Maxwell-Boltzmann distribution of speeds — most probable, mean, rms: Question bank

Reminders you may lean on (built in the parent note):

$f (v) d v$ = ==fraction of molecules with speed between $v$ and $v + d v$ ==; total area $= 1$ , and $f (v) = 0$ for $v < 0$ (speed cannot be negative).
$v_{m p} = \frac{2 R T}{M}$ , $\overset{v}{ˉ} = \frac{8 R T}{π M}$ , $v_{r m s} = \frac{3 R T}{M}$ , fixed ratio $1 : 1.128 : 1.225$ .

True or false — justify

Every molecule in a gas at 300 K moves at

v_{r m s}

False —

v_{r m s}

is just one summary number; molecules are spread across a whole distribution from near-0 to the fast tail, and most sit near

v_{m p}

, not at

v_{r m s}

Raising the temperature increases the area under the

f (v)

curve.

False — the area is always exactly 1 (it counts all molecules); heat only redistributes them, flattening and widening the curve while the peak shifts right.

At any

T > 0

some molecules have essentially zero speed.

True —

f (v) \to 0

v \to 0

but the fraction in the low-speed bins is small-yet-nonzero; the curve touches the origin only exactly at

v = 0

The three speeds always satisfy

v_{m p} < \overset{v}{ˉ} < v_{r m s}

regardless of gas or temperature.

True — they differ only by the fixed constants

2 : 8/ π : 3

, so their order never changes;

T

and

M

scale all three by the same

T / M

Two different gases at the same temperature have the same

v_{r m s}

False — they share the same average kinetic energy

\frac{3}{2} k T

, not the same speed; the heavier gas is slower since

v_{r m s} \propto 1/ M

Two different gases at the same

v_{r m s}

must be at the same temperature.

False — equal

v_{r m s}

means equal

T / M

, so the heavier gas must be hotter to match a lighter one's speed.

Doubling the absolute temperature doubles

v_{r m s}

False —

v_{r m s} \propto T

, so doubling

T

multiplies the speed by

2 \approx 1.41

, not 2.

The distribution

f (v)

is a symmetric bell curve.

False — it is skewed right: speed cannot be negative (hard floor at 0) but has no upper bound, so a long high-speed tail forms.

Spot the error

v_{m p}

is the peak, so it's the average speed."

The peak of a skewed curve is not its mean; because the long right tail has no left counterpart, the mean

\overset{v}{ˉ}

is pulled to the right of the peak

v_{m p}

\overset{v}{ˉ} = 4 R T / π M

, since

2 2 R T / π M = 4 R T / π M

Error:

2 X = 4 X

, so the outer 2 multiplies the inner 2, giving

8 R T / π M

, not

4 R T / π M

"For O₂, use

M = 32

3 R T / M

SI requires kg/mol, so

M = 0.032

; using 32 makes

v_{r m s}

come out

1000 \approx 31.6 \times

too small.

"Just use

v_{r m s}

everywhere since it's close to

\overset{v}{ˉ}

Wrong tool: energy and pressure need

v_{r m s}

(from

\overline{v^{2}}

), but collision frequency, mean free path, and effusion timing need

\overset{v}{ˉ}

— see Mean Free Path and Collision Frequency.

"The factor

4 π v^{2}

f (v)

is just a normalization constant."

No — it is the velocity-shell volume (

4 π v^{2} d v

) that grows with

v

; it is what pushes the peak away from zero, and it stays even after the constant

A

is fixed.

f (v)

is a probability, so

f (v) \leq 1

always."

f (v)

is a probability density (fraction per unit speed), not a probability; it can exceed 1 numerically — only

f (v) d v

and the total area are bounded.

e^{- m v^{2} /2 k T}

alone gives the speed distribution, peaking at

v = 0

The Boltzmann factor (with

m

= single-molecule mass,

k

= Boltzmann constant) alone would peak at 0, but multiplying by the growing

v^{2}

shifts the true peak to

v_{m p} > 0

"Heavier gas ⇒ taller narrower curve ⇒ molecules carry more energy."

The narrower curve means less spread and lower speeds; at the same

T

the average KE

\frac{3}{2} k T

is identical for both gases regardless of mass.

