressure slider: 0.5 to 3.0 atm (step 0.1), labeled "P (atm)" - Volume slider: 1.0 to 20.0 L (step 0.5), labeled "V (L)" - Temperature slider: 100 to 600 K (step 10), labeled "T (K)" - Moles slider: 0.5 to 3.0 mol (step 0.1), labeled "n (mol)" - Below sliders: A "Solve for:" radio button set: [P | V | T | n]. The selected variable is computed from the other three using PV = nRT and displayed in a highlighted result box. - A running calculation display shows the full equation substituted with current values: e.g., "PV = nRT → (1.5)(10.0) = n(0.08206)(300) → n = 0.609 mol"

Interactions: - When any slider moves, the simulation recomputes the "solve for" variable and animates the piston and dot speeds accordingly. - A "Reset" button returns to standard conditions: P = 1.0 atm, V = 22.4 L, T = 273 K, n = 1.0 mol. - A "Hint" button reveals which gas law is being demonstrated (e.g., "You are exploring Boyle's Law — P and V at constant T and n").

Colors: Cylinder outline in dark gray; piston in steel blue; warm molecules in tomato (#FF6347); cool molecules in cornflower blue (#6495ED); result box background in light yellow (#FFFACD).

Pedagogical note: The visual decoupling of molecule speed from dot speed changes lets students see that temperature is proportional to kinetic energy — a core KMT postulate. Connecting the animation to the equation builds intuition before numerical problem-solving (Bloom L2: Understand — explain).

Section 7: Dalton's Law of Partial Pressures

Mixtures of Gases

When two or more gases that do not react chemically are placed in the same container, they behave independently. Each gas exerts its own pressure — called its partial pressure — as if the other gases were not present. Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas:

\[ P_{total} = P_1 + P_2 + P_3 + \cdots \]

or equivalently:

\[ P_{total} = \sum_{i} P_i \]

This law holds because, in an ideal gas, molecules do not interact with each other. Each type of molecule contributes its own collisions with the container walls, independent of the others.

Partial Pressures and Mole Fraction

The partial pressure of each gas is proportional to its mole fraction — the fraction of the total moles that it represents:

\[ \chi_i = \frac{n_i}{n_{total}} \]

\[ P_i = \chi_i \cdot P_{total} \]

Where $ \chi_i $ (chi sub i) is the mole fraction of component $ i $. Notice that the sum of all mole fractions always equals 1: $ \sum_i \chi_i = 1 $.

Example: Air at 1.00 atm is approximately 78% N$_2$, 21% O$_2$, and 1% Ar by mole (mole percent = mole fraction × 100):

$ P_{N_2} = 0.78 \times 1.00 \text{ atm} = 0.78 \text{ atm} $ (= 593 mmHg)
$ P_{O_2} = 0.21 \times 1.00 \text{ atm} = 0.21 \text{ atm} $ (= 160 mmHg)
$ P_{Ar} = 0.01 \times 1.00 \text{ atm} = 0.01 \text{ atm} $ (= 7.6 mmHg)

Collecting Gas Over Water

A common laboratory procedure involves collecting a gas by water displacement. When gas is collected over water, the gas sample is saturated with water vapor. The total pressure in the collection tube is:

\[ P_{total} = P_{gas} + P_{water vapor} \]

To find the partial pressure of the collected gas, you must subtract the vapor pressure of water (which depends on temperature and is found in reference tables) from the total pressure:

\[ P_{gas} = P_{total} - P_{water vapor} \]

This correction is essential for accurate gas stoichiometry calculations involving laboratory gas collection.

Key facts about partial pressures for AP Chemistry problem-solving:

Mole fraction is dimensionless and always between 0 and 1.
The partial pressure of a gas depends only on its mole fraction, not on the identity of the other gases.
Dalton's Law works because ideal gas molecules occupy negligible volume and exert no intermolecular forces on each other.
Always subtract water vapor pressure when working with gas collected by water displacement.

Diagram: Dalton's Law Partial Pressure Infographic

Dalton's Law Partial Pressure Infographic

Type: infographic sim-id: dalton-law-partial-pressure
Library: p5.js
Status: Specified

Learning objective: Students will calculate partial pressures and mole fractions for a gas mixture and verify that the sum of partial pressures equals total pressure (Bloom L3: Apply — calculate, solve).

Canvas size: 800 × 400 px, responsive to window resize.

Layout: Three side-by-side containers connected by an equals sign, styled as laboratory flasks. - Flask 1 (left): Contains only red dots labeled "Gas A: 0.40 atm" - Flask 2 (center): Contains only blue dots labeled "Gas B: 0.35 atm" - Flask 3 (right): Contains a "+" sign then a mixed flask showing red + blue + green dots labeled "Gas A + B + C: 1.00 atm total" - Below the flasks: a pie chart showing mole fractions of each gas component, with the sector sizes proportional to mole fractions. - A table beneath the pie chart summarizes: Gas | Moles | Mole Fraction | Partial Pressure.

