Chapter 18 — Front Matter
Purpose and Overview
Look around you right now. You are surrounded by a gas. The air filling the room, the breath moving in and out of your lungs, the atmosphere stretching dozens of kilometers above your head — all of it is made of particles in constant, rapid, invisible motion. You cannot see the individual molecules of nitrogen and oxygen that make up the air, but their behavior determines the weather, the pressure in your bicycle tires, the way your lungs work, and the altitude at which airplanes can fly.
Gases behave differently from solids and liquids. They expand to fill any container. They can be compressed into a smaller space. They push outward on every surface they touch. They mix completely with other gases. And their behavior changes in predictable, measurable ways when temperature, pressure, or volume changes.
Chemists and physicists have studied gases for centuries, and from that work emerged a set of gas laws — mathematical relationships that connect pressure, volume, temperature, and amount of gas in ways that can be used to make precise predictions. These laws are not just formulas to memorize. They are windows into the behavior of matter at the particle level.
In this chapter, you will learn what gases are made of, how pressure and temperature relate to particle motion, and how to use the gas laws to solve real problems. You will apply algebra to situations involving weather forecasting, breathing, balloons, diving, scuba tanks, aerosol cans, automobile engines, and laboratory gas collection.
Opening Question
A sealed syringe contains a fixed amount of gas at room temperature. If you push the plunger in and cut the volume in half, what happens to the pressure inside? What if you then heat the syringe — what changes? By the end of this chapter, you will be able to calculate the exact answer to both questions, explain the particle-level reason for each, and apply the same reasoning to weather, biology, and industry.
10Sections
6Gas Laws
26Key Terms
16+Worked Examples
Learning Objectives
- 1Identify the four variables that describe a gas sample: pressure, volume, temperature, and amount in moles.
- 2Convert between pressure units (atm, kPa, mmHg, torr) and convert temperatures from Celsius to Kelvin.
- 3Explain gas pressure and gas properties using the kinetic molecular theory.
- 4Apply Boyle's Law to solve for pressure or volume in a changing gas system at constant temperature.
- 5Apply Charles's Law to solve for volume or temperature in a changing gas system at constant pressure.
- 6Apply Gay-Lussac's Law to solve for pressure or temperature in a rigid sealed container.
- 7Apply Avogadro's Law and use molar volume (22.4 L/mol at STP) to convert between moles and volume.
- 8Use the combined gas law to solve problems in which pressure, volume, and temperature all change.
- 9Use the ideal gas law (PV = nRT) to calculate pressure, volume, temperature, or moles for a gas sample.
- 10Describe how real gases deviate from ideal behavior and explain the conditions under which this matters.
Sections 18.1 – 18.3
What Gases Are and How We Measure Them
Properties of Gases · Gas Pressure · Temperature and the Kinetic Molecular Theory
Section 18.1
Properties of Gases
What Makes a Gas Different
Matter exists in three common states: solid, liquid, and gas. The state a substance occupies depends on how strongly its particles attract one another and how much energy those particles have. In solids, particles are locked tightly together. In liquids, particles flow past each other. In gases, particles move freely and independently, separated from one another by large distances compared to the size of the particles themselves.
This simple difference — large separation between particles — explains nearly every property that gases have.
| Gas Property | Explanation | Example |
|---|
| Expand to fill containers | No fixed shape or volume; particles spread in all directions | Perfume diffuses across a room |
| Compressible | Large empty space between particles allows them to be pushed closer | Air pumped into a bicycle tire |
| Low density | Most of the gas volume is empty space | Air: ~1.2 g/L vs. water: ~1000 g/L |
| Exert pressure | Particles collide with container walls billions of times per second | Tire pressure; atmospheric pressure |
| Diffuse and mix | Random, high-speed motion causes gases to mix uniformly | CO₂ and N₂ mixing in air |
The Four Gas Variables
To fully describe any gas sample in chemistry, four variables must be specified. Every gas law in this chapter is a relationship among these four quantities.
| Variable | Symbol | Common Units | Notes |
|---|
| Pressure | P | atm, kPa, mmHg, torr | Force of particle collisions on container walls |
| Volume | V | L, mL | Always the volume of the container |
| Temperature | T | K (Kelvin — always) | Must be in Kelvin for all gas law calculations |
| Amount | n | mol | Number of moles of gas particles present |
Section 18.2
Gas Pressure
Pressure is defined as force divided by area. Gas pressure specifically results from the force of gas particle collisions with surfaces. Every time a gas particle strikes a container wall, it exerts a tiny push. Billions of collisions per second across every square centimeter of the container wall add up to a measurable, steady pressure.
Two things increase gas pressure: more collisions and harder collisions. More particles in the same container means more collisions per second, so more particles means higher pressure. Faster-moving particles (higher temperature) means harder collisions, so higher temperature means higher pressure when volume is fixed.
Pressure Unit Conversions
1 atm = 101.3 kPa = 760 mmHg = 760 torr = 101,325 Pa = 14.7 psi
Memorize the relationships to atm. Use them as conversion factors in dimensional analysis.
| Unit | Abbreviation | Relationship to 1 atm | Common Use |
|---|
| Atmosphere | atm | 1 atm | General chemistry |
| Kilopascal | kPa | 101.3 kPa | SI standard, general chemistry |
| Millimeters of mercury | mmHg | 760 mmHg | Medical, barometers |
| Torr | torr | 760 torr | Laboratory gas measurements |
| Pounds per square inch | psi | 14.7 psi | Tires, everyday use |
Worked Example 18.1
Converting pressure units
- ProblemConvert 2.50 atm to kPa and to mmHg.
