IB Physics B.3 Gas Laws — Notes (SL & HL)

What is pressure?

• \(P\) = pressure (Pa, where 1 Pa = 1 N m⁻²)
• \(F\) = force exerted perpendicular (normal) to the surface (N)
• \(A\) = area of the surface (m²)

This equation is in the data booklet. Pressure is a scalar: a gas at pressure \(P\) pushes outwards with a force \(PA\) on each area \(A\) of its container walls, at right angles to that surface, whichever way the surface faces.

• Atmospheric pressure at sea level is about \(1.01\times10^5\ \text{Pa}\) (101 kPa). The air pushes on every square metre of your body with a force of about 100 kN, balanced by the pressure of the fluids inside you.
• Useful conversions: 1 kPa = 10³ Pa; 1 cm² = 10⁻⁴ m²; 1 cm³ = 10⁻⁶ m³; 1 litre = 10⁻³ m³.

Gas trapped by a piston

Many gas questions use a cylinder closed by a piston that can move without friction. When the piston is at rest, the resultant force on it is zero (A.2). For a vertical cylinder with the piston on top:

If the gas is then heated slowly, the piston rises but the forces on it are unchanged once it settles, so the gas stays at constant pressure. This is the standard way to set up a constant-pressure change.

What is the amount of substance?

• \(n\) = amount of substance (mol)
• \(N\) = number of particles (molecules or atoms) — a pure number, no unit
• \(N_\text{A}\) = Avogadro constant \(=6.02\times10^{23}\ \text{mol}^{-1}\) (data booklet)

The equation is in the data booklet. One mole of any substance contains \(6.02\times10^{23}\) particles. The molar mass \(M\) is the mass of one mole (for example 4.0 g mol⁻¹ for helium, 28 g mol⁻¹ for nitrogen gas). A question will give it to you; then mass \(=nM\), and the mass of one molecule is \(M/N_\text{A}\).

Quick example. A balloon holds 0.25 mol of helium. Number of atoms \(N=nN_\text{A}=0.25\times6.02\times10^{23}=1.5\times10^{23}\). Mass \(=nM=0.25\times4.0=1.0\ \text{g}\).

What is an ideal gas?

The kinetic model pictures the gas as a very large number of molecules in constant random motion. The ideal gas is defined by these assumptions:

• The gas is made of a very large number of identical molecules in random motion, with a range of speeds.
• The molecules are point particles: the total volume of the molecules is negligible compared with the volume of the container.
• There are no intermolecular forces except during collisions, so molecules travel in straight lines at constant velocity between collisions, and the gas has no intermolecular potential energy.
• All collisions, between molecules and with the walls, are elastic: kinetic energy is conserved.
• The duration of a collision is negligible compared with the time between collisions.
• The molecules obey Newton's laws of motion.

No real gas is exactly ideal — real molecules do have size and do attract each other. The model is useful because it is simple, it explains the measured gas laws, and real gases follow it closely under the conditions described in the final section.

What are the three empirical gas laws?

The gas laws were found by experiment (they are empirical) long before kinetic theory explained them. Each holds one variable constant and applies to a fixed amount of gas. Temperatures must be in kelvin: \(T/\text{K}=\theta/^\circ\text{C}+273\) (data booklet).

Constant temperature: Boyle's law

Constant pressure: Charles's law

Plotted against Celsius temperature, the volume of any gas gives a straight line that does not pass through the origin. Extrapolated backwards, the lines for every gas meet the temperature axis at the same point, close to −273 °C. This is absolute zero, 0 K — the temperature at which an ideal gas would have zero volume at constant pressure (or zero pressure at constant volume), and the molecules zero kinetic energy. The kelvin scale starts there, which is why the gas laws only take a simple form in kelvin.

Constant volume: the pressure law

How do the gas laws combine to give PV/T = constant?

