Ideal gas law — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: The ideal gas law formula $PV=nRT$, standard temperature and pressure (STP) definitions, molar volume at STP, rearrangements for gas molar mass and density, Dalton’s law of partial pressures, and AP exam-specific problem-solving for ideal gas questions.

You should already know: Basic empirical gas laws (Boyle's, Charles's, Avogadro's). Unit conversion for temperature, pressure, and volume. Molar mass calculation from chemical formulas.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Ideal gas law?

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 gases at low pressure and high temperature, which covers nearly all problems you will see on the AP exam.

Per the College Board AP Chemistry CED, Unit 3 (Intermolecular Forces and Properties) accounts for 18-22% of total exam weight, and ideal gas law questions appear regularly on both multiple-choice (MCQ) and free-response (FRQ) sections, making up 5-10% of your total exam score directly, with additional connections to other topics like stoichiometry, equilibrium, and thermodynamics. 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.

2. Core Ideal Gas Law ($PV=nRT$) and Unit Requirements

The ideal gas law is derived from combining the three empirical gas laws: Boyle’s law ($P\propto 1/V$ at constant $n,T$), Charles’s law ($V\propto T$ at constant $P,n$), and Avogadro’s law ($V\propto n$ at constant $P,T$). Combining these proportionalities gives the core formula: $$PV=nRT$$ Where $R$ is the universal gas constant. For almost all AP Chemistry problems, you will use $R=0.0821\frac{L\cdot atm}{mol\cdot K}$, which matches the most common units of pressure (atmospheres) and volume (liters) on the exam. If you work with energy units later in the course, you will use $R=8.314\frac{J}{mol\cdot K}$, but this is rarely needed for ideal gas law problems.

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 $R$: if you use $R=0.0821$, pressure must be in atmospheres and volume in liters.

Worked Example

Problem: 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: $T=32.0+273.15=305.15K$
2. List all known values with units matching $R$: $P=1.15atm$, $V=2.50L$, $R=0.0821\frac{L\cdot atm}{mol\cdot K}$, $T=305.15K$
3. Rearrange the ideal gas law to solve for $n$: $n=\frac{PV}{RT}$
4. Plug in values and calculate: $n=\frac{(1.15atm)(2.50L)}{(0.0821\frac{L\cdot atm}{mol\cdot K})(305.15K)}=0.115mol$
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.

3. Derived Relationships: Molar Mass and Gas Density

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. Substitute this into the core ideal gas law: $$PV=\left(\frac{m}{M}\right)RT$$ Rearranging to solve for molar mass gives: $$M=\frac{mRT}{PV}$$ Since density $d=\frac{m}{V}$, 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

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

Exam tip: If you forget the derived formula for molar mass or density, just start from $PV=nRT$ and substitute $n=m/M$ step-by-step. You will never lose points for deriving it yourself, and you avoid memorizing the formula incorrectly.

4. Dalton's Law of Partial Pressures

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. Mathematically: $$P_{total}=P_1+P_2+P_3+...+P_n$$ Since all gases in the same mixture share the same volume and temperature, partial pressure is proportional to the mole fraction of the gas ${\chi }_i=\frac{n_i}{n_{total}}$, 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, so $P_{gas}=P_{total}-P_{water}$.

Worked Example

Problem: 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: $P_{O_2}=P_{total}-P_{H_{2}O}=1.02atm-0.0313atm=0.9887atm$
2. Convert temperature to Kelvin: $T=25+273.15=298.15K$
3. Rearrange the ideal gas law to solve for $n$: $n=\frac{P_{O_2}V}{RT}$
4. Plug in values: $n=\frac{(0.9887atm)(1.50L)}{(0.0821L\cdot atm/(mol\cdot K))(298.15K)}=0.0606mol$

Exam tip: When a problem says a gas is collected over water, always look for the given water vapor pressure and subtract it first. The AP exam always provides the water vapor pressure, you never need to memorize it, but you do need to remember to use it.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using temperature in Celsius instead of Kelvin in $PV=nRT$. 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 $d=PM/RT$ 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 $d=PM/RT$ relationship for calculations, don’t rely on proportionality unless P and T are explicitly held constant.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

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? A) 0.49 atm B) 0.12 atm C) 0.98 atm D) 1.47 atm

Worked Solution: Convert temperature to Kelvin: $25+273.15=298.15K$. Use the ideal gas law directly for helium: $P_{He}=\frac{n_{He}RT}{V}=\frac{(0.10mol)(0.0821L\cdot atm/(mol\cdot K))(298.15K)}{5.0L}\approx 0.49atm$. This can also be verified by calculating total pressure (1.47 atm) and multiplying by the mole fraction of helium (0.10/0.30 = 1/3) to get 0.49 atm. The correct answer is A.

Question 2 (Free Response)

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) The student claims that the gas behaves ideally. State one condition under which the gas would deviate significantly from ideal behavior, and explain why the deviation occurs in terms of ideal gas assumptions.

Worked Solution: (a) Convert units: $V=0.250L$, $T=22+273.15=295.15K$. Use $M=\frac{mRT}{PV}$: $$M=\frac{(0.250g)(0.0821L\cdot atm/(mol\cdot K))(295.15K)}{(1.00atm)(0.250L)}\approx 24.2g/mol$$ (b) Percent error: $$\text{Percent error}=\frac{|\text{experimental}-\text{actual}|}{\text{actual}}\times 100\%=\frac{|24.2-62|}{62}\times 100\%\approx 61\%$$ (c) The gas will deviate significantly from ideal behavior at high pressure or low temperature. At high pressure, gas molecules are packed closely together, so the volume of the molecules themselves is no longer negligible compared to the container volume, violating a core assumption of ideal behavior. At low temperature, intermolecular attractive forces become significant, which also violates the ideal gas assumption of no intermolecular interactions.

Question 3 (Application / Real-World Style)

A standard propane ($C_3{H}_8$) barbecue tank holds 12.0 kg of liquid propane. When connected to a barbecue, the propane vaporizes at 30 °C and 1.00 atm. What volume of gaseous propane is available for combustion at these conditions?

Worked Solution: First, calculate the molar mass of propane: $M=(3\times 12.01)+(8\times 1.008)=44.09g/mol$. Convert mass of propane to moles: $n=\frac{12000g}{44.09g/mol}\approx 272.2mol$. Convert temperature to Kelvin: $T=30+273.15=303.15K$. Rearrange the ideal gas law to solve for volume: $$V=\frac{nRT}{P}=\frac{(272.2mol)(0.0821L\cdot atm/(mol\cdot K))(303.15K)}{1.00atm}\approx 6770L$$ In context, this means a full barbecue tank produces approximately 6770 liters (240 cubic feet) of gaseous propane at room temperature and atmospheric pressure, which matches the large volume expansion observed when liquid propane vaporizes.

7. Quick Reference Cheatsheet

8. 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 $PV=nRT$ 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.

• Kinetic molecular theory
• Deviations from ideal gas behavior
• Gas stoichiometry
• Chemical equilibrium