# Ideal Gas Law

> AP Chemistry · Unit 3: Intermolecular Forces and Properties
> Source: https://www.owlsprep.com/study/ap-chemistry-u3-ideal-gas-law/

Covers the core ideal gas law equation $PV = nRT$, standard temperature and pressure (STP), rearrangements for molar mass and density, Dalton's law of partial pressures, and AP exam problem-solving strategies.

**Prerequisites:** Basic empirical gas laws (Boyle's, Charles's, Avogadro's); Unit conversion for temperature, pressure, and volume; Molar mass calculation from chemical formulas

## Learning objectives

- Relate pressure, volume, temperature and moles of gas using the ideal gas law
- Rearrange the ideal gas law to calculate molar mass and density of unknown gases
- Apply Dalton's law of partial pressures to gas mixtures
- Avoid common exam traps in ideal gas law problems

## Core Ideal Gas Law ($PV = nRT$)

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.

**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 $R = 0.0821 \frac{L \cdot atm}{mol \cdot K}$, 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 $R$: $P = 1.15\ atm$, $V = 2.50\ L$, $R = 0.0821\ \frac{L \cdot atm}{mol \cdot K}$, $T = 305.15\ K$
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.

## 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. 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.

> **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.

**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 $R = 0.0821\ L·atm/(mol·K)$: $V = 273\ mL = 0.273\ L$, $T = 98.0 + 273.15 = 371.15\ K$, $m = 0.512\ g$, $P = 0.980\ atm$
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:

   $$M = 58.4\ g/mol$$

## 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.

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

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:

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

> **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.

**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:

   $$n = \frac{(0.9887\ atm)(1.50\ L)}{(0.0821\ L·atm/(mol·K))(298.15\ K)} = 0.0606\ mol$$

## AP-Style Practice Problems

**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.

## Common pitfalls

- **Wrong:** Using temperature in Celsius instead of Kelvin in $PV = nRT$.
  - Why it fails: Most problems give experimental temperatures in Celsius for realism, and students forget the ideal gas law requires absolute temperature.
  - Correct: Always convert temperature to Kelvin as your first step after writing down known values.
- **Wrong:** Mismatching units of pressure/volume to the gas constant R.
  - Why it fails: 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: After writing down R, explicitly check that your P and V units match R’s units, and convert if necessary before plugging in.
- **Wrong:** Forgetting to subtract water vapor pressure when calculating moles of gas collected over water.
  - Why it fails: Students only use the total pressure given in the problem, ignoring that water contributes to the total pressure.
  - Correct: If the problem states the gas is collected over water, always subtract the given water vapor pressure from total pressure first.
- **Wrong:** Using the old STP molar volume of 22.4 L/mol for problems using the current AP STP definition.
  - Why it fails: Many older textbooks teach the pre-1982 IUPAC STP definition; the AP CED uses the current IUPAC definition.
  - Correct: 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:** Calculating mole fraction from mass percentages directly without converting to moles first.
  - Why it fails: Students confuse mass percent with mole percent and use mass fractions to calculate partial pressure.
  - Correct: Always convert masses of each gas to moles first, then calculate mole fraction from total moles.
- **Wrong:** Assuming density is proportional to molar mass regardless of conditions.
  - Why it fails: 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: Always use the full $d = PM/RT$ relationship for calculations, don’t rely on proportionality unless P and T are explicitly held constant.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Core Ideal Gas Law | $PV = nRT$ | T must be in Kelvin; $R = 0.0821\ L·atm/(mol·K)$ for most AP problems |
| Molar Mass from Ideal Gas Law | $M = \frac{mRT}{PV}$ | m = mass of gas in grams; use when you know mass, P, V, T |
| Gas Density Relationship | $d = \frac{PM}{RT}$ | d = density in g/L; P units must match R |
| Dalton's Law of Partial Pressures | $P_{total} = \sum P_i$ | Sum of partial pressures equals total pressure |
| Partial Pressure from Mole Fraction | $P_i = \chi_i P_{total}$ | $\chi_i = n_i / n_{total}$; 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 | $P_{gas} = P_{total} - P_{H_2O}$ | Water vapor pressure is always given in the problem |
| Mole Fraction Definition | $\chi_i = \frac{n_i}{n_{total}}$ | Sum of all mole fractions in a mixture equals 1 |

## 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](https://www.owlsprep.com/study/ap-chemistry-u3-kinetic-molecular-theory/)
- [Deviation from Ideal Gas Law](https://www.owlsprep.com/study/ap-chemistry-u3-deviation-from-ideal-gas-law/)
- [Mixtures and solutions on the particulate scale](https://www.owlsprep.com/study/ap-chemistry-u3-mixtures-and-solutions-on-the/)

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