- Solve
To kPa: 2.50 atm × (101.3 kPa / 1 atm) = 253 kPa
To mmHg: 2.50 atm × (760 mmHg / 1 atm) = 1900 mmHg
- CheckBoth values are greater than 1 atm equivalent, as expected for 2.50 atm. ✓
Answer: 2.50 atm = 253 kPa = 1900 mmHg
Section 18.3
Temperature and Particle Motion
The Kelvin Scale
Temperature is a measure of the average kinetic energy of particles in a substance. The faster particles move, the higher the temperature. For the gas laws, temperature must always be expressed using the Kelvin scale — an absolute temperature scale that starts at absolute zero, the theoretical temperature at which particles would have no kinetic energy.
Temperature Conversion — Always Required for Gas Laws
T (K) = T (°C) + 273
Absolute zero = 0 K = −273 °C. The Kelvin scale begins at the true absence of thermal energy, which makes proportional relationships in gas laws mathematically valid. Using Celsius in gas law calculations produces wrong answers — no exceptions.
| Reference Point | Temperature (°C) | Temperature (K) |
|---|
| Absolute zero | −273 °C | 0 K |
| Melting point of water | 0 °C | 273 K |
| Room temperature (approx.) | 25 °C | 298 K |
| Body temperature | 37 °C | 310 K |
| Boiling point of water | 100 °C | 373 K |
Worked Example 18.2
Converting temperatures to Kelvin
- ProblemConvert (a) 37 °C, (b) −78 °C, and (c) 0 °C to Kelvin.
- Solve
(a) 37 + 273 = 310 K
(b) −78 + 273 = 195 K
(c) 0 + 273 = 273 K
Answers: (a) 310 K (b) 195 K (c) 273 K
The Kinetic Molecular Theory of Gases
The behavior of gases is explained by the kinetic molecular theory, a model based on five key assumptions about ideal gases:
| # | Assumption | Consequence |
|---|
| 1 | Gas particles are negligibly small compared to the distances between them | Most of a gas is empty space; gases are compressible |
| 2 | Particles move in constant, random, straight-line motion | Gases expand and diffuse; pressure is exerted in all directions |
| 3 | Collisions between particles are elastic (no net energy loss) | Temperature stays constant unless heat is added or removed |
| 4 | Particles do not attract or repel each other between collisions | Gas behavior is independent of chemical identity (ideal model) |
| 5 | Average kinetic energy is directly proportional to temperature in Kelvin | Heating a gas speeds particles; cooling slows them |
Keep the Theory in Mind
As you learn each gas law, connect it back to the kinetic molecular theory. Every mathematical relationship has a particle-level explanation. Understanding both levels — the equation and the model — is the goal of this chapter.
Sections 18.4 – 18.8
The Gas Laws: Boyle, Charles, Gay-Lussac, Avogadro, and Combined
Each law holds two variables constant while relating the other two — then the combined law handles them all at once
Section 18.4
Boyle's Law
Imagine a sealed syringe. The syringe contains a fixed amount of gas at a fixed temperature. When you push the plunger in — reducing the volume — the gas particles are forced into a smaller space. They collide with the walls more frequently. More collisions per second means higher pressure. When volume decreases, pressure increases. Pull the plunger out, and the opposite occurs: more space, fewer collisions per second, lower pressure.
This relationship — pressure and volume change in opposite directions when temperature and amount of gas remain constant — is Boyle's Law, described by Irish scientist Robert Boyle in 1662.
Boyle's Law
At constant temperature and constant amount of gas, pressure and volume are inversely proportional.
P₁V₁ = P₂V₂
When one increases, the other decreases. Their product remains constant. Temperature (T) and moles (n) are held constant.
Worked Example 18.3
Finding new volume when pressure increases
- GivenP₁ = 1.5 atm, V₁ = 4.0 L, P₂ = 3.0 atm. Temperature constant.
- FindV₂
- Solve
P₁V₁ = P₂V₂
(1.5)(4.0) = (3.0)(V₂)
6.0 = 3.0 × V₂
V₂ = 6.0 ÷ 3.0 = 2.0 L
- CheckPressure doubled → volume should halve: 4.0 ÷ 2 = 2.0 L. ✓
Answer: V₂ = 2.0 L
Worked Example 18.4
Finding pressure when volume doubles
- GivenP₁ = 720 mmHg, V₁ = 250 mL, V₂ = 500 mL. Temperature constant.
- Solve
P₁V₁ = P₂V₂
(720)(250) = P₂(500)
180,000 = P₂ × 500
P₂ = 180,000 ÷ 500 = 360 mmHg
- CheckVolume doubled → pressure should halve: 720 ÷ 2 = 360 mmHg. ✓
Answer: P₂ = 360 mmHg
Boyle's Law in Everyday Life
Breathing is Boyle's Law in action. Inhaling expands the chest cavity (volume increases), pressure in the lungs drops below atmospheric, and air flows in. Exhaling compresses the cavity (volume decreases), pressure rises above atmospheric, and air flows out. Scuba divers must also understand Boyle's Law: air compressed in a tank expands as pressure decreases during ascent, which is why breath-holding during a rapid ascent can be dangerous.
