A Nitrogen Molecule's Free Path

solved MeanFreePath

Q. Take the radius of a nitrogen molecule to be roughly 1.0 Å $r = 1.0$ Å. Estimate the mean free path of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm $P = 2.0$ atm and temperature 17 °C $T = 290$ K, in metres?

Inputs

radius

r

1 Å

pressure

P

2 atm

Outputs

path

l

How it's solved

Solved with MeanFreePath

Given

$r = 1.0$ Å
$P = 2.0$ atm
$T = 290$ K from “temperature 17 °C”

Find $l$ .

Working

$r = 1.0$ Å $= 1 \times 10^{−10}$ m
$P = 2.0$ atm $= 202650$ Pa
diameter is twice the radius(nitrogen) $d = 2 \cdot r$
$d = 2 \cdot 1 \times 10^{−10}$
$d = 2 \times 10^{−10}$ m
path between collisions(nitrogen) $l = \frac{R \cdot T}{\sqrt{2} \cdot π \cdot d^{2} \cdot P \cdot N_{A}}$
$l = \frac{8.31446262 \cdot 290}{\sqrt{2} \cdot 3.14159265 \cdot {(2 \times 10^{−10})}^{2} \cdot 202650 \cdot 6.02214076 \times 10^{23}}$
$l = 1.112 \times 10^{−7}$ m
$l \approx 1.1 \times 10^{−7}$ mto 2 significant figures

Conclusion

A molecule sweeps out a cylinder as wide as two molecules and hits whatever lies in it, so the more crowded the gas and the bigger the molecules, the shorter it runs free.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_nitrogen_molecule_s_free_path do4  kind :numerical5  describes "Take the radius of a nitrogen molecule to be roughly %{radius}. %{path} of a nitrogen molecule in a " \6            "cylinder containing nitrogen at %{pressure} and temperature 17 °C, in metres?"7  given radius: "1.0 Å"8  given pressure: "2.0 atm"9  asks path: "Estimate the mean free path"10end

solution.rb

1require_relative "../questions/a_nitrogen_molecule_s_free_path"2require_relative "../science/kinetic_theory/mean_free_path"34solve :a_nitrogen_molecule_s_free_path, using: MeanFreePath do5  objects :nitrogen67  take radius: nitrogen.molecular_radius8  take pressure: nitrogen.pressure910  deduce_from "temperature 17 °C", nitrogen.temperature => "290 K"1112  path = nitrogen.mean_free_path13  answer path:, in: "m", figures: 21415  conclusion "A molecule sweeps out a cylinder as wide as two molecules and hits whatever lies in it, so the more " \16             "crowded the gas and the bigger the molecules, the shorter it runs free."17end

playground.rb

1require_relative "../solutions/a_nitrogen_molecule_s_free_path"23play_with :a_nitrogen_molecule_s_free_path do4  title "A Nitrogen Molecule's Free Path"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	MolecularSize
gas constant	$R$	J mol⁻¹ K⁻¹	—	IdealGas
pressure	$P$	Pa	0 … ∞	IdealGas
volume	$V$	m³	0 … ∞	IdealGas
amount	$n$	mol	—	IdealGas
temperature	$T$	K	0 … 1e+09	IdealGas
molar mass	$M$	kg/mol	—	IdealGas
density	$ρ$	kg/m³	0 … ∞	IdealGas
avogadro constant	$N_{A}$	mol⁻¹	—	Moles
mass	$m$	kg	0 … ∞	Moles
molecules	$N$	—	—	Moles
molecular diameter	$d$	m	0 … ∞	MolecularSize
molecular radius	$r$	m	—	MolecularSize
volume of the molecules	$V_{m}$	m³	—	MolecularSize
fraction filled by molecules	$f_{V}$	—	—	MolecularSize
rms speed	$v_{rms}$	m/s	—	RmsSpeed
average thermal energy	$E_{avg}$	J	—	RmsSpeed
mean free path	$l$	m	—	MeanFreePath
collision frequency	$ν_{c}$	Hz	—	MeanFreePath
pressure of ground	$P_{g}$	Pa	0 … ∞
volume of ground	$V_{g}$	m³	0 … ∞
amount of ground	$n_{g}$	mol	—
temperature of ground	$T_{g}$	K	0 … 1e+09
pressure of nitrogen	$P$	Pa	0 … ∞
volume of nitrogen	$V$	m³	0 … ∞
amount of nitrogen	$n$	mol	—
temperature of nitrogen	$T$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of nitrogen	$ρ$	kg/m³	0 … ∞
molar mass of nitrogen	$M$	kg/mol	—
mass of ground	$m_{g}$	kg	0 … ∞
mass of nitrogen	$m$	kg	0 … ∞
molecules of ground	$N_{g}$	—	—
molecules of nitrogen	$N$	—	—
molecular diameter of ground	$d_{g}$	m	0 … ∞
molecular radius of ground	$r_{g}$	m	—
molecular diameter of nitrogen	$d$	m	0 … ∞
molecular radius of nitrogen	$r$	m	—
volume of the molecules of ground	$V_{m,g}$	m³	—
volume of the molecules of nitrogen	$V_{m}$	m³	—
fraction filled by molecules of ground	$f_{V,g}$	—	—
fraction filled by molecules of nitrogen	$f_{V}$	—	—
rms speed of ground	$v_{rms,g}$	m/s	—
rms speed of nitrogen	$v_{rms}$	m/s	—
average thermal energy of ground	$E_{avg,g}$	J	—
average thermal energy of nitrogen	$E_{avg}$	J	—
mean free path of ground	$l_{g}$	m	—
mean free path of nitrogen	$l$	m	—
collision frequency of ground	$ν_{c,g}$	Hz	—
collision frequency of nitrogen	$ν_{c}$	Hz	—

