A mixture contains 0.300 mol N_2 and 0.100 mol O_2 at a total pressure of 4.00 atm. Calculate the partial pressure of oxygen. [2 points]

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How does the ideal gas law relate the pressure, volume, temperature and amount of a gas?

Topic 3.4 Ideal Gas Law: use the ideal gas law and its partial-pressure and gas-density forms to relate the pressure, volume, temperature and amount of a gas in calculations.

A focused answer to AP Chemistry Topic 3.4, covering the ideal gas law PV equals nRT, the combined gas law, partial pressures and Dalton's law, mole fractions and gas density, with full worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The ideal gas law
Partial pressures and mole fractions
Gas density and molar mass
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What this topic is asking

The College Board (Topic 3.4) wants you to use the ideal gas law, $PV = nRT$ , and its related forms to relate the pressure, volume, temperature and amount of a gas. You should also handle mixtures of gases with Dalton's law of partial pressures and mole fractions, and derive gas density from the law. Calculations on the exam demand the right units, especially absolute (Kelvin) temperature.

The ideal gas law

The law combines the simpler gas relationships: at fixed $n$ and $T$ , $P$ and $V$ are inversely proportional (Boyle's law); at fixed $n$ and $P$ , $V$ and $T$ are directly proportional (Charles's law). For a fixed amount of gas moving between two states, those combine into the combined gas law:

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

which lets you find a new condition without ever computing $n$ .

Partial pressures and mole fractions

Each gas in a mixture behaves independently, exerting the pressure it would if it occupied the container alone. So adding a second gas to a container raises the total pressure but does not change the partial pressure of the gas already there (provided volume and temperature are unchanged). This independence is a direct consequence of the kinetic model, in which gas particles do not interact.

Gas density and molar mass

Because $n = \dfrac{m}{M}$ , substituting into the ideal gas law gives a useful density form:

PV = \frac{m}{M}RT \quad\Rightarrow\quad \rho = \frac{m}{V} = \frac{PM}{RT}

So a gas of higher molar mass is denser at the same conditions, and you can find an unknown molar mass by measuring the density, pressure and temperature of a gas. This connects the gas laws back to the mole concept of Unit 1.

Using the combined gas law

A gas occupies $4.00$ L at $1.00$ atm and $300.$ K. What volume does it occupy at $2.00$ atm and $600.$ K?

step 1 Write the combined gas law

With a fixed amount of gas, $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$ .

step 2 Solve for the new volume

$V_2 = V_1 \times \dfrac{P_1}{P_2} \times \dfrac{T_2}{T_1} = 4.00 \times \dfrac{1.00}{2.00} \times \dfrac{600.}{300.}$ .

step 3 Calculate

$V_2 = 4.00 \times 0.500 \times 2.00 = 4.00$ L.

step 4 Check the reasoning

Doubling the pressure alone would halve the volume, but doubling the absolute temperature doubles it back, so the two effects cancel and the volume is unchanged.

Try this

Q1. Calculate the number of moles of gas in a $10.0$ L container at $2.00$ atm and $400.$ K. [2 points]

Cue. $n = \dfrac{PV}{RT} = \dfrac{(2.00)(10.0)}{(0.08206)(400.)} = 0.609$ mol.

Q2. A mixture contains $0.300$ mol $\text{N}_2$ and $0.100$ mol $\text{O}_2$ at a total pressure of $4.00$ atm. Calculate the partial pressure of oxygen. [2 points]

Cue. $\chi_{\text{O}_2} = \dfrac{0.100}{0.400} = 0.250$ , so $P_{\text{O}_2} = 0.250 \times 4.00 = 1.00$ atm.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). A

2.00

L rigid vessel contains

0.150

mol of nitrogen gas at

300.

K. (a) Calculate the pressure of the gas in atm. (b) An additional

0.0500

mol of oxygen is added at constant temperature. Calculate the new total pressure. (c) Calculate the mole fraction of oxygen and its partial pressure. (d) Justify whether the partial pressure of nitrogen changes when the oxygen is added.

Show worked answer →

A 4-point quantitative FRQ on the ideal gas law and partial pressures ( $R = 0.08206\ \text{L atm mol}^{-1}\text{K}^{-1}$ ).

(a) Pressure (1 point): $P = \dfrac{nRT}{V} = \dfrac{(0.150)(0.08206)(300.)}{2.00} = 1.85$ atm.
(b) Total (1 point): total moles $= 0.150 + 0.0500 = 0.200$ mol, so $P = \dfrac{(0.200)(0.08206)(300.)}{2.00} = 2.46$ atm.
(c) Oxygen (1 point): $\chi_{\text{O}_2} = \dfrac{0.0500}{0.200} = 0.250$ , so $P_{\text{O}_2} = 0.250 \times 2.46 = 0.616$ atm.
(d) Justify (1 point): the partial pressure of nitrogen does not change, because $P_{\text{N}_2}$ depends only on the moles of $\text{N}_2$ , the volume and the temperature, all unchanged; each gas exerts pressure independently.

Markers reward correct use of $PV = nRT$ , summing moles for total pressure, a correct mole fraction and partial pressure, and the reasoning that each gas acts independently.

AP 2021 (style)1 marksSection I (multiple choice). A fixed amount of ideal gas in a rigid container has its absolute temperature doubled. The pressure of the gas will (A) halve (B) stay the same (C) double (D) quadruple. Justify your reasoning.

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A 1-point conceptual MCQ. The answer is (C).

In a rigid container, $V$ and $n$ are constant, so from $PV = nRT$ pressure is directly proportional to absolute temperature: $\dfrac{P}{T} = \dfrac{nR}{V} = \text{constant}$ . Doubling the absolute temperature doubles the pressure. The trap is using Celsius; the proportionality holds only for absolute (Kelvin) temperature.

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)