Section 18.5
Charles's Law
Now keep the pressure on a gas sample constant — use a flexible container like a balloon — but heat the gas. Faster particles hit the walls harder and more often. In a flexible container, the walls move outward until pressure is balanced again. Volume increases when temperature increases. Cool the gas, and the opposite occurs. Volume decreases when temperature decreases.
This relationship — volume and temperature change in the same direction when pressure and amount of gas remain constant — is Charles's Law, named after French scientist Jacques Charles, who described it in 1787.
Charles's Law
At constant pressure and constant amount of gas, volume and temperature (in Kelvin) are directly proportional.
V₁ / T₁ = V₂ / T₂
Temperature must always be in Kelvin. Pressure (P) and moles (n) are held constant.
Worked Example 18.5
Finding new volume when temperature rises
- GivenV₁ = 3.00 L at 20 °C. Temperature rises to 40 °C. Pressure constant.
- ConvertT₁ = 20 + 273 = 293 K T₂ = 40 + 273 = 313 K
- Solve
V₁/T₁ = V₂/T₂
3.00/293 = V₂/313
V₂ = 3.00 × (313/293)
V₂ = 3.00 × 1.068 = 3.20 L
- NoteTemperature did NOT double (in Kelvin, 293 → 313 is only ~7% increase), so volume did not double. Always use Kelvin.
Answer: V₂ = 3.20 L
Worked Example 18.6
Finding the temperature needed for a volume change
- GivenV₁ = 500 mL at 27 °C. Find T₂ when V₂ = 750 mL. Pressure constant.
- ConvertT₁ = 27 + 273 = 300 K
- Solve
500/300 = 750/T₂
T₂ = 750 × (300/500)
T₂ = 450 K
- Convert back450 − 273 = 177 °C
Answer: T₂ = 450 K = 177 °C
Charles's Law in Everyday Life
Hot air balloons demonstrate Charles's Law directly — heating the air inside the envelope expands its volume, making it less dense than the surrounding cool air, and the balloon rises. Bread dough rises in the oven as CO₂ gas from yeast expands with rising temperature. On cold mornings, car tires may appear slightly low because cooler air occupies less volume (or exerts less pressure in a rigid tire).
Section 18.6
Gay-Lussac's Law
Consider a rigid, sealed container — one that cannot change volume. When temperature increases, gas particles move faster and hit the walls harder and more often. But the walls cannot move. The increased force translates directly into increased pressure. When temperature increases in a sealed rigid container, pressure increases. Cool the gas, and pressure decreases.
This relationship — pressure and temperature change in the same direction when volume and amount of gas remain constant — is Gay-Lussac's Law, named after French chemist Joseph Louis Gay-Lussac, who described it in 1808.
Gay-Lussac's Law
At constant volume and constant amount of gas, pressure and temperature (in Kelvin) are directly proportional.
P₁ / T₁ = P₂ / T₂
Temperature must always be in Kelvin. Volume (V) and moles (n) are held constant.
Worked Example 18.7
Aerosol can left in a hot car
- GivenP₁ = 1.2 atm at 25 °C. Temperature rises to 60 °C. Volume constant (rigid can).
- ConvertT₁ = 298 K T₂ = 333 K
- Solve
P₁/T₁ = P₂/T₂
1.2/298 = P₂/333
P₂ = 1.2 × (333/298)
P₂ = 1.34 atm
Answer: P₂ = 1.34 atm — aerosol cans must not be heated for this reason.
Worked Example 18.8
Finding the temperature that produces a target pressure
- GivenP₁ = 3.00 atm at 300 K. Find T₂ when P₂ = 4.50 atm. Volume constant.
- Solve
3.00/300 = 4.50/T₂
T₂ = 4.50 × (300/3.00) = 450 K
- Convert450 − 273 = 177 °C
Answer: T₂ = 450 K = 177 °C
Section 18.7
Avogadro's Law
The three laws above hold the amount of gas constant. What happens when the amount changes? If you pump more gas into a flexible balloon at constant temperature and pressure, the balloon expands. Volume increases when moles of gas increase. Remove gas, and the balloon shrinks. This is Avogadro's Law, named after Italian scientist Amedeo Avogadro.
Avogadro's Law and Molar Volume
At constant temperature and pressure, volume is directly proportional to moles of gas.
V₁ / n₁ = V₂ / n₂
At STP (0 °C and 1 atm), one mole of any ideal gas occupies exactly 22.4 L.
This molar volume is the same regardless of the chemical identity of the gas.
Worked Example 18.9
Using molar volume at STP
- ProblemWhat volume does 0.500 mol of nitrogen gas occupy at STP?
- Solve
0.500 mol × (22.4 L / 1 mol) = 11.2 L
Answer: 11.2 L
Worked Example 18.10
Finding moles from volume at STP
- ProblemA gas sample occupies 8.96 L at STP. How many moles does it contain?
- Solve
n = 8.96 L × (1 mol / 22.4 L) = 0.400 mol
Answer: 0.400 mol
Avogadro's Law and Stoichiometry
When gases react at the same temperature and pressure, their volume ratios equal their mole ratios in the balanced equation. For 2 H₂(g) + O₂(g) → 2 H₂O(g): 2 volumes of H₂ react with 1 volume of O₂ to produce 2 volumes of H₂O. Volume ratios are mole ratios for gases at constant T and P.
Section 18.8
The Combined Gas Law
Boyle's, Charles's, and Gay-Lussac's laws each hold two variables constant and relate the other two. In practice, more than one variable often changes at the same time. The combined gas law merges all three individual laws into a single equation that handles simultaneous changes in pressure, volume, and temperature, while keeping the amount of gas constant.
