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How are the pressure, volume, temperature and amount of a gas tied together in a single law?

Topic 9.3 The Ideal Gas Law: apply PV = nRT (and PV = N k_B T) to relate the state variables of an ideal gas.

A focused answer to AP Physics 2 Topic 9.3, covering the ideal gas law in both molar and molecular forms, the meaning of each state variable, the use of absolute temperature, the special-case proportionalities (Boyle, Charles, Gay-Lussac), and the before-and-after ratio method, with full worked examples.

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Quick answer

The ideal gas law is $PV = nRT$ , where $P$ is pressure, $V$ volume, $n$ the number of moles, $R = 8.31$ J/(mol K) the gas constant and $T$ the absolute (kelvin) temperature. In molecular form it is $PV = N k_B T$ , with $N$ the number of atoms and $k_B$ Boltzmann's constant. The law links all four state variables: fix any two (and the amount) and the others are determined. For a sealed container ( $n$ fixed), it gives the ratio rule $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$ , from which the special cases follow: pressure inversely with volume at fixed temperature (Boyle), volume proportional to temperature at fixed pressure (Charles), and pressure proportional to temperature at fixed volume. Always convert temperatures to kelvin first. The law is the macroscopic summary of the kinetic theory, connecting the motion of atoms to the gas's measurable state.

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What this topic is asking
The ideal gas law
Each variable and the ratio method
Connecting to the kinetic theory
Try this

What this topic is asking

The College Board (Topic 9.3) wants you to apply the ideal gas law, $PV = nRT$ , which ties together the four state variables of a gas: pressure, volume, absolute temperature and amount. You must use it both directly and as a before-and-after ratio, and connect it to the kinetic theory of Topic 9.1.

The ideal gas law

The two forms are the same law counted differently: $nR = N k_B$ , because one mole is $N_A = 6.02 \times 10^{23}$ atoms and $R = N_A k_B$ . Use the molar form when you know moles, and the molecular form when you know the number of atoms. The law is "ideal" because it assumes point atoms with no long-range forces, the same assumptions as the kinetic theory; real gases follow it closely except at very high pressure or very low temperature.

Each variable and the ratio method

The ratio method is the workhorse for "what happens when..." questions: write the combined law before and after, cancel any variable that is held constant, and solve for the unknown. Because temperature appears in the denominator, the absolute scale is essential: a process from $20$ to $40$ degrees Celsius is not a doubling of temperature ( $293$ to $313$ K is only a $7\%$ rise). This single mistake, using Celsius, is the most common error in the topic.

Connecting to the kinetic theory

The ideal gas law is the macroscopic face of the kinetic theory. Combining $PV = N k_B T$ with the kinetic-theory result $\overline{KE} = \tfrac{3}{2}k_B T$ shows that the pressure comes directly from the atoms' motion: more atoms, faster atoms or a smaller volume all raise the pressure, exactly as the collision picture of Topic 9.1 predicts. The strategic payoff is that one law now lets you predict how a gas responds to any change: heat it in a rigid tank and the pressure rises; compress it isothermally and the pressure rises; warm it at constant pressure and it expands. These are precisely the processes drawn on the PV diagrams of the first law (Topic 9.4), where the work done by the gas is the area under the curve. The ideal gas law is the bridge from the microscopic motion of Topic 9.1 to the energy accounting of the rest of the unit.

Heating a gas in a rigid tank

A rigid sealed tank holds gas at $1.0 \times 10^5$ Pa and $250$ K. It is heated to $400$ K. Find the new pressure, then find how many moles it holds if the volume is $0.020$ m cubed. Take $R = 8.31$ J/(mol K).

step 1 Use the constant-volume proportionality

Volume and amount are fixed, so $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$ .

step 2 Solve for the new pressure

$P_2 = P_1 \dfrac{T_2}{T_1} = (1.0 \times 10^5)\dfrac{400}{250} = 1.6 \times 10^5$ Pa.

step 3 Find the moles from the original state

$n = \dfrac{P_1 V}{R T_1} = \dfrac{(1.0 \times 10^5)(0.020)}{(8.31)(250)} = \dfrac{2000}{2078} = 0.96$ mol.

step 4 Check consistency

Using the new state, $n = \dfrac{(1.6 \times 10^5)(0.020)}{(8.31)(400)} = 0.96$ mol, the same value, confirming the gas obeys one consistent equation of state.

Try this

Q1. A gas at constant temperature is compressed from $0.40$ m cubed to $0.10$ m cubed. State the factor by which its pressure changes. [1 point]

Cue. Pressure rises by a factor of $4$ (inverse of the volume change).

Q2. State why temperatures must be in kelvin when using the ideal gas law. [1 point]

Cue. Temperature must be absolute; ratios and proportionalities fail on the Celsius scale (which has an offset zero).

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 2024 (style)6 marksSection II (short FRQ). A sealed container holds

0.50

mol of an ideal gas at a pressure of

1.2 \times 10^5

Pa and a temperature of

300

K. Take

R = 8.31

J/(mol K). (a) Calculate the volume of the gas. (b) The gas is heated at constant volume until its absolute temperature doubles. Calculate the new pressure. (c) State and justify what happens to the average kinetic energy of the atoms during this heating.

Show worked answer →

A 6-point FRQ on the ideal gas law.

(a) Volume (2 points): $V = \dfrac{nRT}{P} = \dfrac{(0.50)(8.31)(300)}{1.2 \times 10^5} = 1.04 \times 10^{-2}$ m cubed.
(b) New pressure (2 points): at constant volume and amount, $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$ . Doubling $T$ doubles $P$ , so $P_2 = 2.4 \times 10^5$ Pa.
(c) Kinetic energy (2 points): average kinetic energy is proportional to absolute temperature, so doubling $T$ doubles the average kinetic energy of the atoms.

Markers reward solving $PV = nRT$ for volume, using the constant-volume proportionality for pressure, and linking temperature to kinetic energy.

AP 2023 (style)1 marksSection I (multiple choice). An ideal gas is compressed to half its volume while its absolute temperature is held constant and no gas escapes. What happens to its pressure? (A) it halves (B) it doubles (C) it is unchanged (D) it quadruples. Justify your reasoning.

Show worked answer →

A 1-point MCQ on Boyle's law (constant temperature). The answer is (B).

At constant temperature and amount, $PV$ is constant, so $P \propto 1/V$ . Halving the volume doubles the pressure. The trap is (A): pressure rises, not falls, when the gas is compressed.

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Sources & how we know this

AP Physics 2: Algebra-Based Course and Exam Description — College Board (2024)