Explain pressure through collision momentum

Establish the model and evidence

Begin with the gas as a collection of moving particles, not as a substance containing a material called heat. A particle approaching a rigid wall with normal velocity vx reverses that component in an elastic collision. Its momentum change is −2 m vx; the wall receives the opposite impulse. Tangential motion does not contribute to that normal impulse. Many such events per second produce an average normal force and therefore pressure.

For an isotropic ideal gas, the three velocity directions contribute equally to the mean squared speed. Thus pressure depends on number density times m⟨vx²⟩, while mean translational kinetic energy is 3kT/2. At fixed number density, increasing absolute temperature raises pressure. Replacing particles by heavier ones at the same temperature lowers their characteristic speeds rather than raising their mean translational energy.

The curve is a calculated speed probability density. A narrow speed interval receives probability from the area under the curve, not from its height alone. Individual particles do not all move at the root-mean-square speed. The most probable, mean and rms speeds answer different statistical questions. The analytic distribution is supplied for visualization; deriving it is not part of the required AP course.

Use the model to connect representations: sketch a collision arrow, annotate the particle and wall impulses, draw a broader distribution for a hotter sample, and state which aggregate quantities remain fixed in your comparison. The number density is held constant here. If a container expands while temperature changes, this fixed-density conclusion cannot simply be reused.

A two-dimensional collision makes the distinction concrete: a particle arriving with components (+200,+100) m/s at a stationary wall normal to x leaves with (−200,+100) m/s in the elastic idealization. Its speed is unchanged, its x-momentum changes by −400m in SI units, and its y-momentum is unchanged. The wall receives the opposite impulse. Average pressure adds many normal momentum transfers rather than all velocity components indiscriminately.

Worked evidence

At the same temperature, particles of mass 4 u and 16 u would have rms speeds in the ratio √(16/4)=2. Their mean translational energies are equal. The mass change does not multiply the pressure if number density and temperature stay fixed.

Predict and investigate

Before switching from 4 u to 40 u at 300 K, predict the rms-speed ratio, pressure ratio and mean-energy ratio.

Capture 300 K with masses 4 u and 40 u. Then retain 4 u and compare 300 K with 600 K. Explain the speed, energy and pressure ratios separately; use the distribution table rather than screen height alone.

Transfer to changed conditions

A sample changes from 250 K to 400 K without changing volume or particle count. Predict pressure and rms-speed ratios.

Compare the worked transfer

Pressure ratio is 400/250=1.60; rms-speed ratio is √1.60≈1.265. The ratios differ because kinetic energy depends on speed squared.

Explain the boundary

Derive the sign of the particle impulse and distinguish a selected collision from average pressure. Explain why a hotter gas can still contain some slowly moving particles.

Pressure inside the gas, not only at its wall

Pressure is a local bulk property as well as a wall-force measurement: imagine an internal dividing plane and count momentum transported across it. In an equilibrium isotropic gas, opposite directions have equal average populations, so there is no preferred bulk-flow direction even though the particles move. A selected collision can transfer energy either way; the hotter-to-colder statement is statistical over many collisions, not a claim about every encounter.

Predict, draw and check: two equal-mass particles in a one-dimensional elastic collision approach with velocities +3 m/s and −1 m/s. They exchange velocities, preserving total momentum m(2 m/s) and total kinetic energy m(5 m²/s²). Draw the before/after arrows and contrast the faster particle losing energy with the slower one gaining it. For a small imaginary plane inside the equilibrium gas, turning the plane does not change pressure in the isotropic model. An anisotropic directed beam would not justify that equal-direction pressure claim. These are virtual collision accounts, not apparatus instructions. (Current clauses 9.1.A and 9.3.A.3.)

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. The numerical cases, text and computed representations are original. All browser evidence is synthetic or calculated, not a record of completed practical work. No hazardous apparatus is required.

Clause-level source/action reconciliation checked 31 October 2026 against the linked current framework and Fall 2026 clarification. Added packets are original virtual comparisons; no new practical or learner-completion claim.