The Combined Gas Law
P₁V₁ / T₁ = P₂V₂ / T₂
Amount of gas (n) must remain constant. Temperature must be in Kelvin.
When any variable is constant, it cancels from both sides, reducing the equation to one of the three individual laws.
Worked Example 18.11
Combined gas law — pressure and temperature both change
- GivenA gas at 2.00 L, 1.00 atm, 25 °C. New conditions: 2.00 atm, 100 °C. Find V₂.
- ConvertT₁ = 298 K T₂ = 373 K
- Solve
P₁V₁/T₁ = P₂V₂/T₂
(1.00)(2.00)/298 = (2.00)(V₂)/373
V₂ = (1.00)(2.00)(373) / (298)(2.00)
V₂ = 746 / 596
V₂ = 1.25 L
- CheckPressure doubled (tends to halve volume). Temperature rose ~25% (tends to increase volume). Net effect: slight reduction. 1.25 L is reasonable. ✓
Answer: V₂ = 1.25 L
Worked Example 18.12
Finding pressure using the combined gas law
- GivenV₁ = 500 mL, P₁ = 0.800 atm, T₁ = 50 °C. New conditions: V₂ = 300 mL, T₂ = 25 °C. Find P₂.
- ConvertT₁ = 323 K T₂ = 298 K
- Solve
P₁V₁/T₁ = P₂V₂/T₂
(0.800)(500)/323 = P₂(300)/298
P₂ = (0.800)(500)(298) / (323)(300)
P₂ = 119,200 / 96,900
P₂ = 1.23 atm
Answer: P₂ = 1.23 atm
Strategy for Combined Gas Law Problems
Always list P₁, V₁, T₁ and P₂, V₂, T₂. Convert all temperatures to Kelvin first. If one variable is constant (not listed as changing), it cancels — and you recognize which simpler law applies. If two or three variables change, use the full combined equation.
Sections 18.9 – 18.10
The Ideal Gas Law and the Limits of Gas Models
PV = nRT · Calculating All Four Variables · Gas Density · Real vs. Ideal Behavior
Section 18.9
The Ideal Gas Law
The combined gas law handles changes in P, V, and T for a fixed amount of gas. But what about situations where the amount of gas — the number of moles — is part of the problem? What if you want to calculate how many moles are present in a sample based on its measured pressure, volume, and temperature?
For that, you need a law that connects all four gas variables in a single equation. That equation is the ideal gas law.
The Ideal Gas Law
PV = nRT
P = pressure in atm V = volume in liters n = moles of gas T = temperature in Kelvin
R = ideal gas constant = 0.08206 L·atm / (mol·K)
When pressure is in kPa, use R = 8.314 L·kPa / (mol·K). Match R to the pressure units used.
Worked Example 18.13
Finding moles using the ideal gas law
- Given5.00 L container, 2.00 atm, 27 °C. Find n.
- ConvertT = 27 + 273 = 300 K
- Solve
PV = nRT → n = PV / (RT)
n = (2.00 atm)(5.00 L) / (0.08206 L·atm/mol·K)(300 K)
n = 10.0 / 24.618
n = 0.406 mol
Answer: n = 0.406 mol
Worked Example 18.14
Finding pressure from moles
- Given0.250 mol CO₂ in a 3.00 L container at 37 °C. Find P.
- ConvertT = 310 K
- Solve
PV = nRT → P = nRT / V
P = (0.250)(0.08206)(310) / (3.00)
P = 6.360 / 3.00
P = 2.12 atm
Answer: P = 2.12 atm
Worked Example 18.15
Starting from grams — converting to moles first
- Given1.50 g of N₂ (M = 28.0 g/mol) in a 500 mL container at 25 °C. Find P.
- Step 1
n = 1.50 g ÷ 28.0 g/mol = 0.0536 mol
- Step 2V = 500 mL = 0.500 L T = 298 K
- Step 3
P = nRT / V
P = (0.0536)(0.08206)(298) / (0.500)
P = 1.311 / 0.500
P = 2.62 atm
Answer: P = 2.62 atm
Gas Density from the Ideal Gas Law
The ideal gas law can be rearranged to find gas density or molar mass. Starting from PV = nRT and substituting n = m/M (where m is mass in grams and M is molar mass in g/mol), density is:
d = PM / (RT)
where d = density in g/L, P = pressure in atm, M = molar mass in g/mol, R = 0.08206 L·atm/mol·K, and T = temperature in Kelvin. Gas density increases with pressure and molar mass, and decreases with temperature.
Worked Example 18.16
Calculating gas density
- ProblemWhat is the density of O₂ (M = 32.0 g/mol) at 1.00 atm and 25 °C?
