The Combined Gas Law and Avogadro's Law

Bubbles grow as they rise

A diver breathes out a small bubble at the bottom of a lake, and it grows as it rises: the shallower the water, the lower the pressure, and the water near the surface is warmer too. Pressure and temperature both change at once, and each law from the last lesson holds one of them fixed. Combining the three gives the combined gas law. Adding the amount of gas gives Avogadro’s law and molar volume.

The combined gas law

When a fixed amount of gas goes from state 1 to state 2, pressure, volume, and temperature can all change:

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

This is the combined gas law综合气体定律; Chinese physics textbooks call it the equation of state of an ideal gas (pV/T = constant). If one quantity stays the same, it cancels and you are back to one of the three laws. To use it:

Convert temperatures to kelvins.
Use the same pressure and volume units on both sides. Matching units cancel, so there is no need to convert to Pa or L.
List the knowns and the unknown, rearrange, then substitute.

Avogadro’s law

In 1811, Amedeo Avogadro proposed that equal volumes of any gases at the same temperature and pressure contain equal numbers of molecules. This is Avogadro’s law阿伏加德罗定律. Put another way, at constant T and P, volume is directly proportional to moles: V₁/n₁ = V₂/n₂.

Why doesn’t the kind of gas matter? Gas molecules are about 10 molecular diameters apart, so a gas’s volume is almost all empty space, and the size of the molecules themselves hardly matters. Equal numbers of molecules at the same temperature and pressure take up the same room.

Molar volume

The volume of one mole of a gas is its molar volume气体摩尔体积, V_m, usually in L/mol:

Conditions	Name	V_m
273.15 K, 1 atm	STP in US textbooks and on the AP equation sheet	22.4 L/mol
0 °C, 101 kPa	标准状况 (“standard conditions”) in Chinese textbooks	22.4 L/mol
0 °C, 100 kPa	IUPAC STP since 1982	22.7 L/mol
25 °C, 1 atm	room conditions	24.5 L/mol

101 kPa is just 1 atm (101.325 kPa), so the Chinese 标准状况 is the same as US STP. Use 22.4 L/mol only at 0 °C and 1 atm. Thermochemistry (Chapter 2.10) has its own standard state, which is a different convention.

Avogadro’s law also means that at the same temperature and pressure, gas densities are in the same ratio as molar masses: ρ₁/ρ₂ = M₁/M₂. Air is about 78% N₂, 21% O₂, and 1% Ar by volume, and at the same T and P volume fractions are mole fractions, so its average molar mass is

\begin{aligned} \bar{M} &= 0.78 \times 28 + 0.21 \times 32 \\ &\quad + 0.01 \times 40 \\ &\approx 29\ \text{g/mol} \end{aligned}

That is why CO₂ (44 g/mol) is denser than air and sinks, while H₂ (2 g/mol) rises.

This lesson uses the atomic masses on the AP periodic table (H 1.008, C 12.01, N 14.01, O 16.00, Cl 35.45).

Solve a few combined gas law problems first. Pick the unknown and fill in both states:

Try it: combined gas law calculator

Solve for

State 1P₁V₁T₁

State 2P₂V₂T₂

Now see how big a mole of gas really is. Does the volume change when you switch gases?

Try it: how big is a mole of gas?

Gas

Conditions

Moles

Example 1: a rising bubble

At the bottom of a lake, the pressure is 3.0 atm and the water is 8 °C. A diver breathes out a 2.0-mL bubble there. What is its volume when it reaches the surface, at 1.0 atm and 25 °C?

Step-by-step solution

Choose the law. The amount of gas in the bubble is fixed while P, V, and T all change, so use the combined gas law, P₁V₁/T₁ = P₂V₂/T₂.
Convert to kelvins.
$\begin{aligned} T_1 &= 8 + 273 = 281\ \text{K} \\ T_2 &= 25 + 273 = 298\ \text{K} \end{aligned}$
Rearrange. Multiply both sides by T₂/P₂:
$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$
Substitute. The atm units cancel, and V stays in mL:
$\begin{aligned} V_2 &= \frac{3.0 \times 2.0 \times 298}{1.0 \times 281} \\ &= 6.36 \approx \boxed{6.4\ \text{mL}} \end{aligned}$
Check. Take it in two steps: the pressure drops to 1/3, so the volume triples to 6.0 mL; warming from 281 K to 298 K adds about 6%, giving 6.4 mL. The data have two significant figures, so the answer has two.

