Kinetic Theory · Lesson 3 of 6
Pressure, Temperature and Molecular Speeds
“Derive gas pressure from molecular impacts, connect temperature to translational energy and apply rms-speed reasoning.”
• Derive P = nm₀〈v²〉/3 from elastic wall collisions, including the collision count and isotropy steps. • Distinguish a molecule’s changing velocity, mean speed, mean-square speed and rms speed. • Derive the temperature relation for average translational kinetic energy and use vᵣₘₛ = √(3 RT/M). • Compare molecular energies and speeds in mixtures without confusing per-particle and total energy. • Explain molecular energy changes at moving pistons and apply the limits of the massive-wall approximation.
The idealised molecular model
A gas presses on a wall because molecules strike it and transfer momentum. Although one impact is brief, a macroscopic wall receives an enormous number of impacts, producing a nearly steady average force. The model uses Newtonian mechanics for particles and statistics for their collective effect. Take velocities relative to the gas’s bulk rest frame; organised flow of the gas as a whole is not the thermal random motion being described.
For a dilute ideal gas, molecular dimensions are small compared with their separation, and intermolecular forces are negligible during most of each flight. Particles move almost freely in straight lines between collisions. Collisions are short compared with the time between them and are modelled as elastic. Energy and momentum are conserved for the complete idealised collision system. Molecule–molecule collisions can change an individual molecule’s speed and direction. At a stationary smooth wall, a perfectly elastic reflection reverses the normal velocity component and leaves tangential components unchanged.
One collision gives an impulse to the wall
Choose a planar wall normal to the x direction and a molecule of mass m₀ arriving with velocity (vx, vy, vz), with vx > 0 directed toward the wall. After reflection its velocity is (−vx, vy, vz). The molecule’s x momentum changes from m₀vx to −m₀vx, a change of −2 m₀vx. Momentum conservation gives the wall the opposite impulse, +2 m₀vx. A force is a rate of momentum transfer, so the next task is to count impacts per unit time.
Count the molecules that can reach a wall patch
Consider a wall patch of area A and a short time Δt. Focus on a symmetric group of molecules whose normal speed magnitude is ux = |vx|. Only molecules within a distance uxΔt can reach the patch in that time, so the relevant slab has volume AuxΔt. Let ng be the number density of the whole group, including both signs of vx. In equilibrium half head toward the wall and half away, giving Nhit = ½ngAuxΔt. This explicit ± grouping is why the factor ½ appears.
Angle brackets〈…〉 denote an average over molecules. Different molecules have different velocities, so the pressure must include all symmetric groups. Summing ngm₀ux² over those groups is the total number density n multiplied by the average squared x velocity: P = nm₀〈vx²〉. The average is weighted by how many molecules belong to each group. Replacing the many speeds by the square of their average would not perform the same sum.
The pressure formula has the right units: n is m⁻³, m₀ is kg and 〈v²〉 is m² s⁻², giving kg m⁻¹ s⁻² = Pa. For the same molecular speed distribution, doubling number density doubles pressure. For the same n and m₀, doubling every molecular speed multiplies pressure by four. These checks connect the formula to its momentum-transfer mechanism.
The cubic container used to imagine axes is not essential. A small patch of any smooth wall is locally planar, and the same normal-impact reasoning applies. Collisions within the gas redistribute velocities but do not destroy a steady equilibrium velocity distribution: molecules leave one velocity group while others enter it. The derivation is an equilibrium average, not a demand that one molecule travel ballistically from wall to wall without ever meeting another molecule.
Pressure also exists on an imaginary surface inside the gas, where molecules crossing it transport momentum. It is not created only at the outer wall. With negligible body forces, a uniform equilibrium gas has the same pressure on both sides of a small internal layer, so there is no net acceleration. In a gravitational field there can be a small pressure gradient balancing weight; exact uniformity over all heights is not required.
Mean velocity, mean speed and rms speed are different
Velocity is a vector and can cancel in an average. In a gas at rest, mean velocity is zero by symmetry, even though molecular speeds are large. Speed is the nonnegative magnitude of velocity. Mean speed 〈v〉 averages those magnitudes. Mean-square speed 〈v²〉 averages their squares, and root-mean-square speed vᵣₘₛ is √〈v²〉. The rms value is convenient because kinetic energy is proportional to v².