- Solve
d = PM / (RT)
d = (1.00)(32.0) / (0.08206)(298)
d = 32.0 / 24.45
d = 1.31 g/L
Answer: d = 1.31 g/L
Section 18.10
Real Gases and Limits of Models
The ideal gas law is a powerful and accurate tool for most chemistry problems. But it is based on a model — the kinetic molecular theory — that makes simplifying assumptions. Real gas particles are not perfectly volumeless. They do attract and repel each other. At high pressure and low temperature, these non-ideal behaviors become significant.
| Condition | Real Gas Behavior | Why It Differs |
|---|
| High pressure | Actual volume is slightly larger than ideal predicts | Particle volume becomes a meaningful fraction of total volume |
| Low temperature | Actual pressure is slightly lower than ideal predicts | Attractive forces between slow-moving particles reduce wall collision force |
| Large, polar molecules | Greater deviation from ideal behavior | Stronger intermolecular attractions (e.g., H₂O, NH₃) |
| Small, nonpolar molecules | Closest to ideal behavior | Weak attractions, negligible particle volume (e.g., He, H₂) |
Using Models Responsibly
The ideal gas model works beautifully for most gases at ordinary laboratory temperatures and pressures — and in those conditions, using PV = nRT produces answers accurate within the precision of typical measurements. When conditions become extreme (very high pressure, very low temperature) or the gas has strong intermolecular forces, the model should be applied with care. Learning to use a model includes knowing when it applies and when it does not. This is part of scientific thinking.
Important Reminder — Always Use Kelvin
For every gas law calculation in this chapter — Boyle's is the only exception, because temperature is constant and cancels — temperature must be in Kelvin. Forgetting to convert from Celsius to Kelvin is the single most common error in gas law problems. Convert before you set up the equation, every time.
Chapter 18 — Back Matter
Chapter Summary
- Sec 18.1Gases expand to fill containers, are compressible, have low density, exert pressure, and diffuse freely. Four variables describe any gas: pressure (P), volume (V), temperature (T in Kelvin), and amount in moles (n).
- Sec 18.2Gas pressure is force per unit area arising from particle collisions with container walls. Standard units: 1 atm = 101.3 kPa = 760 mmHg = 760 torr. Convert between units using these relationships as conversion factors.
- Sec 18.3Temperature must always be in Kelvin for gas law calculations: K = °C + 273. Absolute zero (0 K) is the theoretical temperature of zero kinetic energy. The kinetic molecular theory models gases as small, fast-moving particles with elastic collisions and no intermolecular attractions.
- Sec 18.4Boyle's Law: at constant T and n, pressure and volume are inversely proportional. P₁V₁ = P₂V₂. When pressure increases, volume decreases by the same factor.
- Sec 18.5Charles's Law: at constant P and n, volume and temperature (in Kelvin) are directly proportional. V₁/T₁ = V₂/T₂. When temperature increases, volume increases proportionally.
- Sec 18.6Gay-Lussac's Law: at constant V and n, pressure and temperature (in Kelvin) are directly proportional. P₁/T₁ = P₂/T₂. In a rigid sealed container, heating always increases pressure.
- Sec 18.7Avogadro's Law: at constant T and P, volume is directly proportional to moles. V₁/n₁ = V₂/n₂. At STP (0 °C, 1 atm), one mole of any ideal gas occupies 22.4 L regardless of chemical identity.
- Sec 18.8The combined gas law, P₁V₁/T₁ = P₂V₂/T₂, handles simultaneous changes in P, V, and T at constant n. Constant variables cancel, reducing to one of the three individual laws when applicable.
- Sec 18.9The ideal gas law PV = nRT connects all four gas variables. R = 0.08206 L·atm/mol·K. Use it to calculate any unknown variable from the other three, or to find gas density using d = PM/(RT).
- Sec 18.10Real gases deviate from ideal behavior at high pressure (particle volume matters) and low temperature (intermolecular attractions matter). Ideal gas calculations are accurate for most laboratory conditions. All models have limits.
Formula Review
| Law | Equation | Constant Variables |
|---|
| Pressure conversion | 1 atm = 101.3 kPa = 760 mmHg = 760 torr | — |
| Temperature conversion | T (K) = T (°C) + 273 | — |
| Boyle's Law | P₁V₁ = P₂V₂ | T and n constant |
| Charles's Law | V₁/T₁ = V₂/T₂ | P and n constant |
| Gay-Lussac's Law | P₁/T₁ = P₂/T₂ | V and n constant |
| Avogadro's Law | V₁/n₁ = V₂/n₂ | T and P constant |
| Molar volume at STP | 1 mol = 22.4 L at 0 °C and 1 atm | — |
| Combined Gas Law | P₁V₁/T₁ = P₂V₂/T₂ | n constant |
| Ideal Gas Law | PV = nRT, R = 0.08206 L·atm/mol·K | — |
| Gas density | d = PM / (RT) | — |
Worked Example Review
Example A — Boyle's Law
Gas at 1.50 atm, 8.00 L. New pressure: 3.00 atm. Find V₂.
- Solve
P₁V₁ = P₂V₂ → (1.50)(8.00) = (3.00)(V₂) → V₂ = 12.0/3.00 = 4.00 L
Answer: 4.00 L
Example B — Charles's Law
Gas at 1.50 L, 300 K. New temperature: 450 K. Find V₂.
- Solve
V₁/T₁ = V₂/T₂ → 1.50/300 = V₂/450 → V₂ = 1.50 × (450/300) = 2.25 L
Answer: 2.25 L
Example C — Gay-Lussac's Law
Gas at 2.00 atm, 400 K. New temperature: 600 K. Find P₂.
- Solve
P₁/T₁ = P₂/T₂ → 2.00/400 = P₂/600 → P₂ = 2.00 × (600/400) = 3.00 atm
Answer: 3.00 atm
Example D — Combined Gas Law
Gas at 1.00 atm, 300 K, 4.00 L. New conditions: 2.00 atm, 400 K. Find V₂.
