The ideal gas law

Assumptions of the Ideal Gas Model

The ideal gas model simplifies the behavior of real gases by making four key assumptions. These assumptions let us connect pressure, volume, and temperature through a single equation.

The classical model of an ideal gas rests on four assumptions:

1. The instantaneous velocities of the gas atoms are random in both magnitude and direction.
2. The volumes of the individual atoms are negligible compared to the total volume the gas occupies.
3. All collisions between atoms (and between atoms and container walls) are perfectly elastic — kinetic energy is conserved.
4. The only appreciable forces on the atoms are contact forces that act during collisions. Between collisions, atoms exert no forces on each other.

Because the atoms are modeled as tiny, widely spaced particles with no long-range attractions, an ideal gas has no potential energy stored between particles. All the internal energy of an ideal gas is kinetic energy of its atoms.

Real gases approximate ideal behavior at high temperatures and low pressures, where atoms are far apart and moving fast.

MisconceptionStudents sometimes think gas atoms travel at the same speed. In reality, the velocities are randomly distributed: some atoms move fast, some slow, and the distribution changes with temperature.

Exam TipAlways state that velocities are random, not uniform.

The Ideal Gas Equation

Three equations link the macroscopic properties of a gas — pressure, volume, temperature, and amount of gas. Choosing the right form depends on whether the amount is expressed in moles or in number of atoms.

Key Equations

Ideal gas law (mole form):

$$PV=nRT$$

Variables: $P$ = absolute pressure (Pa), $V$ = volume (m³), $n$ = number of moles (mol), $R$ = universal gas constant = $8.314 \text{J/(mol}\cdot\text{K)}$, $T$ = absolute temperature (K)

SI unit: Each side has units of J (Pa · m³ = J)

Rearrangements:

$$P=\frac{nRT}{V}, V=\frac{nRT}{P}, T=\frac{PV}{nR}, n=\frac{PV}{RT}$$

ProportionalityAt constant $n$ and $T$, pressure is inversely proportional to volume. At constant $n$ and $V$, pressure is directly proportional to temperature. At constant $n$ and $P$, volume is directly proportional to temperature.

Reference sheet status: On the reference sheet.

Ideal gas law (particle form):

$$PV=N{k}_{B}T$$

Variables: $N$ = total number of atoms or molecules, ${k}_{B}$ = Boltzmann constant = $1.38\times {10}^{-23} \text{J/K}$

SI unit: Same as above (J).

Rearrangements:

$$P=\frac{N{k}_{B}T}{V}, T=\frac{PV}{N{k}_{B}}, N=\frac{PV}{{k}_{B}T}$$

ProportionalitySame proportionality relationships as the mole form.

Reference sheet status: On the reference sheet.

Connection between the two forms:

$$N=n{N}_{A} \text{and} {k}_{B}=\frac{R}{{N}_{A}}$$

where ${N}_{A}=6.022\times {10}^{23} {\text{mol}}^{-1}$ (Avogadro’s number).

An ideal gas obeys $PV=nRT=N{k}_{B}T$ exactly. This equation relates the four macroscopic state variables: pressure, volume, temperature, and amount of gas. Temperature must always be in kelvin — never Celsius. Pressure must always be absolute, not gauge.

When a fixed amount of gas changes from one state to another, the equation gives a useful comparison form. Because $n$ and $R$ are constant:

$$\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\frac{{P}_{2}{V}_{2}}{{T}_{2}}$$

This lets you solve for any one unknown after a change, as long as two of the three variables are known in both states.

Examiner InsightAP questions often give temperature in °C. You must convert to kelvin before substituting: $T(\text{K})=T(^{\circ}\text{C})+273$.

Exam TipA single unconverted temperature will make the entire calculation wrong.

Worked Example: Finding the Volume of a Gas

Scenario

A balloon contains 0.050 mol of helium at a pressure of 120 kPa and a temperature of 27 °C. Determine the volume of the balloon.

Step-by-step solution

Equation used

$$PV=nRT$$

Rearranging for volume:

$$V=\frac{nRT}{P}$$

Converting given values to SI:

$$T=27+273=300 \text{K}$$

$$P=120 \text{kPa}\times 1000=1.20\times {10}^{5} \text{Pa}$$

Given

$$n=0.050 \text{mol}$$

$$R=8.314 \text{J/(mol}\cdot\text{K)}$$

Substituting:

$$V=\frac{0.050\times 8.314\times 300}{1.20\times {10}^{5}}$$

$$V=\frac{124.71}{1.20\times {10}^{5}}$$

$$V=1.04\times {10}^{-3} {\text{m}}^{3}$$

Interpretation

The balloon has a volume of about $1.04\times {10}^{-3} {\text{m}}^{3}$, which is roughly one litre — a reasonable size for a small helium balloon.

Graphs of Gas Behavior and Absolute Zero

Graphs of pressure, volume, and temperature reveal the proportional relationships inside the ideal gas law. They also let us identify absolute zero without ever reaching it.

At constant amount of gas, three important graph relationships emerge:

Reading these graphs: On any graph, the slope (gradient) represents the rate of change of the $y$-axis quantity with respect to the $x$-axis quantity. For the $P$ vs $T$ graph at constant volume, the slope equals $\frac{nR}{V}$. A steeper line means fewer moles or a smaller volume (or both). The $y$-intercept of a straight-line gas graph through the origin is zero — but only when temperature is in kelvin.

Absolute zero from a pressure–temperature graph: When pressure is plotted against temperature in degrees Celsius at constant volume, the data form a straight line that does not pass through the Celsius origin. Extrapolating that line backward (to lower temperatures) shows it crosses the temperature axis at approximately $-273 ^{\circ}\text{C}$. This is the temperature at which an ideal gas would have zero pressure — no atomic motion, no collisions with walls. This extrapolated temperature defines absolute zero: $0 \text{K}=-273 ^{\circ}\text{C}$.

No real gas reaches absolute zero — the gas would liquefy first. But the extrapolation from the ideal gas model gives a precise, consistent value.

MisconceptionStudents often plot gas graphs with temperature in °C and expect the line to pass through the origin. A $P$ vs $T$ line only passes through the origin when temperature is in kelvin.

Exam TipAlways check the temperature scale on the axis before interpreting intercepts.