9.2 The Ideal Gas Law

The ideal gas law, $PV = nRT = Nk B T$, connects a gas's pressure, volume, temperature, and amount, as long as you treat the gas with the ideal model assumptions. An ideal gas has tiny particles with random motion, elastic collisions, and no forces between particles except during collisions. These assumptions help you connect equations, graphs, and particle-level explanations in thermodynamics.

Why This Matters for the AP Physics 2 Exam

Thermodynamics is a solid chunk of the AP Physics 2 exam, and the ideal gas law shows up constantly because it ties together almost every gas property you study in Unit 9. You use it on multiple-choice questions to compare gas states and on free-response questions that ask you to predict and justify what happens to a sealed gas when temperature, volume, or pressure changes.

This topic also feeds directly into the Qualitative/Quantitative Translation question, where you describe a scenario in words, derive an equation, and then connect your reasoning back to that equation. Being able to move between graphs, the gas law equation, and plain-language explanations is exactly the kind of skill that question rewards.

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Key Takeaways

• The ideal gas model assumes random particle velocities, negligible particle volume, elastic collisions, and no forces except during collisions.
• One equation runs the show: $PV = nRT = Nk_B T$, where you can use moles with $R$ or particle count with $k_B$.
• Always use absolute temperature in kelvins, never Celsius, when plugging into the gas law.
• On a P-V graph at constant temperature, pressure and volume are inversely related, so the curve bends instead of forming a straight line.
• V-T and P-T graphs at constant pressure or volume are linear and pass through 0 K when extrapolated.
• Extrapolating a pressure vs. temperature graph to zero pressure gives absolute zero, the basis of the Kelvin scale.

Properties of an Ideal Gas

The classical model of an ideal gas is built on a few assumptions that make gas behavior easier to predict.

• Gas particles have random instantaneous velocities and move freely in all directions.
• The volume of individual gas particles is negligibly small compared to the total volume the gas occupies.
• Collisions between gas particles are perfectly elastic, so kinetic energy is conserved.
• Gas particles experience no appreciable forces except during collisions, meaning no intermolecular attractions to worry about.

Heating a gas or doing work on it can change its energy, but those energy transfers are not part of the microscopic assumptions that define an ideal gas. Keep that separate in your head.

These assumptions let you connect the large-scale properties of a gas with one equation:

$PV = nRT = Nk_B T$

Where:

• $P$ = pressure (in pascals, Pa)
• $V$ = volume (in cubic meters, m³)
• $n$ = number of moles
• $R$ = ideal gas constant (8.31 J/mol·K)
• $T$ = temperature (in Kelvin, K)
• $N$ = number of gas particles
• $k_B$ = Boltzmann constant (1.38 × 10⁻²³ J/K)

The two forms are the same relationship written differently. Use $nRT$ when you know moles and $Nk_B T$ when you know the number of particles, since $N = n N_A$.

Graphs of Gas Behavior

Graphs of gas variables let you describe and determine properties of an ideal gas without always solving the full equation. 📈

On a pressure-volume (P-V) graph at fixed temperature, pressure and volume have an inverse relationship, so the graph curves rather than running straight. A high-pressure, low-volume point and a low-pressure, high-volume point can represent the same gas sample at the same temperature.

On a volume-temperature (V-T) graph at constant pressure, volume is directly proportional to absolute temperature in kelvins, so the graph is linear and passes through 0 K when extrapolated.

On a pressure-temperature (P-T) graph at constant volume, pressure is directly proportional to absolute temperature in kelvins, so the graph is also linear. You can use it to find an unknown pressure or temperature from the slope or from interpolation and extrapolation.

When you extrapolate a pressure-temperature graph to where pressure would equal zero, you reach absolute zero (0 K or -273.15°C). 🥶

How to Use This on the AP Physics 2 Exam

Problem Solving

When pressure, volume, or temperature changes for a fixed amount of gas, set up a ratio version of the gas law instead of solving for $R$ every time. Since $n$ and $R$ stay constant, $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$. Cancel whatever is held constant and solve for the unknown. Convert every temperature to kelvins first.

Free Response

For questions that ask you to predict and justify, name which variables are held constant, then explain the direct or inverse relationship that follows from $PV = nRT$. For example, at constant volume, raising temperature raises pressure because the equation makes $P$ proportional to $T$.

Common Trap

Watch the units. Pressure should be in pascals, volume in cubic meters, and temperature in kelvins when using $R = 8.31$ J/mol·K. If you keep pressure in atm or temperature in Celsius, your numbers will be off.

Practice Problem 1: Ideal Gas Law Application

Solution

Use the ideal gas law and recognize that when volume and number of moles stay constant, pressure and temperature are directly proportional.

Starting with the ideal gas law: $PV = nRT$

Since volume and number of moles are constant, you can write: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Rearranging to solve for the new pressure: $P_2 = P_1 \times \frac{T_2}{T_1}$

Substituting the given values: $P_2 = 1.5 \text{ atm} \times \frac{450 \text{ K}}{300 \text{ K}} = 1.5 \text{ atm} \times 1.5 = 2.25 \text{ atm}$

Therefore, the new pressure is 2.25 atm.

Practice Problem 2: Absolute Zero Extrapolation

Solution

This problem requires extrapolating to absolute zero using Charles' Law, which states that volume is directly proportional to temperature when pressure is constant.

According to Charles' Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

First, convert the temperature to Kelvin: $T_1 = 27°C + 273.15 = 300.15 \text{ K}$

You want the temperature at which the volume becomes zero: $V_2 = 0 \text{ L}$

Rearranging the equation: $T_2 = T_1 \times \frac{V_2}{V_1} = 300.15 \text{ K} \times \frac{0 \text{ L}}{2.0 \text{ L}} = 0 \text{ K}$

Therefore, the volume would theoretically become zero at 0 K (or -273.15°C), which is absolute zero. This illustrates why absolute zero is the lowest possible temperature: the volume of a gas cannot be negative.

Common Misconceptions

• Temperature must be in kelvins. Plugging in Celsius breaks the gas law because the relationships are built on absolute temperature. Always convert first.
• Real gases are not perfectly ideal. The model works best at low pressure and high temperature, where particle volume and intermolecular forces stay negligible. Near condensation, real gases deviate.
• A P-V graph at constant temperature is not a straight line. Because $P$ and $V$ are inversely related, the curve bends; assuming it is linear leads to wrong predictions.
• Absolute zero is a theoretical limit, not a normal lab temperature. It is where an ideal gas would extrapolate to zero pressure, not a point you can actually reach.
• The two forms of the equation are not different laws. $PV = nRT$ and $PV = Nk_B T$ describe the same gas; you just switch between moles and particle count using Avogadro's number.
• More particles or higher temperature does not automatically mean higher pressure. It depends on what is held constant. Always check which variables are fixed before predicting the change.

Related AP Physics 2 Guides

• 9.1 Kinetic Theory of Temperature and Pressure
• 9.3 Thermal Energy Transfer and Equilibrium
• 9.6 Entropy and the Second Law of Thermodynamics
• 9.4 The First Law of Thermodynamics
• 9.5 Specific Heat and Thermal Conductivity