- Solve
P₁V₁/T₁ = P₂V₂/T₂
(1.00)(4.00)/300 = (2.00)(V₂)/400
V₂ = (1.00)(4.00)(400) / (300)(2.00) = 1600/600 = 2.67 L
Answer: 2.67 L
Example E — Ideal Gas Law
Gas at 2.50 atm, 10.0 L, 27 °C. Find moles.
- Solve
T = 300 K
n = PV/(RT) = (2.50)(10.0) / (0.08206)(300) = 25.0 / 24.618 = 1.02 mol
Answer: 1.02 mol
Key Terms Review
absolute zero
The theoretical temperature at which gas particles would have zero kinetic energy; equal to 0 K or −273 °C.
atmosphere (atm)
A unit of pressure equal to 101.3 kPa or 760 mmHg; defined as standard atmospheric pressure at sea level.
Avogadro's Law
At constant T and P, the volume of a gas is directly proportional to the number of moles present. V₁/n₁ = V₂/n₂.
Boyle's Law
At constant T and n, pressure and volume are inversely proportional. P₁V₁ = P₂V₂.
Charles's Law
At constant P and n, volume is directly proportional to temperature in Kelvin. V₁/T₁ = V₂/T₂.
combined gas law
P₁V₁/T₁ = P₂V₂/T₂ — combines Boyle's, Charles's, and Gay-Lussac's laws to handle simultaneous changes in P, V, and T at constant n.
compressible
Capable of being reduced in volume by applying pressure; a key property of gases due to large spaces between particles.
diffusion
The gradual movement and spreading of gas particles through space until they are uniformly distributed.
elastic collision
A collision in which kinetic energy is conserved overall; assumed in the kinetic molecular theory of gases.
Gay-Lussac's Law
At constant V and n, pressure is directly proportional to temperature in Kelvin. P₁/T₁ = P₂/T₂.
ideal gas
A hypothetical gas that perfectly follows all assumptions of kinetic molecular theory and obeys all gas laws exactly under all conditions.
ideal gas constant (R)
The proportionality constant in the ideal gas law; R = 0.08206 L·atm/(mol·K) when using atmospheres and liters.
ideal gas law
PV = nRT — relates pressure, volume, temperature, and moles for an ideal gas in a single equation.
Kelvin (K)
The SI unit of absolute temperature; calculated as K = °C + 273. Required for all gas law calculations.
kinetic molecular theory
A model of gas behavior: small particles in constant random motion with elastic collisions, negligible volume, no attractions, and average kinetic energy proportional to temperature in Kelvin.
manometer
A laboratory instrument used to measure the pressure of a gas sample in an enclosed container.
molar volume
The volume occupied by one mole of an ideal gas at STP; equal to 22.4 L/mol regardless of chemical identity.
pressure
Force per unit area; for a gas, the collective result of particle collisions with container walls.
real gas
An actual gas whose behavior deviates from ideal behavior, especially at high pressure and low temperature.
STP
Standard temperature and pressure: defined as 0 °C (273 K) and 1 atm; used as a reference point for gas law calculations and molar volume.
torr
A unit of pressure equal to 1 mmHg; 760 torr = 1 atm. Named after Evangelista Torricelli, inventor of the barometer.
Concept Check
Answer each question in one or two sentences before moving to the longer practice problems.
- Why must temperature be converted to Kelvin before using gas law equations?
- A gas sample is at fixed temperature in a flexible balloon. If you squeeze the balloon to reduce its volume, what happens to the pressure inside? Which gas law describes this?
- Which gas variable — pressure, volume, or temperature — is being changed when you pump more air into an already-inflated tire at constant temperature?
- Why do gas particles exert pressure on the walls of a container?
- What does it mean to say that one mole of any ideal gas occupies 22.4 L at STP?
- A rigid sealed steel canister is placed in a campfire. Temperature inside rises dramatically. Volume cannot change. What happens to pressure inside, and why could this be dangerous?
- Name one difference between an ideal gas and a real gas, and state the conditions under which this difference matters most.
- If you know P, V, and T for a gas sample, which equation would you use to find the number of moles?
Gas Law Practice
- Convert 2.75 atm to kPa.
- Convert 608 mmHg to atm.
- Convert 152 kPa to atm and to mmHg.
- A gas exerts 1.25 atm. Express this in torr and in kPa.
- Convert 950 mmHg to atm and to kPa.
Convert each temperature to Kelvin: (a) 0 °C (b) 25 °C (c) 100 °C (d) −40 °C (e) −196 °C (liquid nitrogen).
Convert each Kelvin temperature to Celsius: (a) 373 K (b) 500 K (c) 200 K (d) 77 K (e) 1000 K.
- A gas at 1.00 atm occupies 6.00 L. What is its volume at 3.00 atm? (Temperature constant.)
- A gas is compressed from 12.0 L to 3.00 L at constant temperature. Initial pressure was 0.50 atm. What is the final pressure?
- A sample of argon at 2.40 atm occupies 500 mL. At what pressure will it occupy 1.20 L?
- A flexible container holds 800 mL at 760 mmHg. The pressure increases to 1140 mmHg. What is the new volume?
- A diver takes a breath occupying 2.50 L at the surface (1.00 atm). The diver descends to where pressure is 3.00 atm. What is the new volume of air in the lungs? Explain why this calculation is relevant to a breath-holding diver who ascends too quickly.
- A balloon at 300 K holds 5.00 L. Temperature rises to 450 K at constant pressure. What is the new volume?
- A gas at 27 °C occupies 2.00 L. It is cooled to −73 °C at constant pressure. What is the new volume?
