General Chemistry · Unit 12

Gas Laws: Boyle's, Charles's, and the Ideal Gas Law

Gas laws come down to one key question first: what stays constant, what changes, and do you need a simple gas law or PV = nRT? This unit builds from thermochemistry before you move into solutions.

What you'll learn

Apply Boyle's, Charles's, and Gay-Lussac's Laws to single-variable gas problems. Solve combined and ideal gas law problems using PV = nRT. Use Avogadro's Law and molar volume at STP for gas stoichiometry. Calculate partial pressures with Dalton's Law, including gases collected over water.

12.1 Start Here: How Gases Behave and What Pressure Really Means

Gases are the most spread-out state of matter. Their particles are far apart, move quickly in random directions, and keep colliding with the walls of the container.

Pressure is force per unit area (P = F/A). In a gas, that force comes from particle collisions with the container walls. If you keep that particle picture in mind, the gas laws make much more sense.

This is also why a sharp knife cuts better than a dull one: the same force applied over a smaller area creates greater pressure.

In chemistry, 1 atmosphere (1 atm) is the standard reference pressure. Real atmospheric pressure can change with weather and altitude, but many chemistry problems use 1 atm as the starting reference. A barometer measures atmospheric pressure by the height of mercury it can support, and the SI unit of pressure is the pascal (Pa = N/m²).

Pressure Conversion Factors — Memorize These

1 atm = 760 mmHg = 760 torr 1 atm = 101.325 kPa = 101,325 Pa 1 atm = 14.696 psi = 1.01325 bar

These five pressure units all measure the same thing. Recognize which unit a problem gives you and convert before setting up any gas law equation.

Unit	Symbol	= 1 atm	Common Use
Atmosphere	atm	1.000	Standard chemistry problems
Millimeters mercury	mmHg	760.0	Barometers, medical settings
Torr	torr	760.0	Equivalent to mmHg
Kilopascal	kPa	101.325	SI unit, international use
Pascal	Pa	101,325	Formal SI calculations

12.2 Boyle's Law: Pressure and Volume Move Opposite Ways

At constant temperature and constant amount of gas, pressure and volume are inversely proportional.

Do not miss the condition words here. If temperature is not constant, Boyle's Law is not the right shortcut anymore.

If pressure doubles, volume is cut in half. The product P × V stays constant.

Boyle's Law (T and n constant)

P₁V₁ = P₂V₂ P × V = k (constant)

Required Conditions

Constant temperature (T) and constant amount of gas (n).

State 1

Large Volume, Low Pressure

Molecules have more space, so they hit the container walls less often.

P1 is lower because wall collisions are less frequent.

Inverse Relationship

P1V1 = P2V2

Volume halved

Pressure doubles

At constant T and n, squeezing the gas into less space raises the pressure.

State 2

Small Volume, High Pressure

The same number of particles are forced into less space, so wall collisions increase.

P2 is higher because the particles strike the walls more often.

The particle model shows why Boyle's Law works: with the same number of gas particles at constant temperature, squeezing the gas into less space increases wall collisions and raises pressure.

Real-world example: A scuba diver exhales air bubbles at depth (~3 atm). As the bubbles rise toward the surface (~1 atm), the pressure drops to about one-third, so the volume triples.

The P vs. V graph for Boyle's Law always forms a hyperbola.

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Exam Tip

Boyle's Law only works when T and n are both constant.
The phrase "constant temperature" or "isothermal" is your signal to use Boyle's Law.
A pressure increase must compress the gas (decrease volume), and vice versa.

12.3 Charles's Law: Volume and Temperature Rise Together

At constant pressure and constant amount of gas, volume is directly proportional to absolute temperature in Kelvin. Double the Kelvin temperature → double the volume. The V vs. T graph is a straight line through the origin when T is in Kelvin.

The most common mistake in this whole unit happens here: leaving temperature in Celsius. Do not do that. Gas-law temperature must be in Kelvin.

Charles's Law (P and n constant) — T ALWAYS in Kelvin!

