Lesson 9.1 · Unit 9 · CED topic 9.1
Kinetic theory of temperature and pressure
Open a bottle of perfume at one end of a room and, within minutes, someone at the other end smells it. Nothing pushed the scent across the room; the molecules did it themselves, moving at several hundred meters per second and colliding constantly. Physics 1 treated a few objects at a time. A gas has about \(10^{22}\) atoms in a thimble, so we cannot follow them one by one. Instead we ask what the crowd does on average, and two measurable quantities appear: pressure and temperature. This lesson explains both in terms of atomic motion, which is the heart of thermodynamics.
Describe the pressure a gas exerts on its container in terms of atomic motion within that gas.
- Atoms in a gas collide with and exert forces on other atoms in the gas and with the container in which the gas is contained. Collisions can be analyzed using conservation of momentum.
- The pressure exerted by a gas on a surface is the ratio of the sum of the magnitudes of the perpendicular components of the forces exerted by the gas’s atoms on the surface to the area of the surface: \(P = F_\perp/A\).
- Pressure exists throughout the gas itself, not just at the boundary between the gas and the container.
Describe the temperature of a system in terms of the atomic motion within that system.
- The temperature of a system is characterized by the average kinetic energy of the atoms within that system.
- The Maxwell–Boltzmann distribution is a graphical representation of the energies and speeds of atoms at a given temperature.
- The root-mean-square speed corresponding to the average kinetic energy of an ideal gas is related to the temperature by \(K_{avg} = \tfrac{1}{2}m v_{rms}^2 = \tfrac{3}{2}k_BT\).
Pressure from collisions
Pressure is force per unit area, \(P = F_\perp/A\), measured in pascals (1 Pa = 1 N/m\(^2\)). The key word is perpendicular: only the part of the force pressing straight into the surface counts. Where does that force come from in a gas? From atoms bouncing off the walls.
Consider one atom of mass \(m\) hitting a wall head-on at speed \(v\) and rebounding at the same speed (an elastic collision). Take the direction toward the wall as positive. Its momentum changes from \(+mv\) to \(-mv\), so \(\Delta p = -2mv\). By conservation of momentum the wall receives \(+2mv\), the same size and opposite in sign. One bounce is a tiny impulse; an average force appears because there are astronomically many bounces. At room temperature, a wall of area 1 cm\(^2\) is struck by roughly \(10^{23}\) atoms every second. The impacts blur into a steady push: \(F_{avg} = N_{hits}\cdot 2mv/\Delta t\).
Three more ideas follow from this picture, and each is a common exam question.
- More atoms means more pressure. Doubling the number of atoms in the same container doubles the number of hits per second, so \(P\) doubles (at fixed temperature).
- Smaller volume means more pressure. Squeeze the same atoms into half the volume and they reach any wall twice as often, so \(P\) doubles.
- Faster atoms mean more pressure. A faster atom hits harder (larger \(\Delta p\)) and hits more often. Both effects raise the pressure, so raising the temperature raises \(P\) at fixed volume.
The pressure is not special to the walls. Imagine a thin imaginary sheet in the middle of the gas. Atoms cross it from both sides, and the momentum exchanged with the neighbors is the same kind of exchange that happens at a wall. That is why a pressure gauge or a sealed balloon registers the same pressure at any depth inside a gas (ignoring the small difference due to gravity). Pressure exists throughout the gas.
Temperature as average kinetic energy
Temperature is not the same as heat or energy. For a gas of atoms, temperature measures the average translational kinetic energy per atom. For an ideal monatomic gas, \[ K_{avg} = \tfrac{1}{2}m v_{rms}^2 = \tfrac{3}{2}k_BT, \] where \(k_B = 1.38\times10^{-23}\) J/K is Boltzmann’s constant and \(T\) is the absolute (kelvin) temperature. The \(rms\) speed is the square root of the mean of the squares of the speeds, a kind of typical speed that weights fast atoms a little more than slow ones. Solving for it gives \[ v_{rms} = \sqrt{\frac{3k_BT}{m}}. \] Notice what this does and does not say. Two gases at the same temperature have the same average kinetic energy per atom, but the lighter atoms are faster because the same energy is shared by less mass. Doubling the kelvin temperature doubles the average kinetic energy but multiplies the speed by only \(\sqrt{2}\), because the speed depends on \(\sqrt{T}\).
The kelvin scale starts at absolute zero, where the average kinetic energy would be zero. A temperature in kelvin is the Celsius temperature plus 273 (more precisely 273.15). Using Celsius in \(\tfrac{3}{2}k_BT\) is a mistake: 0\(^\circ\)C does not mean zero energy. Doubling the Celsius temperature from 20\(^\circ\)C to 40\(^\circ\)C raises the kelvin temperature only from 293 K to 313 K, a 7% increase.
The Maxwell–Boltzmann distribution
The atoms in a gas do not all move at the same speed. Collisions constantly pass energy around, so at any instant some atoms are slow, most are near the middle, and a few are very fast. The Maxwell–Boltzmann distribution is the graph of how many atoms have each speed (or each energy) at a given temperature. You do not need its equation. You do need to read it.