Why questions

Why is there a

v^{2}

factor and not, say,

v^{3}

Speed near

v

lives on a sphere in velocity space; a sphere's surface area (number of velocity directions) scales as

4 π v^{2}

, so exactly two powers of

v

appear.

Why does the exponential use

- m v^{2} /2 k T

and not

- m v / k T

The Boltzmann weight depends on energy, and kinetic energy is

\frac{1}{2} m v^{2}

(with

m

the single-molecule mass); energy — not speed — sets how "affordable" a state is at temperature

T

Why does squaring make

v_{r m s}

the largest of the three?

Squaring over-weights fast molecules (twice the speed counts four times), so the fast tail drags

\overline{v^{2}}

above the plain mean and the peak.

Why is

v_{r m s}

(not

\overset{v}{ˉ}

) the speed tied to temperature?

Because

\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k T

links the mean square speed to

T

;

v_{r m s} = \overline{v^{2}}

inherits that clean link, which underpins Average Kinetic Energy and Temperature.

Why do all three speeds scale as

T / M

Each comes from the same

f (v)

whose only speed scale is

k T / m

(equivalently

R T / M

); the different numerical constants (2,

8/ π

, 3) do not change the

T / M

dependence.

Why does hydrogen escape Earth's atmosphere but nitrogen does not?

At the same

T

v_{r m s} \propto 1/ M

, so light H₂ has a much faster tail exceeding escape velocity, while heavy N₂'s tail rarely does — the same mass effect behind Graham's Law of Effusion.

Why does the curve flatten and widen as

T

rises even though area stays 1?

Higher

T

spreads molecules over a broader range of speeds, so to keep the fixed area of 1 the peak must drop as the base broadens.

Why does effusion rate depend on

\overset{v}{ˉ}

rather than

v_{r m s}

Effusion counts how often molecules cross a hole per second, which scales with the plain average speed

\overset{v}{ˉ}

, not the energy-weighted

v_{r m s}

Edge cases

What is

f (v)

for

v < 0

Exactly zero everywhere — speed is a magnitude and cannot be negative, so the distribution is defined only on

v \geq 0

and the whole area-of-1 lives to the right of the origin.

What is

f (v)

at exactly

v = 0

Exactly zero, because the

v^{2}

factor vanishes there — no molecule sits at a mathematically perfect standstill within the distribution's weighting.

What happens to

f (v)

v \to \infty

It decays to 0, but only after the exponential

e^{- m v^{2} /2 k T}

overpowers the growing

v^{2}

; the tail is arbitrarily long yet its area vanishes.

T \to 0

, where do all three speeds go?

All

\to 0

since each

\propto T

; the curve collapses toward a spike at the origin — a classical picture that real gases abandon (they liquefy/solidify, see Real Gases and van der Waals Equation).

T \to \infty

, what happens to the curve?

All three speeds grow without bound (

\propto T

) and the curve spreads ever wider and flatter while keeping area 1; long before this the ideal-gas picture breaks (molecules dissociate/ionize), so it is a mathematical, not physical, limit.

For a mixture of two gases at the same

T

, do they share one curve?

No — each species has its own

f (v)

with its own peak (heavier one shifted left); they only share the temperature and hence the same average KE.

Does the distribution apply to a single molecule at one instant?

No —

f (v)

describes the ensemble (many molecules, or one molecule sampled over long time); a single molecule at one instant has just one definite speed.

\overline{v^{2}}

the same as

(\overset{v}{ˉ})^{2}

No —

\overline{v^{2}} > (\overset{v}{ˉ})^{2}

always (their gap is the variance), which is exactly why

v_{r m s} > \overset{v}{ˉ}

What does the underlying Boltzmann Distribution reduce to if you ignore the

v^{2}

shell factor?

You'd get the velocity-component distribution (e.g. of

v_{x}

), which is a symmetric Gaussian peaked at 0 — the skew appears only when you go from a component to the full speed.

Recall Quick self-test

Order of the three speeds, smallest first? ::: $v_{m p} < \overset{v}{ˉ} < v_{r m s}$ , always. Density vs probability — which can exceed 1? ::: The density $f (v)$ can exceed 1; the fraction $f (v) d v$ and total area cannot. Same $T$ for two gases guarantees same what? ::: Same average kinetic energy $\frac{3}{2} k T$ .