V₁T₁ = V₂T₂ K = °C + 273

Required Conditions

Constant pressure (P) and constant amount of gas (n). Temperature must be in Kelvin.

State 1

Low Temperature

Cooler gas particles move more slowly, so the volume stays smaller when pressure remains constant.

At the lower Kelvin temperature, the gas occupies the smaller volume V1.

Direct Relationship

V1T1 = V2T2

Temperature increases

Volume increases

When pressure stays constant, hotter gas expands because the particles move faster and need more space.

Always convert °C to K before using the equation.

State 2

High Temperature

Hotter gas particles move faster, so the piston rises and the volume increases while the outside pressure stays the same.

At the higher Kelvin temperature, the gas expands to the larger volume V2.

The particle picture, equation, and volume change all show the same rule: at constant pressure, hotter gas occupies more space.

Why Kelvin? At absolute zero (0 K = −273°C), gas volume theoretically reaches zero. Kelvin starts at this true zero point, making the proportional relationship mathematically valid. Using Celsius gives wrong answers.

12.4 Gay-Lussac's Law: Pressure Changes When a Rigid Container Is Heated

At constant volume and constant amount of gas, pressure is directly proportional to absolute temperature in Kelvin. This applies to any rigid container. Heating → higher pressure.

Start here when the problem mentions a rigid can, sealed tank, or fixed-volume container. That is the clue that volume is constant and pressure is the thing that responds.

Gay-Lussac's Law (V and n constant) — T ALWAYS in Kelvin!

P₁T₁ = P₂T₂

Required Conditions

Constant volume (V) and constant amount of gas (n). The container must be rigid, and temperature must be in Kelvin.

State 1

Low Temperature, Low Pressure

Cooler particles move more slowly, so their wall collisions are less frequent and less forceful inside the fixed-volume tank.

1 = 300 K

P1 is lower because the slower particles do not strike the walls as hard.

Direct Relationship

P1T1 = P2T2

Temperature increases

Pressure increases

Because volume cannot expand, heating the gas raises particle speed and pressure instead.

Rigid container = volume unchanged. Always convert °C to K.

State 2

High Temperature, High Pressure

Hotter particles move faster, so the same tank experiences more frequent and more forceful wall collisions.

2 = 600 K

P2 is higher because the faster particles hit the walls harder and more often.

In a rigid container, volume cannot expand. Heating the gas increases particle speed, so pressure rises instead.

Real-world example: An aerosol spray can in a hot car can explode. The rigid metal walls prevent volume change, so the rising temperature directly drives up the internal pressure — sometimes past the structural limit of the can.

12.5 Avogadro's Law and Molar Volume at STP

At constant temperature and pressure, volume is directly proportional to moles. At the same temperature and pressure, equal volumes of gases contain the same number of molecules.

Notice the wording carefully: equal volumes mean equal numbers of molecules, not automatically equal numbers of atoms.

Avogadro's Law (T and P constant)

V₁n₁ = V₂n₂ Molar volume at STP (0°C = 273 K, exactly 1 atm): 1 mol any ideal gas = 22.4 L

Required Conditions

Constant temperature (T) and constant pressure (P). At STP, 1 mol of any ideal gas occupies 22.4 L.

State 1

Fewer Moles, Smaller Volume

With fewer gas molecules present, the sample occupies the smaller volume when temperature and pressure stay fixed.

At fixed T and P, the smaller amount of gas corresponds to the smaller volume V1.

Direct Relationship

V1n1 = V2n2

Moles increase

Volume increases

If temperature and pressure stay constant, adding more gas molecules requires more space.

Equal volumes at the same T and P contain equal numbers of molecules.

State 2

More Moles, Larger Volume

With more gas molecules under the same conditions, the sample occupies a larger volume.

At the same T and P, the larger amount of gas occupies the larger volume V2.

Avogadro's Law says the amount of gas and the volume rise together when temperature and pressure stay fixed.

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Classic trap

"Does 1 L of CH4 or 1 L of H2 contain more hydrogen atoms at STP?" Both contain the same number of molecules by Avogadro's Law.
But CH4 has 4 H atoms per molecule while H2 has only 2.
So 1 L of CH4 contains twice as many hydrogen atoms.
Equal molecules does not mean equal atoms. Always distinguish atoms from molecules.