Problem
Two representative particles have speeds 200 m s⁻¹ and 400 m s⁻¹. Find mean speed and rms speed. Also consider equal particles with opposite velocity vectors of magnitude 300 m s⁻¹.
- 1.Mean speed is (200+400)/2 = 300 m s⁻¹.
- 2.Mean-square speed is (200²+400²)/2 = 100000 m² s⁻², giving rms speed √100000 ≈ 316 m s⁻¹.
- 3.The square of the mean speed is 90000 m² s⁻², which is not the mean-square speed 100000 m² s⁻².
- 4.For velocities +300 and −300 m s⁻¹ along one line, mean velocity is zero but mean speed and rms speed are both 300 m s⁻¹. Zero average velocity does not mean zero thermal motion.
Temperature measures average translational kinetic energy
Multiply the pressure result by volume and use N = nV. Then PV = (1/3) Nm₀〈v²〉. The total translational kinetic energy is Etr = N(½m₀〈v²〉), so PV = (2/3) Etr. Independently, the ideal-gas equation gives PV = NkBT. Equating the two expressions relates a bulk thermodynamic measurement to microscopic average energy.
For a monatomic ideal gas, translational energy supplies the relevant internal energy in this model. A molecule with active rotation or vibration can also store energy in those modes. Ideal-gas assumptions neglect intermolecular interaction energy, not necessarily the internal vibrational potential energy of a molecule’s bonds. Therefore Etr = (3/2) NkBT is universal for classical translational motion, while total U can be larger and will be examined with equipartition.
At one equilibrium temperature, a light molecule and a heavy molecule have the same average translational energy, 3 kBT/2. Their typical speeds differ because ½m₀v² must give the same average energy. The heavier mass requires a smaller mean-square speed. Temperature is neither the speed of one selected molecule nor an amount of total energy independent of sample size.
Problem
Estimate vᵣₘₛ for nitrogen at 300 K with M = 0.028 kg mol⁻¹ and R = 8.314 J mol⁻¹ K⁻¹.
- 1.Use vᵣₘₛ = √(3 RT/M) = √[(3)(8.314)(300)/0.028].
- 2.The mean-square speed is about 2.67×10⁵ m² s⁻², so vᵣₘₛ ≈ 517 m s⁻¹.
- 3.Rounded molecular constants give a nearby value such as 516 m s⁻¹. This is an order-of-hundreds-of-metres-per-second thermal speed, not a wind speed of the whole gas.
- 4.The fact that room air does not cross a room in a straight flight at this speed will be explained by repeated collisions and mean free path.
Mixtures share a temperature, not a common speed
Problem
Argon and chlorine are mixed in the mass ratio 2:1 at 27 °C. Find (i) the ratio of average translational kinetic energies per particle and (ii) the ratio vᵣₘₛ,Ar/vᵣₘₛ,Cl₂. Use masses 39.9 u for Ar and 70.9 u for Cl₂.
- 1.(i) Both species share one temperature, so each has average translational energy 3 kBT/2. The per-particle energy ratio is 1:1.
- 2.(ii) Since vᵣₘₛ ∝ 1/√M, the ratio is √(70.9/39.9) ≈ 1.333, or about 1.33:1.
- 3.The supplied mass proportion does not affect either per-particle result at the same T. It changes particle counts and therefore total energies of each species.
- 4.Chlorine’s rotational or vibrational energy does not alter the universal translational-energy relation. “Average kinetic energy” in this speed calculation means the translational part.
Problem
Estimate the average translational thermal energy of a helium particle at (i)27 °C, (ii)6000 K and (iii)10⁷ K, using kB = 1.38×10⁻²³ J K⁻¹ and 27 °C ≈ 300 K.
- 1.Use the per-particle relation 〈εtr〉 = 3 kBT/2, rather than multiplying by Avogadro’s constant.
- 2.(i) At 300 K, 〈εtr〉 = (1.5)(1.38×10⁻²³)(300) = 6.21×10⁻²¹ J.
- 3.(ii) At 6000 K, 〈εtr〉 = 1.242×10⁻¹⁹ J.