Interactions: - Sliders below the diagram let the student adjust the moles of Gas A, B, and C (each 0.5 to 3.0 mol). - All values (mole fractions, partial pressures, pie chart sectors) update in real time. - Color coding: Gas A = tomato red, Gas B = cornflower blue, Gas C = forest green. - Total pressure is fixed at 1.00 atm and displayed prominently.

Pedagogical note: Seeing the pie chart update as mole fractions change helps students internalize that partial pressure is proportional to composition, not molecular identity. This supports Bloom L2 (Understand) and L3 (Apply).

Section 8: Real Gases and Deviations from Ideality

When Ideal Gas Law Breaks Down

The ideal gas law is a remarkably useful model, but it rests on two assumptions that are never perfectly true:

Gas molecules occupy no volume (they are point masses).
Gas molecules exert no intermolecular forces on one another.

Real gases deviate from ideal behavior most significantly under conditions where these assumptions are most incorrect:

High pressure: At high pressures, molecules are forced close together, and their actual finite volume becomes significant. The measured volume is larger than what the ideal gas law predicts.
Low temperature: At low temperatures, molecules move slowly and their IMFs become significant. The gas may be attracted inward and exert less pressure on the walls than expected. This effect can cause condensation to a liquid.

The identity of the gas also matters. Gases with stronger IMFs (e.g., CO$_2$, NH$_3$, H$_2$O vapor) show greater deviations than gases with weak IMFs (e.g., H$_2$, He, N$_2$). Gases with larger molecular volumes also deviate more.

Deviations from ideality are often expressed using the compressibility factor $ Z $:

\[ Z = \frac{PV}{nRT} \]

For an ideal gas, $ Z = 1 $ exactly. When $ Z < 1 $, attractive forces dominate and the gas occupies less volume than expected. When $ Z > 1 $, molecular volume dominates and the gas is less compressible than expected.

The Van der Waals Equation

The Van der Waals equation is a modified ideal gas law that corrects for the two sources of non-ideal behavior:

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

The two correction terms are:

Pressure correction $ \frac{an^2}{V^2} $: Added to the measured pressure to account for the fact that attractive IMFs reduce the force with which molecules hit the walls. The constant $ a $ is larger for gases with stronger intermolecular attractions (e.g., $ a = 3.64 \text{ L}^2\text{·atm·mol}^{-2} $ for N$_2$; $ a = 6.49 $ for CO$_2$; $ a = 5.54 $ for NH$_3$).
Volume correction $ nb $: Subtracted from the total volume to account for the space actually occupied by the gas molecules themselves. The constant $ b $ (in L/mol) represents the excluded volume per mole of gas; it is related to the physical size of the molecules.

The Van der Waals equation is not just a correction formula — it carries physical meaning. Comparing $ a $ values tells you which gas has stronger IMFs; comparing $ b $ values tells you which molecule is physically larger. For the AP exam, you are expected to understand qualitatively how real gases deviate from ideality and why, and to interpret Van der Waals constants conceptually. You are not expected to solve the cubic form of the Van der Waals equation algebraically.

Conditions for significant deviations from ideal behavior:

Very high pressures (above roughly 10 atm)
Very low temperatures (near the boiling point of the gas)
Gases with large molecules or strong IMFs
Polar or associating molecules (like NH$_3$ or H$_2$O vapor)

Conditions where ideal gas law works well:

Low to moderate pressures (up to a few atm)
High temperatures (well above the boiling point of the substance)
Small, nonpolar molecules (H$_2$, He, N$_2$)

Section 9: Connecting IMFs to Physical Properties — Putting It All Together

A Unified View

The concepts of this chapter form a coherent story: molecular structure determines polarity, polarity determines IMF type and strength, and IMF strength determines physical properties.

Consider three molecules with nearly the same molar mass (~46 g/mol):

Ethane (C$_2$H$_6$), 30 g/mol — nonpolar; only London dispersion forces; bp = $-89^\circ$C
Dimethyl ether (CH$_3$OCH$_3$), 46 g/mol — polar (dipole-dipole); bp = $-24^\circ$C
Ethanol (C$_2$H$_5$OH), 46 g/mol — polar with O–H group; hydrogen bonding; bp = $78^\circ$C

The 167°C difference in boiling points between dimethyl ether and ethanol — two molecules of identical molecular formula C$_2$H$_6$O — is explained entirely by the presence or absence of O–H hydrogen bonding. Ethanol's O–H group can form strong hydrogen bonds with other ethanol molecules; dimethyl ether cannot because its oxygen has no H attached.

This kind of analysis — reasoning from structure to IMF to property — is exactly what the AP Chemistry exam tests. Practice by asking: What bonds does this molecule have? Is it polar? Does it have N–H, O–H, or F–H groups? How does that compare to a similar molecule?