12.6 The Combined Gas Law: One Gas Sample, Two Sets of Conditions

Use the Combined Gas Law when one sample of gas changes from an initial state to a final state and the amount of gas stays constant. This law is useful when pressure, volume, and temperature are connected across two sets of conditions.

If you are confused, look for the story structure: before and after, same gas sample, no gas added or lost. That is your cue.

Combined Gas Law (n constant) — T in Kelvin, P in same units both sides

P₁ × V₁T₁ = P₂ × V₂T₂

Use this table to see all five laws side by side. The 'Held Constant' column is what you check first when reading a gas law problem.

Law	Held Constant	Equation	Graph of	Shape
Boyle's	T, n	P₁V₁ = P₂V₂	P vs. V	Hyperbola
Charles's	P, n	V₁T₁ = V₂T₂	V vs. T	Straight line
Gay-Lussac's	V, n	P₁T₁ = P₂T₂	P vs. T	Straight line
Combined	n only	P₁V₁T₁ = P₂V₂T₂	—	—
Avogadro's	T, P	V₁n₁ = V₂n₂	V vs. n	Straight line

12.7 The Ideal Gas Law: When PV = nRT Is the Better Tool

Combines all four gas variables into one equation for one set of conditions. Unlike the Combined Gas Law, you do not need before-and-after conditions — just what is true right now. Solve for whichever variable is unknown.

The temptation is to avoid PV = nRT or reach for it too early. Use it when you have one state, not two, and when pressure, volume, temperature, and moles all matter together.

Ideal Gas Law — T in Kelvin; choose R to match your pressure unit

PV = nRT R = 0.08206 L·atm·mol⁻¹·K⁻¹ when P is in atm R = 8.314 L·kPa·mol⁻¹·K⁻¹ when P is in kPa R = 8.314 J·mol⁻¹·K⁻¹ for energy problems

Derived From Ideal Gas Law

Molar mass from a gas sample: M = mRTPV (m = mass of sample in grams) Gas density: d = PMRT (d in g/L when V is in liters)

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Exam Tip

Ideal gas behavior assumes no intermolecular forces and negligible molecular volume.
Real gases deviate most at very high pressure or very low temperature.
For introductory chemistry, always assume ideal behavior unless the problem states otherwise.

12.8 Dalton's Law: Total Pressure and Gas Collected Over Water

In a mixture of non-reacting gases, each gas exerts its own pressure independently. The partial pressure of each gas equals the pressure it would exert alone at the same T and V. Total pressure is the sum of all partial pressures.

Do not miss the classic trap here: if the gas is collected over water, the measured pressure is not just your gas. You have to subtract water vapor first.

When a gas is collected over water, the measured pressure includes both the gas you want and water vapor. Subtract the water vapor pressure first to get the pressure of the dry gas. Then use that dry gas pressure in the Ideal Gas Law.

Dalton's Law of Partial Pressures

P_total = P₁ + P₂ + P₃ + … Mole fraction: χ_a = n_an_total Partial pressure: P_a = χ_a × P_total Gas collected over water: P_{dry gas} = P_barometer − P_{H2O vapor}

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Water vapor pressure at common temperatures: 20°C → 17.5 mmHg | 25°C → 23.8 mmHg | 30°C → 31.8 mmHg | 37°C → 47.1 mmHg.
These must be looked up in a table — they cannot be calculated from the gas laws.

12.9 Kinetic Molecular Theory: Why the Gas Laws Work

The gas laws arise from the behavior of molecules described by the Kinetic Molecular Theory (KMT):

This section ties the whole unit together. If the formulas start to feel memorized instead of understood, come back here and connect each law to particle motion.

Gas molecules are tiny relative to the space between them

The actual volume of molecules is negligible compared to the container. This is why gases are almost entirely empty space and are easily compressed.

Gas molecules are in constant, rapid, random motion

They travel in straight lines between collisions. This motion produces pressure when molecules hit container walls.

