Ideal Gas Law

AP Chemistry· AP Chemistry CED — Intermolecular Forces and Properties· 14 min read

1. Core Ideal Gas Law ($PV = nRT$)★★☆☆☆⏱ 4 min

The ideal gas law is a combined equation of state that relates the four measurable properties of an ideal gas: pressure ( $P$ ), volume ( $V$ ), temperature ( $T$ ), and amount of gas in moles ( $n$ ). An ideal gas is defined as a gas where intermolecular forces are negligible and gas molecules have no volume relative to their container, an approximation that holds for most AP exam problems.

Unlike individual empirical gas laws that only relate two variables while holding others constant, the ideal gas law relates all four variables at once, making it applicable to nearly all introductory gas problems. It accounts for 5-10% of your total AP Chemistry exam score directly.

📘 Definition

Ideal Gas

A hypothetical gas that follows two core assumptions: 1) gas molecules have negligible volume compared to the container volume, and 2) there are no intermolecular attractive or repulsive forces between molecules. This approximation holds for most gases at low pressure and high temperature.

PV = nRT

Where $R$ is the universal gas constant. For almost all AP Chemistry problems, you will use $R = 0.0821 \frac{L \cdot a t m}{m o l \cdot K}$ , which matches the most common units of pressure (atmospheres) and volume (liters) on the exam. The non-negotiable requirement of the ideal gas law is that temperature must be in Kelvin, the absolute temperature scale. All other units must match the units of $R$ .

📐 Worked Example

A 2.50 L sample of nitrogen gas is held at 32.0 °C and 1.15 atm. How many moles of nitrogen gas are in the sample?

1
Convert temperature from Celsius to Kelvin:
$T = 32.0 + 273.15 = 305.15\ K$
2
List all known values with units matching $R$ : $P = 1.15 a t m$ , $V = 2.50 L$ , $R = 0.0821 \frac{L \cdot a t m}{m o l \cdot K}$ , $T = 305.15 K$
3
Rearrange the ideal gas law to solve for $n$ :
$n = \frac{PV}{RT}$
4
Plug in values and calculate:
$n = \frac{(1.15\ atm)(2.50\ L)}{(0.0821\ \frac{L \cdot atm}{mol \cdot K})(305.15\ K)} = 0.115\ mol$
5
Verify units: All units cancel except moles, which matches the question request.

Exam tip:

Always write down units for every variable and cancel units as you go. If your final units do not match what the question asks for, you know you rearranged the formula incorrectly before you calculate a wrong numerical answer.

2. Derived Relationships: Molar Mass and Gas Density★★★☆☆⏱ 4 min

The ideal gas law can be rearranged to solve for two extremely useful properties for unknown gases: molar mass ( $M$ ) and density ( $d$ ). Recall that moles are defined as $n = \frac{m}{M}$ , where $m$ is mass of the gas in grams, and $M$ is molar mass in g/mol. Substituting into the core ideal gas law gives:

PV = \left(\frac{m}{M}\right)RT

M = \frac{mRT}{PV}

Since density $d = \frac{m}{V}$ , we can substitute this into the equation to get a direct relationship between density, molar mass, pressure, and temperature:

d = \frac{PM}{RT}

These derivations are very common on AP FRQs, where you may be asked to derive the relationship yourself or use it to find the molar mass of an unknown gas from experimental data.

📐 Worked Example

A 0.512 g sample of an unknown volatile liquid is vaporized at 98.0 °C and 0.980 atm. The volume of the vapor is measured as 273 mL. What is the molar mass of the unknown liquid?