P_{g} \cdot V_{g} = n_{g} \cdot R \cdot T_{g}

ideal gas law

ρ_{g} = \frac{n_{g} \cdot M_{g}}{V_{g}}

density of a gas

m_{g} = n_{g} \cdot M_{g}

mass of a gas

N_{g} = n_{g} \cdot N_{A}

molecules in a gas

d_{g} = 2 \cdot r_{g}

diameter is twice the radius

V_{m,g} = \frac{N_{g} \cdot π \cdot {d_{g}}^{3}}{6}

molecules as little spheres

f_{V,g} = \frac{V_{m,g}}{V_{g}}

share of the space the molecules fill

n_{g} \neq 0

holds some gas?

v_{rms,g} = {(\frac{3 \cdot R \cdot T_{g}}{M_{g}})}^{0.5}

rms speed of a gas

E_{avg,g} = \frac{3 \cdot R \cdot T_{g}}{2 \cdot N_{A}}

three halves k t for each molecule

l_{g} = \frac{R \cdot T_{g}}{\sqrt{2} \cdot π \cdot {d_{g}}^{2} \cdot P_{g} \cdot N_{A}}

path between collisions

ν_{c,g} = \frac{v_{rms,g}}{l_{g}}

collisions each second

P_{g} \cdot V_{g} = n_{g} \cdot R \cdot T_{g}

ideal gas law(ground)

P \cdot V = n \cdot R \cdot T

ideal gas law(nitrogen)

ρ_{g} = \frac{n_{g} \cdot M_{g}}{V_{g}}

density of a gas(ground)

ρ = \frac{n \cdot M}{V}

density of a gas(nitrogen)

m_{g} = n_{g} \cdot M_{g}

mass of a gas(ground)

m = n \cdot M

mass of a gas(nitrogen)

N_{g} = n_{g} \cdot N_{A}

molecules in a gas(ground)

N = n \cdot N_{A}

molecules in a gas(nitrogen)

d_{g} = 2 \cdot r_{g}

diameter is twice the radius(ground)

d = 2 \cdot r

diameter is twice the radius(nitrogen)

V_{m,g} = \frac{N_{g} \cdot π \cdot {d_{g}}^{3}}{6}

molecules as little spheres(ground)

V_{m} = \frac{N \cdot π \cdot d^{3}}{6}

molecules as little spheres(nitrogen)

f_{V,g} = \frac{V_{m,g}}{V_{g}}

share of the space the molecules fill(ground)

f_{V} = \frac{V_{m}}{V}

share of the space the molecules fill(nitrogen)

v_{rms,g} = {(\frac{3 \cdot R \cdot T_{g}}{M_{g}})}^{0.5}

rms speed of a gas(ground)

v_{rms} = {(\frac{3 \cdot R \cdot T}{M})}^{0.5}

rms speed of a gas(nitrogen)

E_{avg,g} = \frac{3 \cdot R \cdot T_{g}}{2 \cdot N_{A}}

three halves k t for each molecule(ground)

E_{avg} = \frac{3 \cdot R \cdot T}{2 \cdot N_{A}}

three halves k t for each molecule(nitrogen)

l_{g} = \frac{R \cdot T_{g}}{\sqrt{2} \cdot π \cdot {d_{g}}^{2} \cdot P_{g} \cdot N_{A}}

path between collisions(ground)

l = \frac{R \cdot T}{\sqrt{2} \cdot π \cdot d^{2} \cdot P \cdot N_{A}}

path between collisions(nitrogen)

ν_{c,g} = \frac{v_{rms,g}}{l_{g}}

collisions each second(ground)

ν_{c} = \frac{v_{rms}}{l}

collisions each second(nitrogen)

n \neq 0

holds some gas?(nitrogen)