Collisions are perfectly elastic

No kinetic energy is lost in collisions. Total kinetic energy stays constant at constant temperature.

No intermolecular forces between molecules

Ideal gas molecules move completely independently between collisions.

Average kinetic energy is proportional to Kelvin temperature

KE_avg = 3/2k_BT. This directly explains Charles's Law (T↑ → molecules move faster → push walls harder → volume expands at constant P) and Gay-Lussac's Law (T↑ → harder wall collisions → P increases at constant V).

Use these like quick gas-law checks. Pick the law, setup, or pressure first, then reveal the reasoning. For more reps, use the Unit 12 Practice page before moving into solutions.

Which Gas Law Fits?

Read This First

Choose the gas law that fits this situation.

Look for the condition that does not change before you choose an answer.

Scenario: A sealed syringe of gas is compressed from 2.0 L to 1.0 L while the temperature stays constant and no gas escapes.

Choose one law, then click Check.

Valid Setup for PV = nRT

Read This First

Choose the setup you could use correctly in PV = nRT.

Check temperature units and pressure units before you choose an answer.

Problem: A gas sample has P = 755 mmHg, V = 0.500 L, and T = 25°C. You want to solve for moles.

Choose one setup, then click Check.

Avogadro: Molecules or Atoms?

Read This First

Choose which sample has more hydrogen atoms.

Start with molecules. Then think about atoms in each molecule before you answer.

Compare the two samples below at the same temperature and pressure.

Choose one answer, then click Check.

Collected Over Water: Which Pressure Do You Use?

Read This First

Choose the pressure that belongs in PV = nRT for hydrogen gas.

The measured pressure includes more than one gas, so read the setup before you choose.

Scenario: A gas sample is collected over water. Use the values below to decide which pressure belongs to the dry gas.

Measured pressure: 755 mmHg

Water vapor pressure: 23.8 mmHg

Gas collected: H2

Choose one pressure, then click Check.

Ideal Gas Isothermal Curves

Read This First

At this same volume, which point has the higher pressure?

Compare the temperatures of the two curves before you choose.

Scenario: Point A and Point B sit at the same volume on different isothermal curves. Decide which point must have the higher pressure.

▶ Interactive: PV = nRT — Compare Two Points Before You Reveal

Choose Point A or Point B, then click Check.

Example 1 · Boyle's Law — Syringe Compression

Problem: Find the new volume after the pressure changes.

Given: V₁ = 4.50 L, P₁ = 1.20 atm, P₂ = 3.00 atm, constant temperature

Find: V₂

Step 1 — Identify the law: T and n constant → Boyle's Law

P₁ = 1.20 atm V₁ = 4.50 L P₂ = 3.00 atm V₂ = ?

Step 2 — Apply Boyle's Law and solve for V₂

P₁V₁ = P₂V₂ → V₂ = P₁V₁P₂ = 1.20 atm × 4.50 L3.00 atm

atm cancels on top and bottom → unit remaining: L ✓

V₂ = 5.403.00 = 1.80 L

Sanity check: pressure × 2.5 increase → volume ÷ 2.5. 4.50 ÷ 2.5 = 1.80 ✓

Example 2 · Charles's Law — Balloon Heating

Problem: Find the new volume after heating a balloon at constant pressure.

Given: V₁ = 2.50 L, T₁ = 25.0°C, T₂ = 100.0°C, constant pressure

Find: V₂

Step 1 — Use Charles's Law, then convert both temperatures to Kelvin first

T₁ = 25.0 + 273 = 298 K T₂ = 100.0 + 273 = 373 K

Step 2 — Apply Charles's Law

V₁T₁ = V₂T₂ → V₂ = V₁ × T₂T₁ = 2.50 L × 373 K298 K = 3.13 L

Sanity check: T increased → V must increase. 3.13 > 2.50 ✓

Example 3 · Gay-Lussac's Law — Spray Can in Fire

Problem: Find the new pressure when the gas is heated in a rigid can.