This equation is in the data booklet. It follows from the empirical laws by taking the gas from state 1 \((P_1,V_1,T_1)\) to state 2 \((P_2,V_2,T_2)\) in two steps:

1. Constant temperature \(T_1\): change the pressure from \(P_1\) to \(P_2\). The volume becomes \(V'\). Boyle's law: \(P_1V_1=P_2V'\), so \(V'=\dfrac{P_1V_1}{P_2}\).
2. Constant pressure \(P_2\): change the temperature from \(T_1\) to \(T_2\). The volume goes from \(V'\) to \(V_2\). Charles's law: \(\dfrac{V'}{T_1}=\dfrac{V_2}{T_2}\).
3. Substitute for \(V'\): \(\dfrac{P_1V_1}{P_2T_1}=\dfrac{V_2}{T_2}\Rightarrow\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\).

Any two of the three laws are enough; the third then follows. Each single law is just this equation with one variable cancelled.

What is the ideal gas equation?

• \(P\) = pressure (Pa); \(V\) = volume (m³); \(T\) = temperature (K)
• \(n\) = amount of substance (mol); \(R\) = molar gas constant \(=8.31\ \text{J K}^{-1}\text{mol}^{-1}\)
• \(N\) = number of molecules; \(k_\text{B}\) = Boltzmann constant \(=1.38\times10^{-23}\ \text{J K}^{-1}\)

Both forms, and both constants, are in the data booklet. The constant in \(PV/T=\text{constant}\) is proportional to the amount of gas: experiment shows it is \(nR\) for any gas that behaves ideally. Since \(n=N/N_\text{A}\),

The Boltzmann constant is the gas constant per molecule. Notice that \(PV\) has the unit of energy: Pa × m³ = N m⁻² × m³ = N m = J.

• Equal volumes of different ideal gases at the same temperature and pressure contain the same number of molecules — the type of gas does not appear in \(PV=Nk_\text{B}T\).
• Density form (not in the data booklet — derive it from \(PV=nRT\) when needed): \(\rho=\dfrac{nM}{V}=\dfrac{PM}{RT}\). At constant pressure, doubling the kelvin temperature halves the density.

Showing changes of state on a pressure–volume diagram

Every equilibrium state of a fixed amount of gas is a point on a \(P\)–\(V\) diagram, and \(T=PV/nR\) can be read from that point. A change of state is a line:

HL students meet the same diagrams again in B.4 Thermodynamics, where the area under the line gives the work done by the gas.

How do molecular collisions produce gas pressure?

Step 1 — one collision

A molecule of mass \(m\) hits a wall with velocity component \(u_x\) perpendicular to the wall. The collision is elastic, so it rebounds with \(-u_x\); the parallel component is unchanged.

Step 2 — one molecule in a cube

The molecule hits face X once every \(\Delta t=2L/u_x\). By Newton's second law (A.2), the average force on the face is the rate of change of momentum:

Step 3 — all N molecules

Adding the forces from all \(N\) molecules and dividing by the area \(L^2\):

The motion is random, so no direction is special: \(\overline{u_x^2}=\overline{u_y^2}=\overline{u_z^2}\). Since \(v^2=u_x^2+u_y^2+u_z^2\), \(\overline{u_x^2}=\tfrac13\overline{v^2}\). The total mass of gas is \(Nm\), so \(Nm/V=\rho\):

• \(P\) = pressure (Pa); \(\rho\) = density of the gas (kg m⁻³)
• \(\overline{v^2}\) = mean of the squares of the molecular speeds, the mean square speed (m² s⁻²)
• \(v_\text{rms}=\sqrt{\overline{v^2}}\) is the root mean square speed — the measure of average translational speed used with this equation

The equation is in the data booklet (as \(P=\tfrac13\rho v^2\), where \(v^2\) is understood as the mean square speed). The derivation itself is not; you should be able to explain its key steps in words: momentum change per collision, rate of collisions, force = rate of change of momentum, pressure = force ÷ area, and the factor \(\tfrac13\) from three equivalent directions.