- 4.(iii) At 10⁷ K, 〈εtr〉 = 2.07×10⁻¹⁶ J.
- 5.These are classical translational energy scales. At very high stellar temperatures helium is ionised, so the calculation is not a claim that neutral helium atoms survive unchanged or that their full internal energy is just this translational term.
Problem
At what temperature does an argon particle have the same rms speed as helium at −20 °C? Use MAr = 39.9 g mol⁻¹ and MHe = 4.0 g mol⁻¹.
- 1.Equal √(3 RT/M) values imply TAr/MAr = THe/MHe.
- 2.Convert helium temperature to about 253 K. Then TAr = 253×39.9/4.0 ≈ 2524 K ≈ 2.52×10³ K.
- 3.The common factor 3 R cancels; both molar masses may be in the same units because only their ratio is used.
- 4.The heavier argon needs a much higher absolute temperature to have the same characteristic speed. Multiplying −20 °C by a mass ratio would be meaningless.
Problem
Three equal-capacity vessels contain neon, chlorine and uranium hexafluoride respectively at the same pressure and temperature. Compare molecular counts and rms speeds.
- 1.PV = NkBT gives equal N because each vessel has the same P, V and T under the ideal-gas approximation.
- 2.The rms speed still depends on each species’ molecular mass through √(3 kBT/m₀). Equal counts do not make the masses equal.
- 3.Neon is the lightest of the three, so it has the largest rms speed. Chlorine is slower, and the much heavier polyatomic gas is slower still.
- 4.The translational energy per particle is nevertheless equal for all three. Total internal energy need not be equal because active internal modes differ.
A small isotope mass difference gives a small speed difference
Problem
Compare the characteristic thermal speeds of molecules with masses 235+6×19 and 238+6×19 atomic mass units at the same temperature. Identify which is faster and estimate its percentage speed advantage.
- 1.The total molecular masses are 349 u and 352 u. Include all six fluorine atoms rather than comparing 235 with 238 alone.
- 2.The lighter 349 u molecule is faster. Its rms-speed ratio relative to 352 u is √(352/349) ≈ 1.004289.
- 3.The fractional advantage relative to the heavier species is 1.004289−1 = 0.004289, or about 0.43%.
- 4.The same inverse-square-root mass scaling also governs corresponding characteristic speeds of a classical equilibrium gas distribution. The numerical difference is small because most of the mass is shared between the two molecules.
In comparable molecular-flow conditions, particles that move faster encounter small openings more frequently, so a lighter gas can pass through faster on average. The simple mass comparison gives a rate proportional to 1/√M in that regime. Diffusion through a surrounding gas also involves collisions and concentration gradients, so not every macroscopic spreading time is determined solely by a molecular-mass ratio. No preferred bulk direction is implied for an equilibrium gas before such a boundary or gradient is introduced.
A moving wall can change molecular energy
Elastic reflection preserves a molecule’s speed relative to a massive wall in the wall’s rest frame. If the wall itself moves, that is not the same as preserving speed in the laboratory. A Galilean change of reference frame first subtracts the wall velocity; reflection reverses the relative normal velocity; adding the wall velocity back gives the outgoing laboratory velocity. Work transferred by a moving boundary can therefore change the gas’s internal energy.
Problem
A ball approaches a very massive bat with laboratory velocity−u. The bat moves toward the ball with velocity+V. Explain (a) the rebound speed, (b) gas warming during compression, (c) cooling during expansion with no heat input, and (d) the possible advantage of a heavy bat at a given swing speed.
- 1.(a) In the bat frame, the incoming velocity is−u−V. Elastic reflection gives outgoing relative velocity+(u+V). Adding the bat velocity gives laboratory rebound velocityu+2 V.
- 2.For example, u = 10 m s⁻¹ and V = 5 m s⁻¹ give rebound speed 20 m s⁻¹ in the massive-bat approximation. The ball’s energy increase comes from work by the moving bat.
- 3.(b) A piston moving inward toward approaching molecules similarly tends to increase their kinetic energies. Compression with negligible heat loss raises the gas’s internal energy and temperature after collisions redistribute the energy.
- 4.(c) A receding piston takes energy from particles striking it, so gas doing expansion work without replacement heat loses internal energy and cools. Heat exchange can offset this effect in an isothermal expansion.