Given: P₁ = 1.44 atm, T₁ = 23°C, T₂ = 475°C, constant volume

Find: P₂

Step 1 — Use Gay-Lussac's Law, then convert both temperatures to Kelvin

T₁ = 23 + 273 = 296 K T₂ = 475 + 273 = 748 K

Step 2 — Apply Gay-Lussac's Law

P₁T₁ = P₂T₂ → P₂ = P₁ × T₂T₁ = 1.44 atm × 748 K296 K = 3.64 atm

The pressure more than doubled — enough to rupture the can.

Example 4 · Combined Gas Law — Weather Balloon

Problem: Find the balloon's new volume after both pressure and temperature change.

Given: V₁ = 8.50 L, P₁ = 750 mmHg, T₁ = 22°C, P₂ = 365 mmHg, T₂ = −15°C

Find: V₂

Step 1 — Choose the law: this is the same gas sample, the amount of gas stays constant, and both pressure and temperature change. Use the Combined Gas Law.

T₁ = 295 K T₂ = 258 K P in mmHg both sides → no conversion needed

Step 2 — Solve the Combined Gas Law for V₂

V₂ = P₁V₁T₂T₁P₂ = 750 × 8.50 × 258295 × 365 = 15.3 L

Sanity check: P dropped by ~half (big ↑ volume) while T dropped slightly (small ↓ volume). Net result = volume increase. ✓

Example 5 · Ideal Gas Law — Finding Moles

Problem: Use the ideal gas law to determine how many moles of gas are present.

Given: V = 8.25 L, P = 3.75 atm, T = 127°C

Find: n

Step 1 — Use PV = nRT, then convert temperature and choose the matching gas constant

T = 127 + 273 = 400 K P in atm → R = 0.08206 L·atm/mol·K

Step 2 — Solve PV = nRT for n

n = PVRT = 3.75 atm × 8.25 L0.08206 L·atm·mol⁻¹·K⁻¹ × 400 K = 30.9432.82 = 0.943 mol

Units: atm·LL·atm·mol⁻¹·K⁻¹·K → mol ✓

Example 6 · Ideal Gas Law — Molar Mass of Unknown Gas

Problem: Find the molar mass of the unknown gas.

Given: mass = 0.422 g, V = 0.250 L, T = 45°C, P = 740 mmHg

Find: molar mass

Plan: first find the moles of gas using PV = nRT. Then use molar mass = mass ÷ moles.

Step 1 — Use PV = nRT for moles first, so convert the temperature and pressure units

T = 45 + 273 = 318 K P = 740 ÷ 760 = 0.9737 atm

Step 2 — Find moles via PV = nRT

n = PVRT = 0.9737 atm × 0.250 L0.08206 L·atm·mol⁻¹·K⁻¹ × 318 K = 0.243426.10 = 0.009328 mol

Step 3 — Molar mass = mass ÷ moles

M = 0.422 g0.009328 mol = 45.2 g/mol

Example 7 · Dalton's Law — Gas Collected Over Water

Problem: Find the pressure of dry hydrogen and then the number of moles collected.

Given: total pressure = 755 mmHg, water vapor pressure = 23.8 mmHg, T = 25°C, V = 0.380 L

Find: P(H2) and mol H2

Step 1 — Isolate the dry gas pressure: the total pressure includes both hydrogen gas and water vapor, so subtract the water vapor pressure first.

P(H2) = P(total) − P(H2O) = 755 − 23.8 = 731.2 mmHg

Step 2 — Use only the dry hydrogen pressure in PV = nRT.

P = 731.2 ÷ 760 = 0.9621 atm T = 298 K

n = PVRT = 0.9621 × 0.3800.08206 × 298 = 0.365624.45 = 0.01495 mol H2

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Next step after Unit 12

Gas laws connect particles to measurable changes in pressure, volume, and temperature. The next move is to study solutions, where concentration and dissolved particles become the new focus. Keep Unit 12 active with the Unit 12 Practice page and the larger practice hub, then pair it with Why Practice Tests Beat Rereading if picking the right gas-law setup still feels shaky.

General Chemistry · Unit 12 · Gas Laws