The link to temperature

Write \(P=\tfrac13\rho\overline{v^2}\) as \(PV=\tfrac13Nm\overline{v^2}\) and compare it with \(PV=Nk_\text{B}T\):

This is the B.1 result: the kelvin temperature tells you the mean translational kinetic energy per molecule. At the same temperature, molecules of every gas have the same average kinetic energy, so lighter molecules move faster: \(v_\text{rms}=\sqrt{3k_\text{B}T/m}\) (not in the data booklet — it follows from the line above). For nitrogen at 300 K, \(v_\text{rms}\approx520\ \text{m s}^{-1}\).

Explaining the gas laws with molecules

You need a qualitative explanation of each law. Every good answer has the same chain: what happens to the molecules → frequency and/or size of the momentum change at the walls → force on the walls → pressure.

What is the internal energy of an ideal gas?

• \(U\) = internal energy (J); \(N\), \(n\), \(k_\text{B}\), \(R\), \(T\) as before
• Both forms are in the data booklet.

Internal energy is, in general, the total kinetic energy plus intermolecular potential energy of the particles (B.1). For an ideal gas:

• there are no intermolecular forces, so the potential energy is zero;
• a monatomic gas (He, Ne, Ar) has only translational kinetic energy, \(\tfrac32k_\text{B}T\) per atom on average;
• so \(U=N\times\tfrac32k_\text{B}T\). Using \(Nk_\text{B}=nR\) gives the molar form.

Consequences you will be tested on:

• For a fixed amount of ideal gas, \(U\) depends only on temperature — not on pressure or volume. In an isothermal change \(\Delta U=0\).
• \(U\propto T\) in kelvin: doubling \(T\) doubles \(U\). Two samples at the same temperature have internal energies in the ratio of their numbers of particles.
• Combining with \(PV=nRT\): \(U=\tfrac32PV\). The internal energy can be read straight from a point on a \(P\)–\(V\) diagram.

When does a real gas behave like an ideal gas?

A real gas departs from ideal behaviour when it is compressed to high density or cooled towards liquefaction. Intermolecular attractions then pull molecules back as they approach a wall, reducing the pressure, and the molecules' own volume reduces the space they can move in. Ideal-gas extrapolations such as \(V\to0\) at 0 K never happen for a real gas: it liquefies first.

Under everyday conditions (about 10⁵ Pa and 300 K), air, nitrogen, oxygen, helium and similar gases obey \(PV=nRT\) to within a fraction of a per cent — which is why the ideal-gas model is used throughout this course.

Exam tips for B.3

Common misconceptions

How this topic connects

Gas pressure as molecular collisions: Newton's laws inside a container

The pressure of a gas is pure A.2 mechanics applied to molecules. Each collision with a wall changes a molecule's momentum by \(2mu_x\); Newton's second law turns the rate of change of momentum into a force, and Newton's third law says the wall feels an equal and opposite force. Dividing by area gives pressure. Thermodynamic quantities — pressure, temperature, internal energy — are therefore averages of mechanical quantities over enormous numbers of particles, and the result \(P=\tfrac13\rho\overline{v^2}\) is the bridge between them.

Temperature as molecular kinetic energy: the microscopic story behind the gas laws

Comparing the kinetic-theory result \(PV=\tfrac13Nm\overline{v^2}\) with the empirical law \(PV=Nk_\text{B}T\) shows that \(\tfrac12m\overline{v^2}=\tfrac32k_\text{B}T\): temperature measures average molecular kinetic energy. Every gas law then has a molecular reading — heating raises molecular speeds, so at constant volume the pressure rises (pressure law) and at constant pressure the gas must expand (Charles's law); at constant temperature the kinetic energy is unchanged, so only the collision rate per unit area changes with volume (Boyle's law).

The ideal gas and other useful simplifications in physics

The ideal gas ignores molecular size and attractions yet predicts real-gas behaviour well at low density. Physics uses the same strategy everywhere: point masses and frictionless surfaces in mechanics (A.1–A.3), the rigid body (A.4, HL), the black body and the simple energy-balance model of the Earth (B.2), the nuclear model of the atom (E.1) and the model of a star as a sphere of hot gas (E.5). A good model keeps only what matters for the question, and its limits — here, high pressure and low temperature — are as important to state as its predictions.

Key equations

Check your understanding

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