Master AP Chemistry Topic 3.4 with this free practice quiz. Test your skills on ideal gas law, partial pressures, molar mass of gases, and gas density to prepare for your AP exam.
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Introduction to Gas Variables
Gases differ from solids and liquids because their particles are widely separated and move freely. Gas particles have relatively large amounts of empty space between them, allowing gases to expand to fill their containers. Their behavior can be described using four macroscopic variables: pressure, volume, temperature, and amount of gas.
Pressure (P) is the force exerted per unit area by gas particles colliding with the walls of their container. More frequent or more forceful collisions increase pressure. The SI unit is the pascal (Pa), although atmospheres (atm) and kilopascals (kPa) are commonly used in chemistry.
Volume (V) is the space occupied by a gas, usually measured in liters (L). A gas expands to fill its container, so its volume is generally the volume of the container.
Temperature (T) measures the average kinetic energy of gas particles. Gas law calculations require absolute temperature in kelvin (K), calculated using:
T(K) = T(°C) + 273.15
The amount of gas (n) is measured in moles. One mole contains Avogadro’s number of particles, approximately 6.022 × 10²³. The number of moles connects the measurable properties of a gas to the number of particles present.
These variables are related because changing temperature, volume, or the number of particles changes the frequency and force of collisions that produce gas pressure.
The Ideal Gas Equation
The ideal gas law combines the relationships among pressure, volume, temperature, and amount of gas into one equation:
PV = nRT
Here, P is pressure, V is volume, n is the number of moles, T is absolute temperature in kelvin, and R is the ideal gas constant.
The value of R depends on the units used for pressure and volume. Common values include:
- R = 0.08206 L·atm·mol⁻¹·K⁻¹
- R = 8.314 L·kPa·mol⁻¹·K⁻¹
- R = 8.314 J·mol⁻¹·K⁻¹
Always choose an R value consistent with the units in the question. For example, when pressure is given in atmospheres and volume in liters, use R = 0.08206 L·atm·mol⁻¹·K⁻¹.
The equation shows how gas variables depend on one another. For a fixed amount of gas, increasing temperature while maintaining constant volume increases pressure because particles move faster and collide with the container walls more forcefully. Increasing volume at constant temperature decreases pressure because particles collide with the walls less frequently.
At constant pressure, increasing temperature causes a gas to expand if the amount of gas remains constant. At constant temperature, increasing the amount of gas in a fixed volume increases pressure because more particles are present to collide with the walls.
These relationships explain why the ideal gas law is useful for predicting how gases respond to changes in experimental conditions.
The Kinetic Molecular Model and Ideal Gas Behavior
The ideal gas law is based on a simplified model of gas particles. According to the kinetic molecular model, ideal gas particles have negligible individual volume compared with the volume of their container, experience no significant intermolecular attractions or repulsions, and undergo elastic collisions. Their average kinetic energy depends only on absolute temperature.
Real gases approximate ideal behavior when particles are far apart and intermolecular attractions are relatively unimportant. This is generally the case at high temperatures and low pressures. High temperature increases particle kinetic energy, making attractive forces less significant relative to particle motion. Low pressure corresponds to greater average separation between particles, reducing the importance of particle volume and intermolecular interactions.
Real gases deviate more noticeably from ideal behavior at high pressures and low temperatures. At high pressure, particles are crowded together, so their own volume becomes significant. At low temperature, particles move more slowly, allowing intermolecular attractions to have a greater effect.
For AP Chemistry, students should be able to explain these deviations using particle-level reasoning rather than treating the ideal gas equation as universally exact.
Calculations Using the Ideal Gas Law
The ideal gas equation can be rearranged to solve for any one of its four variables:
P = nRT / V
V = nRT / P
n = PV / RT
T = PV / nR
A useful strategy is to identify the unknown, convert all quantities into compatible units, substitute the known values, and check whether the answer has a reasonable magnitude.
Worked example: Calculating gas volume
A sample contains 0.500 mol of an ideal gas at 300 K and 1.20 atm. Calculate its volume.
Start with:
PV = nRT
Rearrange for volume:
V = nRT / P
Substitute the values:
V = (0.500 mol)(0.08206 L·atm·mol⁻¹·K⁻¹)(300 K) / 1.20 atm
V = 10.3 L
The sample occupies approximately 10.3 L. The units of moles, kelvin, and atmospheres cancel appropriately, leaving liters.