1
Convert units to match $R = 0.0821 L \cdot a t m / (m o l \cdot K)$ : $V = 273 m L = 0.273 L$ , $T = 98.0 + 273.15 = 371.15 K$ , $m = 0.512 g$ , $P = 0.980 a t m$
2
Use the rearranged formula for molar mass:
$M = \frac{mRT}{PV}$
3
Plug in values:
$M = \frac{(0.512\ g)(0.0821\ \frac{L \cdot atm}{mol \cdot K})(371.15\ K)}{(0.980\ atm)(0.273\ L)}$
4
Cancel units: atm, L, and K cancel, leaving g/mol (the correct unit for molar mass).
5
Calculate the final result:
$M = 58.4\ g/mol$

3. Dalton's Law of Partial Pressures★★☆☆☆⏱ 3 min

For mixtures of non-reacting ideal gases, the ideal gas law applies to the mixture as a whole and to each individual gas in the mixture. Dalton’s law of partial pressures states that the total pressure of a mixture is equal to the sum of the partial pressures of each individual gas, where the partial pressure of a gas is the pressure it would exert if it occupied the entire container alone.

P_{total} = P_1 + P_2 + P_3 + ... + P_n

Since all gases in the same mixture share the same volume and temperature, partial pressure is proportional to the mole fraction of the gas $χ_{i} = \frac{n _{i}}{n _{t o t a l}}$ , giving the useful relationship:

P_i = \chi_i P_{total}

One of the most common AP exam applications of Dalton’s law is for gases collected over water: the total pressure in the collection vessel is the sum of the pressure of the collected gas and the vapor pressure of water at the experimental temperature:

P_{gas} = P_{total} - P_{water}

📐 Worked Example

Oxygen gas is collected over water at 25 °C. The total pressure in the collection vessel is 1.02 atm, and the vapor pressure of water at 25 °C is 0.0313 atm. If the volume of the vessel is 1.50 L, how many moles of oxygen gas were collected?

1
Calculate the partial pressure of oxygen using Dalton’s law:
$P_{O_2} = P_{total} - P_{H_2O} = 1.02\ atm - 0.0313\ atm = 0.9887\ atm$
2
Convert temperature to Kelvin:
$T = 25 + 273.15 = 298.15\ K$
3
Rearrange the ideal gas law to solve for $n$ :
$n = \frac{P_{O_2}V}{RT}$
4
Plug in values and calculate:
$n = \frac{(0.9887\ atm)(1.50\ L)}{(0.0821\ L·atm/(mol·K))(298.15\ K)} = 0.0606\ mol$

4. AP-Style Practice Problems★★★☆☆⏱ 3 min

📐 Worked Example

A rigid 5.0 L cylinder contains 0.10 mol of helium gas and 0.20 mol of neon gas at 25 °C. What is the partial pressure of helium in the cylinder? Options: A) 0.49 atm, B) 0.12 atm, C) 0.98 atm, D) 1.47 atm

1
Convert temperature to Kelvin:
$25 + 273.15 = 298.15\ K$
2
Use the ideal gas law directly for helium to find partial pressure:
$P_{He} = \frac{n_{He}RT}{V} = \frac{(0.10\ mol)(0.0821\ L·atm/(mol·K))(298.15\ K)}{5.0\ L} ≈ 0.49\ atm$
3
The correct answer is A, which can be verified by multiplying total pressure (1.47 atm) by helium's mole fraction (1/3) to get the same result.

📐 Worked Example

A student performs an experiment to determine the molar mass of an unknown gas. The student measures 0.250 g of the gas in a 250 mL flask at 22 °C and 1.00 atm of pressure. (a) Calculate the molar mass of the unknown gas from the data. (b) If the actual molar mass of the gas is 62 g/mol, calculate the percent error in the student's experiment. (c) State one condition where the gas would deviate significantly from ideal behavior, and explain why.

1
Part (a): Convert units to match $R$ : $V = 0.250 L$ , $T = 22 + 273.15 = 295.15 K$
2
Calculate molar mass:
$M = \frac{mRT}{PV} = \frac{(0.250\ g)(0.0821\ L·atm/(mol·K))(295.15\ K)}{(1.00\ atm)(0.250\ L)} ≈ 24.2\ g/mol$
3
Part (b): Calculate percent error:
$\text{Percent error} = \frac{|\text{experimental} - \text{actual}|}{\text{actual}} \times 100\% = \frac{|24.2 - 62|}{62} \times 100\% ≈ 61\%$
4
Part (c): The gas deviates significantly at high pressure or low temperature. At high pressure, molecules are packed closely, so their own volume is no longer negligible, violating an ideal gas assumption. At low temperature, intermolecular forces become significant, also violating ideal assumptions.

