Lesson 9.2 · Unit 9 · CED topic 9.2

The ideal gas law

A bicycle tire, a helium balloon and the air in a classroom all obey nearly the same equation, even though they hold different gases at different pressures. That is a remarkable fact, and it is possible because, at ordinary conditions, the atoms of a gas are so far apart that each one hardly notices the others. Physicists capture this with a simple model, the ideal gas, and one equation that links pressure, volume, amount and temperature. This lesson states the model, introduces the mole, and practices using the law to find a missing variable or a factor of change.

The ideal gas model

A model is a simplification chosen so that we can calculate. The ideal gas model makes four assumptions, which together say that the atoms behave like tiny, perfectly elastic billiard balls that interact only when they touch.

• Random motion. At any instant the velocities of the atoms point in random directions with no preferred direction.
• Negligible size. The atoms take up almost none of the container’s volume; nearly all of the volume is empty space between them.
• Elastic collisions. No kinetic energy is lost when atoms collide with each other or with the walls.
• No forces between collisions. The atoms do not attract or repel one another except at the instants of impact. (Gravity on a single atom is negligible.)

Real gases fit the model best when they are hot (atoms are fast and the weak attractions do not matter) and dilute (low pressure, so the atoms are far apart). Helium and other monatomic gases at room temperature and atmospheric pressure fit very well. On the AP exam, unless a problem says otherwise, ideal gases are monatomic.

The law

The pressure, volume, temperature and amount of an ideal gas are related by \[ PV = nRT = Nk_BT. \] Each symbol needs care.

• \(P\) is the absolute pressure, in pascals. A tire gauge reads gauge pressure, the amount above atmospheric pressure (1 atm = \(1.01\times10^5\) Pa), so add atmospheric pressure before using the law.
• \(V\) is the volume of the container, in cubic meters (1 L = \(10^{-3}\) m\(^3\)).
• \(n\) is the amount in moles. One mole contains Avogadro’s number of particles, \(N_A = 6.02\times10^{23}\), so \(N = nN_A\).
• \(T\) is the absolute temperature in kelvin.
• \(R = 8.31\) J/(mol·K) is the universal gas constant and \(k_B = 1.38\times10^{-23}\) J/K is Boltzmann’s constant. They are related by \(R = N_Ak_B\), which is why the two forms of the law are the same equation: \(nRT = nN_Ak_BT = Nk_BT\).

Use the \(n\) form when the problem gives moles and the \(N\) form when it counts atoms or molecules. Using the form \(PV = Nk_BT\) you can see the physical content: pressure is proportional to the number of atoms per unit volume and to the temperature. The law also holds for mixtures, and the same law applies to each gas in the mixture independently.

Factors of change

Many exam questions ask what happens to one variable when another changes. The approach is always the same: write the law, decide which quantities are held fixed, and let the others vary.

• Fixed \(n\) and \(T\): \(PV = \) constant. Halving \(V\) doubles \(P\).
• Fixed \(n\) and \(V\): \(P \propto T\). Doubling the kelvin temperature doubles \(P\).
• Fixed \(n\) and \(P\): \(V \propto T\). Doubling the kelvin temperature doubles \(V\).
• Fixed \(P\) and \(T\): \(V \propto n\). Adding gas expands a balloon in proportion.

For a fixed amount of gas that changes from one state to another, divide the law for the final state by the law for the initial state. This gives the combined gas law, \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}, \] in which any unit may be used for \(P\) and \(V\) (it cancels) but \(T\) must be in kelvin.