9478 / 2027
Temperature and Ideal Gases overview

Topic 4 of 5

Pressure from particle collisions

A wall experiences pressure because gas particles repeatedly transfer momentum to it. Find the impulse from one collision, divide by the interval between repeated impacts, then sum and average over the gas.

State the ideal-gas assumptions

Consider N particles of the same mass m in a stationary container at equilibrium. The kinetic model assumes:

  • Very many particles obey Newton's laws and move continually and randomly. Motion is isotropic: there is no preferred direction and no bulk drift.
  • The particles' own volume is negligible compared with the gas volume.
  • Forces between particles are negligible except during collisions. Between collisions, each particle travels at constant velocity.
  • Collisions between particles and with stationary walls are elastic. Collision durations are negligible compared with travel times.

Particles need not all have the same speed. Elastic collisions conserve the relevant total kinetic energy, but can change individual velocities. These assumptions describe a model; they are not equally accurate for every real gas state.

Find the wall impulse

Let the box length normal to a selected pair of walls be L, and let each wall's area be A, so V = AL. Take +x towards the right wall. For the incoming particle, let cx > 0 be the magnitude of its normal velocity component. The outgoing x-component is -cx.

A smooth stationary wall reverses the normal component while leaving tangential velocity components unchanged. Using momentum change = final momentum - initial momentum:

Impulse on particle
= (-mcx) - (+mcx)
= -2mcx
Impulse on wall = +2mcx

The wall pushes the particle inward; the particle pushes the wall outward. These are equal and opposite forces on different bodies. The wall impulse is positive in the chosen x direction.

Build the wall force from impulses and their rate

Take +x towards the right wall, with cx the positive incoming normal component. The smooth stationary wall reverses that component while leaving the tangential component unchanged.

One elastic collision and two force recipients

The particle reverses normal velocity and gives the wall an outward impulseThe upper view shows a particle touching the left face of a smooth vertical wall. Positive x points right, normal to the wall. Its incoming velocity vector has drawing components plus 133 horizontally and plus 75 upward; its outgoing vector has minus 133 horizontally and the same plus 75 upward, so speed is unchanged. Dashed component arrows identify the reversed normal component and unchanged positive tangential component. The same circle represents the particle at contact, not two particles. The separately placed lower contact view shows equal-length, opposite horizontal force arrows: force on the particle points left and force on the wall points right. Those forces act on different bodies during the collision. They are not extra velocity vectors or the long-time average force.Smooth stationary wall+x+cx-cx+cy+cyOutgoingIncoming velocityDuring contact: two recipientsOn particleOn wall

Blue solid arrows are velocities; dashed arrows are their components. Purple arrows in the lower view are forces during contact. The particle's momentum change is -2mcx; the wall receives the opposite impulse, +2mcx.

Return to the same wall: two normal-direction legs

The normal projection travels L away and L back between successive right-wall impactsTwo parallel walls have a normal separation L, represented by 240 drawing units. Positive x points right. Immediately after impact one at the right wall, the upper arrow points left across length L with x velocity minus c sub x. The lower arrow returns right across the same length L with x velocity plus c sub x, ending at the same right wall for impact two. The two horizontal lanes are separated only to make the two legs readable; their vertical offset is not physical tangential travel. This is a projection onto the wall-normal coordinate, not an oblique path. Total normal distance is 2L, so the elapsed return interval is 2L divided by the positive component c sub x. It is a flight interval between impacts, not the short duration of either collision.+xOppositewallSamewall12-cx+cxNormal x-projection onlyL

The vertical separation on the page only separates the two legs. Between successive impacts on this wall, Δt = 2L/cx. Averaging its +2mcx impulse over that interval gives mcx2/L, not the force during the brief collision.

The one-third factor is an ensemble average

Varied particle velocities belong to a three-dimensional gasA perspective box contains a small illustrative subset of equal-sized particle symbols, with velocity arrows in varied projected directions and of varied lengths. A separate coordinate triad names x, y and z. This is a schematic view of a three-dimensional ensemble, not a literal particle count or a numerical demonstration that this handful of arrows averages to zero. In the stated isotropic equilibrium model there is no preferred direction and no bulk drift, so the mean squared x, y and z components are equal. That equality applies to ensemble averages; it does not give every particle equal components or the same speed.Varied directions and speedsxyzSchematic three-dimensional ensemble

Isotropy gives mean(cx2) = mean(cy2) = mean(cz2). Since c2 is the sum of the three squared components, each component's mean square is mean(c2)/3. A zero mean signed velocity does not make the mean square zero.

One collision reverses the normal velocity component. The separate round-trip panel is an x-projection with two length-L legs, not an oblique trajectory. The ensemble sketch represents varied velocities; it does not show the literal particle count or equal components for each particle.

Use the same-wall return time

In the simple bouncing-particle construction, the particle crosses to the opposite wall and returns before striking the original wall again. Using the positive normal speed cx:

Time between impacts on the same wall = 2L/cx
Mean force contribution
= impulse / interval
= (2mcx)/(2L/cx)
= mcx2/L

L/cx accounts for only one crossing. The brief collision duration gives the much larger force during contact, not the force averaged over the intervening flight time.

Optional check In the simple bouncing-particle model, a particle of mass m has normal speed c_x between opposite walls separated by L. Which calculation gives its time-averaged force contribution on the same selected wall?
In the simple bouncing-particle model, a particle of mass m has normal speed c_x between opposite walls separated by L. Which calculation gives its time-averaged force contribution on the same selected wall?

Sum the normal contributions

The N particles share mass m but have different velocity components. In the following equations, mean(cx2) means the average of the squared x-components over the particles:

Fwall = (m/L)(cx12 + cx22 + ... + cxN2)
= Nm mean(cx2)/L

Pressure is the time-averaged normal force per unit area:

p = Fwall/A
= Nm mean(cx2)/(AL)
Therefore pV = Nm mean(cx2)

Use three-dimensional isotropy

Isotropic motion gives equal mean-square components along any three perpendicular axes:

mean(cx2) = mean(cy2) = mean(cz2)

For each particle, the square of its speed is the sum of the squared components. Taking the mean gives:

c2 = cx2 + cy2 + cz2
mean(c2) = 3 mean(cx2)
pV = ⅓Nm mean(c2)

The factor of one third applies to the mean-square components, not to every particle's individual components. A random gas can have zero mean velocity while mean(c2) is positive, so it still exerts pressure.

The simple return-time construction follows uninterrupted flights between walls. In a gas with elastic interparticle collisions, velocities are redistributed between particles; averaging the equilibrium ensemble retains the same isotropic pressure result. Do not claim that every actual particle always follows one uninterrupted round trip.