5. Common Pitfalls

Wrong move:

Using temperature in Celsius instead of Kelvin in $P V = n R T$ .

Why:

Most problems give experimental temperatures in Celsius for realism, and students forget the ideal gas law requires absolute temperature.

Correct move:

Always convert temperature to Kelvin as your first step after writing down known values.

Wrong move:

Mismatching units of pressure/volume to the gas constant R.

Why:

Students memorize $R = 0.0821$ but forget it requires liters and atmospheres; they leave volume in mL or pressure in kPa and get the wrong order of magnitude.

Correct move:

After writing down R, explicitly check that your P and V units match R’s units, and convert if necessary before plugging in.

Wrong move:

Forgetting to subtract water vapor pressure when calculating moles of gas collected over water.

Why:

Students only use the total pressure given in the problem, ignoring that water contributes to the total pressure.

Correct move:

If the problem states the gas is collected over water, always subtract the given water vapor pressure from total pressure first.

Wrong move:

Using the old STP molar volume of 22.4 L/mol for problems using the current AP STP definition.

Why:

Many older textbooks teach the pre-1982 IUPAC STP definition; the AP CED uses the current IUPAC definition.

Correct move:

Remember current AP STP is 1 bar (0.9869 atm) and 273.15 K, with a molar volume of 22.7 L/mol; confirm which STP the problem specifies.

Wrong move:

Calculating mole fraction from mass percentages directly without converting to moles first.

Why:

Students confuse mass percent with mole percent and use mass fractions to calculate partial pressure.

Correct move:

Always convert masses of each gas to moles first, then calculate mole fraction from total moles.

Wrong move:

Assuming density is proportional to molar mass regardless of conditions.

Why:

Students memorize $d = P M / R T$ but forget density depends on P and T, so two gases with different molar masses can have the same density at different conditions.

Correct move:

Always use the full $d = P M / R T$ relationship for calculations, don’t rely on proportionality unless P and T are explicitly held constant.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Core Ideal Gas Law	$P V = n R T$	T must be in Kelvin; $R = 0.0821 L \cdot a t m / (m o l \cdot K)$ for most AP problems
Molar Mass from Ideal Gas Law	$M = \frac{m R T}{P V}$	m = mass of gas in grams; use when you know mass, P, V, T
Gas Density Relationship	$d = \frac{P M}{R T}$	d = density in g/L; P units must match R
Dalton's Law of Partial Pressures	$P_{t o t a l} = \sum P_{i}$	Sum of partial pressures equals total pressure
Partial Pressure from Mole Fraction	$P_{i} = χ_{i} P_{t o t a l}$	$χ_{i} = n_{i} / n_{t o t a l}$ ; mole fraction is unitless
AP Current STP	1 bar (0.9869 atm), 273.15 K	Molar volume = 22.7 L/mol at STP
Gas collected over water	$P_{g a s} = P_{t o t a l} - P_{H_{2} O}$	Water vapor pressure is always given in the problem
Mole Fraction Definition	$χ_{i} = \frac{n _{i}}{n _{t o t a l}}$	Sum of all mole fractions in a mixture equals 1

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Calculate partial pressure of gas mixture
2022 · FRQ
Find molar mass of unknown gas

What's Next

Mastering the ideal gas law is a critical prerequisite for the next topics in Unit 3: deviation from ideal gas behavior and kinetic molecular theory (KMT). Without a solid understanding of how to manipulate $P V = n R T$ and solve for unknown gas properties, you will struggle to explain why real gases deviate from ideal behavior and connect KMT molecular predictions to measurable gas properties. The ideal gas law also connects to later topics across the AP Chemistry curriculum: it is used to calculate pressure changes in equilibrium problems, find molar masses of gaseous products in reaction stoichiometry, and relate gas properties to thermodynamics problems involving vaporization. Next, you will build on this foundation to explain gas behavior at the molecular level.