- A gas sample at 400 K has a volume of 3.60 L. At what temperature will it occupy 2.40 L at the same pressure?
- A gas at 500 K occupies 12.0 L. At what temperature (in °C) will it occupy 8.00 L?
- A gas at 20 °C occupies 750 mL at constant pressure. At what Celsius temperature will it occupy 1.00 L?
- A sealed container holds gas at 1.20 atm and 300 K. Temperature increases to 450 K at constant volume. What is the new pressure?
- A gas at 27 °C exerts 2.00 atm in a rigid container. What pressure will it exert at 127 °C?
- A sealed aerosol can has pressure 1.40 atm at 25 °C. The can is heated to 80 °C. What is the final pressure?
- A gas in a rigid vessel at 400 K exerts 3.00 atm. At what temperature (in °C) will it exert 1.50 atm?
- A gas exerts 1.00 atm at −73 °C. At what Celsius temperature will it exert 2.00 atm in the same rigid container?
- A gas at 1.00 atm, 300 K, and 6.00 L is changed to 2.00 atm and 600 K. What is the new volume?
- A gas occupies 250 mL at 740 mmHg and 20 °C. What volume does it occupy at 760 mmHg and 0 °C?
- A gas at 2.50 atm, 400 K, and 3.00 L is compressed to 5.00 atm at 500 K. What is the new volume?
- A gas sample at STP occupies 4.48 L. What volume will it occupy at 2.00 atm and 100 °C?
- A gas is changed from 1.50 atm and 27 °C in 8.00 L to 3.00 atm in 3.00 L. What is the final temperature in °C?
- What volume does 2.00 mol of an ideal gas occupy at STP?
- A gas sample at STP has a volume of 11.2 L. How many moles does it contain?
- How many grams of oxygen gas (O₂) are present in 44.8 L at STP?
- A balloon at constant T and P holds 2.50 L when it contains 0.100 mol. How many moles would fill it to 5.00 L under the same conditions?
- At STP, what volume is occupied by 64.0 g of SO₂ (molar mass = 64.1 g/mol)?
- Calculate the pressure exerted by 2.00 mol of an ideal gas in a 10.0 L container at 25 °C.
- A 5.00 L container holds gas at 3.00 atm and 127 °C. How many moles of gas does it contain?
- What volume does 0.400 mol of N₂ occupy at 1.50 atm and 30 °C?
- A 3.00 g sample of helium (He, M = 4.00 g/mol) is placed in a 10.0 L container at 27 °C. What is the pressure?
- Find the molar mass of a gas if 1.50 g occupies 800 mL at 1.20 atm and 27 °C.
- Calculate the density of carbon dioxide (CO₂, M = 44.0 g/mol) at 1.00 atm and 25 °C.
Error Analysis
Each problem contains an error. Identify it, explain what went wrong, and show the correct solution.
Problem A — Boyle's Law
Gas at 2.00 atm, 5.00 L. Pressure decreases to 1.00 atm. P₁V₁ = P₂V₂ → (2.00)(5.00) = (1.00)(V₂) → V₂ = 2.50 L
Is the answer reasonable? Pressure decreased, so what should happen to volume? Identify the calculation error and give the correct answer.
Problem B — Charles's Law
Gas at 27 °C, 4.00 L. Heated to 54 °C. Temperature doubled (27 × 2 = 54), so volume doubles: V₂ = 8.00 L.
Explain what is wrong with this reasoning. Convert both temperatures to Kelvin and solve correctly.
Problem C — Ideal Gas Law
Find moles in a 2.00 L container at 1.00 atm and 27 °C. n = PV/RT = (1.00)(2.00) / (0.08206)(27) = 0.902 mol
Identify the error. What temperature should be used? Solve correctly.
Problem D — Combined Gas Law
Gas at 800 mmHg, 300 K, 3.00 L changed to 400 mmHg, 600 K. Student gets V₂ = 12.0 L.
Check all steps. Is the reasoning and arithmetic correct? Verify whether the answer makes physical sense: pressure decreased and temperature increased — which directions do those push volume?
Real-World Applications
- Breathing mechanics. When you inhale, the volume of your chest cavity increases. Using Boyle's Law, explain why this causes air to flow into your lungs. What happens to chest volume and lung pressure when you exhale?
- Hot air balloons. The burner heats air inside the balloon envelope. Use Charles's Law to explain why this increases volume, and explain how lower density creates lift.
- Scuba diving. A scuba diver carries compressed air. As the diver descends, water pressure increases. The regulator delivers air at the pressure of the surrounding water. Why is it dangerous for a breath-holding diver to ascend rapidly? Use Boyle's Law in your explanation.
- Tire pressure and temperature. A car tire is inflated to 35.0 psi on a summer day at 35 °C. On a winter morning the temperature is −10 °C. Assuming the volume does not change, use Gay-Lussac's Law to calculate the winter morning pressure. (Work as a ratio directly — you do not need to convert psi to atm for this calculation.)
- Aerosol cans. An aerosol can at 22 °C contains gas at 1.30 atm. It is placed in a fire and reaches 400 °C. Using Gay-Lussac's Law, calculate the pressure inside. Does this help explain the warning "Do not expose to heat or flame"?
- Weather balloons. A weather balloon is released at sea level with a volume of 1.00 m³ at 1.00 atm and 15 °C. The balloon rises until pressure outside is 0.350 atm and temperature is −45 °C. Using the combined gas law, calculate the balloon's new volume at altitude.