- Shape. It rises from zero (very few atoms are nearly at rest), reaches a peak, and has a long tail toward high speed.
- Higher temperature. The whole curve shifts to the right, the peak lowers, and the curve broadens. The area under the curve is the same because it counts the same number of atoms.
- Rms speed and peak. The rms speed lies to the right of the peak, because of the tail.
- Heavier atoms. At the same temperature, heavier atoms have a curve shifted to the left (slower) and narrower.
The tail matters: at any temperature a few atoms are much faster than average, which is why a puddle can evaporate well below its boiling point and why chemical reactions speed up when warmed.
A helium atom (\(m = 6.64\times10^{-27}\) kg) strikes a wall head-on at 1300 m/s and rebounds with the same speed. Find the impulse it delivers to the wall, and the average force if \(2.0\times10^{23}\) such collisions happen on 1.0 cm\(^2\) each second.
One atom. \(\Delta p = 2mv = 2(6.64\times10^{-27})(1300) = 1.73\times10^{-23}\) kg·m/s.
Many atoms. The momentum delivered each second is \(2.0\times10^{23}\times1.73\times10^{-23} = 3.5\) kg·m/s per second, so \(F_{avg} = 3.5\) N.
Pressure. \(P = F/A = 3.5\ \text{N}/(1.0\times10^{-4}\ \text{m}^2) = 3.5\times10^{4}\) Pa. Real atoms strike at many angles and speeds, so this is a model, but the order of magnitude shows how a steady pressure arises from impulses.
Find the rms speed of helium (\(m = 6.64\times10^{-27}\) kg) and of nitrogen molecules (\(m = 4.65\times10^{-26}\) kg) at 300 K. Treat nitrogen as a single particle.
Equation. \(v_{rms} = \sqrt{3k_BT/m}\).
Helium. \(v_{rms} = \sqrt{3(1.38\times10^{-23})(300)/6.64\times10^{-27}} = 1370\) m/s.
Nitrogen. \(v_{rms} = \sqrt{3(1.38\times10^{-23})(300)/4.65\times10^{-26}} = 517\) m/s.
Check. The mass ratio is 7.0, so the speed ratio should be \(\sqrt{7.0} = 2.65\): \(1370/517 = 2.65\). ✓ Both gases have the same average kinetic energy, \(\tfrac{3}{2}k_BT = 6.21\times10^{-21}\) J per particle.
A sealed rigid container of gas is heated. Explain why the pressure on the walls rises.
Argument. Heating raises the temperature, and temperature measures the average kinetic energy of the atoms, so the atoms move faster on average. The volume and the number of atoms are unchanged. Each atom therefore hits a wall more often (it crosses the container faster) and with a larger momentum change per hit (\(\Delta p = 2mv\) is larger). Both effects increase the total perpendicular force per unit area, so the pressure increases.
- Pressure. \(P = F_\perp/A\); it comes from collisions, analyzed with momentum and impulse, and it exists throughout the gas.
- Temperature. Proportional to the average kinetic energy of the atoms: \(K_{avg} = \tfrac{3}{2}k_BT\), with \(T\) in kelvin.
- Speed versus temperature. \(v_{rms} \propto \sqrt{T/m}\). Same \(T\), same average kinetic energy; lighter atoms are faster.
- Distribution sketches. Higher \(T\): peak moves right and lower, curve wider, area unchanged. Heavier atoms: curve moves left.
- Common error. Doubling the Celsius temperature does not double the energy or the pressure.
- An atom of mass \(5.0\times10^{-26}\) kg hits a wall head-on at 400 m/s and rebounds elastically. What impulse does the wall receive?
Show answer
\(\Delta p = 2mv = 2(5.0\times10^{-26})(400) = 4.0\times10^{-23}\) kg·m/s, directed into the wall.
- The temperature of a gas rises from 300 K to 1200 K. By what factor does the rms speed of its atoms change?
Show answer
\(v_{rms} \propto \sqrt{T}\), and \(T\) increases by a factor of 4, so the speed increases by a factor of 2.
- Neon (20 u) and argon (40 u) are at the same temperature. Which has the greater average kinetic energy per atom, and which is faster on average?
Show answer
Their average kinetic energies are equal, since \(K_{avg} = \tfrac{3}{2}k_BT\). Neon is lighter, so it is faster; the ratio of rms speeds is \(\sqrt{40/20} = 1.41\).
- Calculate the average kinetic energy of an atom in a gas at 27\(^\circ\)C.
Show answer
\(T = 300\) K. \(K_{avg} = \tfrac{3}{2}(1.38\times10^{-23})(300) = 6.21\times10^{-21}\) J.
- A student says the pressure in a closed jar is highest at the walls and zero in the middle. Correct this.
Show answer
Pressure exists throughout the gas. Atoms in the middle exchange momentum with their neighbors just as atoms at the wall do, so the pressure is (nearly) the same everywhere in the gas. The walls are just where we usually measure it.
- Sketch how the Maxwell–Boltzmann curve for a gas changes when it is cooled. State what happens to the peak position, the peak height, and the area.
Show answer
The peak moves to lower speed, the peak becomes taller and the curve narrower, and the area under the curve stays the same because the number of atoms has not changed.