- 5.(d) At the same pre-impact swing speed, a heavier bat has greater momentum and is less slowed by the impact, approaching the massive-bat limit more closely. A real player’s ability to swing the bat also matters.
- 6.A finite-mass bat does not imply that the ball’s rebound must be slower than its incoming speed. Recoil changes the result, and the initial velocities and masses together determine it. The u+2 V result specifically assumes negligible change in the bat’s velocity.
Why room air does not simply settle on the floor
Random molecular motion and incessant collisions keep gas particles moving throughout a room. Gravity still has an effect: at equilibrium density is slightly larger at lower height, with a pressure gradient supporting the gas’s weight. Over a few metres, the gravitational energy change m₀gh is tiny compared with the room-temperature translational scale 3 kBT/2. For nitrogen at 300 K over 3 m, the ratio is only about 2×10⁻⁴. Thus room-scale variations are small, though atmospheric height changes over kilometres are significant.
Quick Quiz
Quiz
In a stationary smooth-wall elastic collision, which velocity change is correct?
The wall’s normal impulse reverses vx. Smoothness leaves tangential vy and vz unchanged.
For an isotropic ideal gas, pressure is
The factor 1/3 follows from equal average squared components in three directions.
At the same temperature, argon and chlorine have what ratio of average translational energy per particle?
Average translational energy is 3 kBT/2, independent of molecular mass or mixture proportion.
If molar mass is quadrupled at the same temperature, rms speed becomes
vᵣₘₛ is proportional to 1/√M, so multiplying M by 4 divides speed by 2.
A ball arrives at−10 m s⁻¹ while a massive bat moves at+5 m s⁻¹. In an elastic head-on reflection the outgoing laboratory speed is
In the bat frame the incoming speed is 15 m s⁻¹. Reflection and adding the bat’s+5 m s⁻¹ give 20 m s⁻¹.
Practice Problems
- Derive gas pressure from impacts on a wall patch of area A. Explain the factors 2 in impulse,½ in incoming-particle count and⅓ from isotropy. Define the velocity group whose number density you count.
- Two equal particles have speeds 100 m s⁻¹ and 500 m s⁻¹. Find mean speed, mean-square speed and rms speed. Explain why squaring the mean speed gives a different result.
- Calculate rms speeds of helium and nitrogen at 300 K using molar masses 0.0040 and 0.028 kg mol⁻¹ and R = 8.314 J mol⁻¹ K⁻¹. Compare their translational energies per particle.
- A gas has mass density 1.2 kg m⁻³ and rms speed 500 m s⁻¹. Find its kinetic-theory pressure. State the isotropy and equilibrium assumptions.
- Find the average translational energy of one molecule at 400 K using kB = 1.38×10⁻²³ J K⁻¹. Then find the total translational energy of 2.0 mol using R = 8.314 J mol⁻¹ K⁻¹.
- At what temperature does nitrogen of molar mass 28 g mol⁻¹ have the same rms speed as oxygen of molar mass 32 g mol⁻¹ at 320 K? Show the equality involving T/M.
- A ball with incoming laboratory velocity−12 m s⁻¹ reflects elastically from a massive surface moving at+4 m s⁻¹. Find its outgoing velocity using the surface frame. Explain the corresponding molecular mechanism of piston compression.
- Explain why equal-capacity vessels of neon and chlorine at the same P and T contain equal particle numbers but have different rms speeds. Distinguish total internal energy from total translational energy.
- For a nitrogen molecule of mass 4.65×10⁻²⁶ kg at 300 K, compare m₀gh over a height of 3 m with 3 kBT/2, using g = 9.8 m s⁻² and kB = 1.38×10⁻²³ J K⁻¹. Explain the room-air observation.
Key Takeaways
• Gas pressure is momentum transfer per area per time; isotropy gives P = nm₀〈v²〉/3. • Temperature sets average translational energy 3 kBT/2, not each molecule’s instantaneous speed. • Rms speed is √〈v²〉 = √(3 RT/M); mean velocity, mean speed and rms speed are different averages. • Equal-temperature species have equal per-particle translational energy but different characteristic speeds. • A moving piston transfers work through molecular collisions; total U can include rotation and vibration as well as translation.