- Industrial gas storage. A factory produces 450 L of nitrogen gas at 25 °C and 1.00 atm. The gas is compressed into a 45.0 L steel cylinder at 25 °C. Calculate the pressure in the cylinder in atm and in kPa.
- Stoichiometry and gas volumes. Methane burns: CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g). At STP, how many liters of O₂ are needed to react with 44.8 L of CH₄? How many liters of CO₂ are produced? Use Avogadro's Law reasoning.
Chapter 18 Checkpoint
Chapter 18 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 18
Section 1 — Vocabulary and Concepts
- Name the four variables that describe a gas sample and give the correct unit for each in gas law calculations.
- State Boyle's Law, Charles's Law, and Gay-Lussac's Law in words. For each, identify which two variables are being related and which two are held constant.
- Explain why a real gas might deviate from ideal behavior at high pressure. What assumption of the kinetic molecular theory breaks down?
Section 2 — Conversions
- Convert 3.50 atm to kPa and to mmHg.
- Convert 912 mmHg to atm.
- Convert each temperature to Kelvin: (a) 0 °C (b) 37 °C (c) −100 °C
Section 3 — Gas Law Calculations
- A gas at 2.00 atm and 8.00 L is expanded to 16.0 L at constant temperature. Find the new pressure.
- A gas at 250 K occupies 4.00 L. At what temperature will it occupy 6.00 L at constant pressure?
- A sealed rigid container holds gas at 1.50 atm and 300 K. The temperature is raised to 450 K. Find the new pressure.
- A gas at 1.00 atm, 300 K, and 3.00 L is changed to 2.50 atm and 400 K. Find the new volume using the combined gas law.
- How many moles of gas are in a 2.00 L container at 1.50 atm and 27 °C?
- What volume does 0.200 mol of gas occupy at STP?
Section 4 — Applications
- A 10.0 g sample of CO₂ (M = 44.0 g/mol) is placed in a 5.00 L container at 25 °C. Calculate the pressure using the ideal gas law.
- A weather balloon released at 1.00 atm and 15 °C with volume 2.00 m³ rises to where pressure is 0.400 atm and temperature is −35 °C. Find the new volume.
Section 5 — Error Analysis
- A student solves a Charles's Law problem using temperatures of 25 °C and 50 °C and concludes that the volume doubled because "the temperature doubled." Explain the error and calculate the correct volume ratio.
- A student calculates n = PV/(RT) for a gas at 1.00 atm, 3.00 L, and 25 °C and gets n = (1.00)(3.00) / (0.08206)(25) = 1.46 mol. Identify the error and give the correct answer.
Mixed Review
These problems draw on skills from Chapters 15, 16, and 17 as well as Chapter 18, keeping earlier material sharp.
- How many particles are in 0.500 mol of nitrogen gas? (Chapter 15)
- Balance the equation: __ C₃H₈ + __ O₂ → __ CO₂ + __ H₂O. (Chapter 13)
- How many grams of CO₂ are produced when 44.0 g of propane (C₃H₈, M = 44.1 g/mol) burns completely? Use the equation from above. (Chapter 16)
- What volume of CO₂ gas at STP is produced from that combustion? Use your answer from above and the molar volume. (Chapters 16 and 18)
- A gas at STP has a density of 1.43 g/L. Use molar volume (22.4 L/mol) to calculate its molar mass. Can you identify this gas?
- A gas occupies 11.2 L at STP. How many moles is this? How many molecules? (Chapters 15 and 18)
- A student collects 250 mL of hydrogen gas over water at 25 °C and a total pressure of 755 mmHg. The vapor pressure of water at 25 °C is 23.8 mmHg. What is the partial pressure of the dry hydrogen gas? (Hint: P_total = P_H₂ + P_water_vapor.)
- A student prepares 0.200 mol of NaCl by reacting sodium with chlorine: 2 Na(s) + Cl₂(g) → 2 NaCl(s). What volume of Cl₂ at 1.00 atm and 25 °C is consumed? (Chapters 16 and 18)
Looking Ahead
Chapter 19: Liquids, Solids, and Phase Changes
In Chapter 18, you studied gases — the state of matter in which particles move most freely. In Chapter 19, you will turn to liquids and solids, where particles are far closer together and far more strongly attracted to each other. You will explore how intermolecular forces determine whether a substance is solid, liquid, or gas at a given temperature and pressure, and you will study the energy changes involved in melting, freezing, vaporization, condensation, and sublimation. Heating curves and cooling curves will let you read these transitions from a graph, and phase diagrams will show you how temperature and pressure together determine which state any substance occupies. You will discover why water boils at a lower temperature on a mountain, why dry ice sublimes without becoming liquid, and how evaporation cools your skin. Chapter 19 connects directly to the gas laws of Chapter 18 and prepares you for the thermochemistry of Chapter 20.
MGU Library Connection
- Gas Laws Formula Sheet — Appendix H of this textbook
- SI Units and Measurement Reference — Appendix B of this textbook
- Chapter 15: The Mole and Molar Mass — foundational for mole calculations in gas law problems
- Chapter 16: Stoichiometry — connects mole ratios to gas volume ratios via Avogadro's Law
- Chapter 19: Liquids, Solids, and Phase Changes — extends particle model from gases to condensed states
- MGU Dictionary entries: pressure, temperature, kinetic energy, Boyle's Law, Charles's Law, ideal gas law, molar volume