1.7.11 · D5Thermodynamics
Question bank — Mean free path, mean speed, RMS speed — derivations
True or false — justify
Each statement is either subtly wrong or right-for-a-non-obvious-reason. Say which, and why.
At the same temperature, a heavier gas has a lower RMS speed than a lighter gas
True — depends on mass, and same means same energy , so a bigger must be paired with a smaller .
At the same temperature, every gas molecule has the same average kinetic energy regardless of mass
True — this is equipartition: has no on the right, so KE depends only on , not on which molecule it is.
and are equal because both are "the average speed"
False — averages , while averages then square-roots; squaring over-weights fast molecules, so for any real spread of speeds.
If you doubled every molecule's speed, and would both double
True — both are homogeneous of degree one in the speeds ( linear, a root of a quadratic average), so scaling all speeds by 2 scales both by exactly 2.
Doubling the absolute temperature doubles
False — , so doubling multiplies by ; you must quadruple to double the speed.
The mean free path depends on how fast the molecules move
False — contains no speed; speed cancels because both the distance flown and the number of collisions grow proportionally with speed.
At fixed volume, heating a gas increases its mean free path
False — at fixed the number density is unchanged, and has no , so stays the same; only the collision rate rises (via ).
At fixed pressure, heating a gas increases its mean free path
True — at fixed , ; raising spreads molecules apart (lower ), so each flies further between hits.
The in comes from the geometry of the collision cylinder
False — the cylinder gives ; the comes from the relative speed between two moving molecules averaging , which raises the collision rate.
Collision cross-section is , not , because molecules are large
False — it is because a collision happens when centres come within one diameter (the sum of two radii), so the effective target radius is , not the molecular radius.
Spot the error
Each item has a plausible-sounding claim or step. Name the flaw.
"Since g/mol for N₂, ."
Wrong units — is in SI, so must be in kg/mol (), not ; using grams makes the speed too large.
"Pressure is because molecules move in the -direction."
The factor is wrong — the correct relation uses the full speed: , and enters only through isotropy of the three directions.
"The most probable speed is the average speed, since it's where most molecules are."
The peak of is the mode, not the mean; the distribution has a long high-speed tail, so (the tail drags the average above the peak).
"Since KE , the molecule with the mean speed carries the mean energy."
No — mean energy involves , but ; the energy-carrying speed is , which is strictly larger than .
"To get , divide total distance by collisions: , giving — done."
Incomplete — this treats targets as stationary; accounting for their motion replaces by relative speed in the collision count, adding the missing .
" is odd, so it vanishes."
False — over nothing is forced to vanish by parity; only symmetric integrals over kill odd integrands, and is even anyway. The integral is .
"Because has no pressure in it, RMS speed is independent of everything but and gas type."
Correct as stated for , but the trap is thinking the same of — does depend on (via ), even though does not.
Why questions
Answer with the mechanism, not the formula.
Why is always, for any spread of speeds?
Squaring before averaging gives fast molecules extra weight (pushing up), while the long fast tail pulls above the peak ; equality would require every molecule to have identical speed.
Why does the speed cancel out of the mean free path?
A faster molecule flies further and hits targets more often in the same proportion, so distance-per-collision — the ratio — is unaffected by speed.
Why do we use (not ) when computing collision rates and ?
Collision frequency counts encounters per second, governed by the ordinary average travel rate ; energy per collision is a separate question handled by .
Why is ?
No direction is special (isotropy), so ; since is their sum, each equals one third of it.
Why does the momentum change at a wall equal , not ?
The wall reverses the -velocity from to ; momentum is a vector, so the change in magnitude is the difference .
Why is the collision cross-section built from diameter rather than radius?
Two spheres touch when their centres are separated by the sum of radii ; the moving molecule's centre must miss every target centre by more than , so the "danger disc" has radius .
Why does the relative speed of two molecules average ?
Relative velocity is the vector difference of two independent velocities of equal mean speed; the mean square of a difference of independent equal-variance vectors is twice one variance, so the typical relative speed scales as .
Why does squeezing a gas (raising at fixed ) shorten the mean free path?
Higher at fixed means higher number density ; more targets packed in means a molecule collides sooner, so shrinks.
Why does let us connect the invisible molecular motion to measurable pressure?
Pressure comes from momentum delivered by molecular impacts, which depends on ; matching to (see Ideal gas law) extracts , hence , from macroscopic data.
Edge cases
Boundary and degenerate scenarios — the ones that break a memorised formula.
What happens to as pressure approaches zero (near-vacuum)?
as — molecules almost never meet, and eventually exceeds container size so the "collide with each other" picture breaks and wall collisions dominate.
What happens to as ?
; classically all motion ceases, though real matter would have liquefied/solidified and quantum zero-point motion (outside this model) would remain.
If the gas is a single molecule in a box, is "" or "" even meaningful?
Not as ensemble averages — the whole Maxwell–Boltzmann picture needs a large ; with one molecule these become that molecule's own speed and its square-root-of-square, which coincide, so the ordering collapses.
For a hypothetical gas where every molecule had exactly the same speed, what is ?
Exactly 1 — with zero spread, , so the strict inequality becomes equality; the gap between them is a pure measure of the speed spread.
When becomes comparable to the container size, does the mean-free-path formula still apply?
No — the derivation assumes many molecule–molecule collisions before hitting a wall; once box size, transport becomes ballistic (effusion-like) and the collisional formula no longer describes the flights.
At extremely high density (liquid-like packing), does still hold?
No — it assumes molecules are point-sized targets in mostly empty space; when packing is tight, molecules overlap in "reach," multiple simultaneous encounters occur, and the dilute-gas assumption fails.
If two directions somehow had different mean-square velocities (anisotropic gas), does survive?
No — that identity rests entirely on isotropy; break the symmetry (e.g. by a directional flow or field) and each direction must be handled separately.
Recall One-line summary of the traps
Mass lowers speed but not energy; from the spread; ignores speed but not density; the is relative motion, and the cross-section uses diameter, not radius.