Using the combined gas law
When the amount of gas remains constant, the ideal gas equation can compare two sets of conditions:
P₁V₁ / T₁ = P₂V₂ / T₂
This is the combined gas law. It is useful when a gas changes pressure, volume, and temperature without gaining or losing particles. Temperatures must be in kelvin.
For example, if a gas is heated in a rigid, sealed container, its volume and amount remain constant. The pressure-temperature relationship simplifies to:
P₁ / T₁ = P₂ / T₂
This proportionality follows directly from PV = nRT.
Dalton’s Law of Partial Pressures
A mixture of nonreacting gases exerts a total pressure equal to the sum of the pressures each gas would contribute independently under the same temperature and volume conditions. This is Dalton’s law of partial pressures.
P_total = P₁ + P₂ + P₃ + …
A partial pressure is the pressure contributed by one gas in a mixture. For an ideal gas mixture, each component contributes according to its amount in moles.
The mole fraction of a gas is the fraction of the total number of gas particles represented by that gas:
X_A = n_A / n_total
For an ideal gas mixture, the partial pressure of gas A is:
P_A = X_A × P_total
Therefore:
P_A = (n_A / n_total) × P_total
Mole fractions add up to 1, and the sum of all partial pressures equals the total pressure.
Worked example: Gas mixture
A container holds 2.0 mol of N₂ and 1.0 mol of O₂ at a total pressure of 900 kPa. Assuming ideal behavior, calculate the partial pressure of O₂.
First, calculate the total amount:
n_total = 2.0 + 1.0 = 3.0 mol
Then calculate the mole fraction of oxygen:
X_O₂ = 1.0 / 3.0 = 0.333
Calculate its partial pressure:
P_O₂ = 0.333 × 900 kPa = 300 kPa
The partial pressure of N₂ is 600 kPa, and the two partial pressures sum to the total pressure of 900 kPa.
Dalton’s law is also useful when gases are collected over water. The measured pressure includes both the gas of interest and water vapor:
P_total = P_gas + P_H₂O
Thus:
P_gas = P_total − P_H₂O
The water vapor pressure must be obtained for the experimental temperature.
Gas Density and Molar Mass
The ideal gas law can connect gas pressure, temperature, density, and molar mass. Density is mass per unit volume:
d = m / V
Molar mass (M) is the mass of one mole of a substance, usually expressed in g/mol. Since the number of moles is mass divided by molar mass:
n = m / M
Substituting this into the ideal gas equation gives:
PV = (m / M)RT
Rearranging and using d = m/V gives the ideal gas density equation:
d = PM / RT
This relationship shows that gas density increases with pressure and molar mass but decreases with temperature, assuming the gas behaves ideally. At the same pressure and temperature, a gas with a greater molar mass has a greater density because each mole has more mass while occupying the same molar volume.
The equation can also be rearranged to calculate molar mass:
M = dRT / P
This is useful when an unknown gas can be weighed and its pressure, temperature, and volume measured.
Worked example: Determining molar mass
An ideal gas has a density of 1.80 g/L at 1.00 atm and 300 K. Determine its molar mass.
Start with:
M = dRT / P
Substitute the known values:
M = (1.80 g/L)(0.08206 L·atm·mol⁻¹·K⁻¹)(300 K) / 1.00 atm
M = 44.3 g/mol
The calculated molar mass is approximately 44.3 g/mol. This value is close to the molar mass of CO₂, approximately 44.01 g/mol, making carbon dioxide a plausible identification if other experimental evidence supports it.
Applying Gas Laws to Chemical Reactions
Gas calculations often appear in stoichiometry problems. A balanced chemical equation gives mole ratios, while the ideal gas law converts between the moles of a gas and its measurable pressure, volume, and temperature.
For example, the decomposition of calcium carbonate can be represented by:
CaCO₃(s) → CaO(s) + CO₂(g)
One mole of CaCO₃ produces one mole of CO₂ when the reaction proceeds completely. If the amount of CO₂ produced is known, its volume under specified conditions can be calculated using V = nRT/P.
Similarly, if a gas volume is measured experimentally, the ideal gas law can determine the number of moles produced. That amount can then be compared with the theoretical yield calculated from the balanced equation to determine reaction yield.
For AP Chemistry, the key connection is that gas laws describe the physical behavior of gaseous substances, while stoichiometry describes the quantitative relationships in chemical reactions. Combining them allows experimental gas measurements to reveal how much reaction occurred or to help identify an unknown gas.