Example 2: carbon dioxide at STP

What are the moles and mass of 11.2 L of CO₂ at STP? What volume of H₂ at the same conditions contains the same number of molecules? Is CO₂ denser or less dense than air?

Step-by-step solution

STP. At 0 °C and 1 atm, one mole of any ideal gas takes up 22.4 L.
Moles.
$\begin{aligned} n &= 11.2\ \text{L} \times \frac{1\ \text{mol}}{22.4\ \text{L}} \\ &= \boxed{0.500\ \text{mol}} \end{aligned}$
Mass. M(CO₂) = 12.01 + 2(16.00) = 44.01 g/mol:
$\begin{aligned} m &= 0.500 \times 44.01 \\ &= \boxed{22.0\ \text{g}} \end{aligned}$
Avogadro’s law. Equal numbers of molecules at the same T and P take up equal volumes, so the answer is 11.2 L of H₂. Its mass is only 0.500 mol × 2.016 g/mol = 1.01 g.
Densities. At STP, density = M ÷ 22.4 L/mol. Air averages about 29.0 g/mol (from 78% N₂, 21% O₂, 1% Ar):
$\begin{aligned} \rho(\ce{CO2}) &= 44.01 / 22.4 \\ &= 1.96\ \text{g/L} \\ \rho(\text{air}) &= 29.0 / 22.4 \\ &= 1.29\ \text{g/L} \end{aligned}$
CO₂ is denser, so it collects near the floor. A faster way: at the same T and P, densities are in the same ratio as molar masses, and 44 > 29.
A common mistake. Using 22.4 L/mol for a gas that is not at STP, or for a liquid such as water. It applies only to gases at 0 °C and 1 atm.

Practice: compare the two quantities.

Same T and P: molecules in 1 L of H₂ vs. 1 L of O₂
Same T and P: mass of 1 L of H₂ vs. 1 L of O₂
Same T and P: volume of 1 g of H₂ vs. 1 g of O₂
Same T and P: volume of 1 mol of N₂ vs. 1 mol of CO₂
Same T and P: density of CH₄ vs. O₂
Molecules in 22.4 L of N₂ at STP vs. 22.4 L of N₂ at 25 °C and 1 atm
Volume of 1 mol of gas at STP vs. at 25 °C and 1 atm
Same T and P: mass of 1 L of air vs. 1 L of CO₂

Common mistakes

Mistake 1: “One mole of anything takes up 22.4 L.” 22.4 L/mol applies only to gases, and only at 0 °C and 1 atm. One mole of water at STP is liquid (or ice), with a volume of only about 18 mL.

Mistake 2: “A mole of gas at room temperature is also 22.4 L.” Gases expand when warmed. At 25 °C and 1 atm, a mole of gas takes up about 24.5 L. Unless the problem says STP, don’t use 22.4 L/mol.

Mistake 3: “Equal volumes of gases at the same T and P have equal masses.” Avogadro’s law is about equal numbers of molecules. Different molecules have different masses, so the masses are in the ratio of the molar masses.

Quick check

How many moles of O₂ are in 5.6 L at STP?

At the same temperature and pressure, equal masses of which gas take up the largest volume?

Summary

The combined gas law: P₁V₁/T₁ = P₂V₂/T₂. Use kelvins and matching pressure and volume units.
Avogadro’s law: equal volumes of gases at the same T and P contain equal numbers of molecules, so V ∝ n. At the same T and P, densities are in the ratio of molar masses.
Molar volume is 22.4 L/mol at STP (0 °C and 1 atm, the same as the Chinese 标准状况) and about 24.5 L/mol at 25 °C.

On to AP (for later)

The AP equation sheet lists STP as 273.15 K and 1.0 atm and gives the molar volume of an ideal gas at STP as 22.4 L/mol. AP problems more often use PV = nRT (next lesson) because it works at any temperature and pressure. AP also uses Avogadro’s law in particle diagrams: two containers of equal volume at the same T and P must show the same number of gas particles.

The Combined Gas Law and Avogadro's Law综合气体定律与阿伏加德罗定律

Learning goals

Before you start

Bubbles grow as they rise