Key Vocabulary Review

The following list summarizes the essential vocabulary from this chapter:

Intermolecular forces (IMFs): Attractions between separate molecules
Intramolecular forces: Bonds within a molecule (covalent, ionic)
London dispersion forces: Temporary dipole attractions; universal; increase with molar mass
Polarizability: Ease of electron cloud distortion; determines LDF strength
Dipole-dipole forces: Attractions between permanent dipoles in polar molecules
Hydrogen bonding: Strong dipole-dipole interaction; requires H bonded to N, O, or F
Ion-dipole forces: Attraction between ions and polar molecules; strongest IMF
Crystalline solid: Particles in ordered repeating lattice; sharp melting point
Amorphous solid: Disordered particle arrangement; gradual softening
Surface tension: Energy cost to expand liquid surface; related to IMF strength
Viscosity: Resistance to flow; increases with IMF strength and molecular size
Capillary action: Liquid flow in narrow tubes; driven by adhesion vs. cohesion
Partial pressure: Pressure a gas exerts in a mixture; $ P_i = \chi_i P_{total} $
Mole fraction: $ \chi_i = n_i / n_{total} $; dimensionless
Ideal Gas Law: $ PV = nRT $; relates all four gas variables
Molar volume at STP: 22.4 L/mol
Van der Waals equation: Corrected gas law for real gases; accounts for IMFs and molecular volume
Deviations from ideality: Occur at high P, low T, or for gases with strong IMFs

Chapter Review: Practice Problems

The following practice problems span the full range of concepts in this chapter.

Conceptual / Short Answer:

Rank the following molecules in order of increasing boiling point and explain your ranking: CH$_4$, NH$_3$, N$_2$, H$_2$O.
Explain why a glass capillary tube produces a concave meniscus with water but a convex meniscus with mercury.
A student claims that CO$_2$ is a polar molecule because C=O bonds are polar. Is the student correct? Explain.
Under what conditions does a real gas deviate most significantly from ideal behavior? Explain the molecular basis.
Why is the Van der Waals $ a $ constant for NH$_3$ larger than for N$_2$?

Quantitative Problems:

A 2.50 L sample of gas at 1.20 atm is compressed at constant temperature to 0.80 L. What is the new pressure?

Use Boyle's Law: $$ P_1V_1 = P_2V_2 \rightarrow P_2 = \frac{P_1V_1}{V_2} = \frac{(1.20)(2.50)}{0.80} = 3.75 \text{ atm} $$
A gas occupies 4.00 L at $27^\circ$C. What volume will it occupy at $127^\circ$C at constant pressure?

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \rightarrow V_2 = \frac{V_1 T_2}{T_1} = \frac{(4.00)(400)}{300} = 5.33 \text{ L} \]
How many grams of CO$_2$ are present in a 5.00 L sample at $25^\circ$C and 2.50 atm?

\[ n = \frac{PV}{RT} = \frac{(2.50)(5.00)}{(0.08206)(298)} = 0.512 \text{ mol} \rightarrow 0.512 \times 44.01 = 22.5 \text{ g} \]
A mixture of gases contains 0.60 mol N$_2$, 0.20 mol O$_2$, and 0.20 mol CO$_2$ at a total pressure of 2.00 atm. Find the partial pressure of each gas.

\[ \chi_{N_2} = \frac{0.60}{1.00} = 0.60 \Rightarrow P_{N_2} = 1.20 \text{ atm} $$ $$ \chi_{O_2} = 0.20 \Rightarrow P_{O_2} = 0.40 \text{ atm} $$ $$ \chi_{CO_2} = 0.20 \Rightarrow P_{CO_2} = 0.40 \text{ atm} \]
What volume of O$_2$ gas at STP is produced by the decomposition of 9.80 g of H$_2$O$_2$?

\[ 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 $$ $$ n_{H_2O_2} = \frac{9.80}{34.02} = 0.288 \text{ mol} \rightarrow n_{O_2} = 0.144 \text{ mol} $$ $$ V = 0.144 \text{ mol} \times 22.4 \text{ L/mol} = 3.23 \text{ L} \]

End of Chapter 6

References

See Annotated References

Law	Constant Variables	Relationship	Equation	Physical Example
Boyle's Law	\( T, n \)	\( P \) and \( V \) inversely proportional	\( P_1V_1 = P_2V_2 \)	Compressing a syringe
Charles's Law	\( P, n \)	\( V \) and \( T \) directly proportional	\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)	Balloon in cold air
Avogadro's Law	\( T, P \)	\( V \) and \( n \) directly proportional	\( \frac{V_1}{n_1} = \frac{V_2}{n_2} \)	Inflating a tire
Ideal Gas Law	—	All four variables	\( PV = nRT \)	Any ideal gas calculation