A Nitrogen Molecule's Collisions

solved MeanFreePath

Q. Take the radius of a nitrogen molecule to be roughly 1.0 Å $r = 1.0$ Å. The molecular mass of N2 is 28.0 g/mol $M = 28.0$ g/mol. Estimate the collision frequency 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 collisions per second?

Inputs

radius

r

1 Å

molar mass

M

28 g/mol

pressure

P

2 atm

Outputs

frequency

ν_{c}

How it's solved

Solved with MeanFreePath

Given

$r = 1.0$ Å
$M = 28.0$ g/mol
$P = 2.0$ atm
$T = 290$ K from “temperature 17 °C”

Find $ν_{c}$ .

Working

$r = 1.0$ Å $= 1 \times 10^{−10}$ m
$M = 28.0$ g/mol $= 0.028$ kg/mol
$P = 2.0$ atm $= 202650$ Pa
rms speed of a gas(nitrogen) $v_{rms} = {(\frac{3 \cdot R \cdot T}{M})}^{0.5}$
$v_{rms} = {(\frac{3 \cdot 8.31446262 \cdot 290}{0.028})}^{0.5}$
$v_{rms} = 508.273776$ m/s
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) $l = \frac{R \cdot T}{\sqrt{2} \cdot π \cdot d^{2} \cdot P \cdot N_{A}}$
$l = \frac{8.31446262 \cdot 290}{\sqrt{2} \cdot 3.14159265 \cdot {(2 \times 10^{−10})}^{2} \cdot 202650 \cdot 6.02214076 \times 10^{23}}$
$l = 1.11175684 \times 10^{−7}$ m
collisions each second(nitrogen) $ν_{c} = \frac{v_{rms}}{l}$
$ν_{c} = \frac{508.273776}{1.11175684 \times 10^{−7}}$
$ν_{c} = 4.572 \times 10^{9}$ s⁻¹
$ν_{c} = 4.57180704 \times 10^{9}$ Hz $= 4.57180704 \times 10^{9}$ s⁻¹
$ν_{c} \approx 4.6 \times 10^{9}$ s⁻¹to 2 significant figures

Conclusion

Each collision comes after a run of one mean free path, so a molecule moving at its rms speed collides as often as that speed covers the path.

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_collisions do4  kind :numerical5  describes "Take the radius of a nitrogen molecule to be roughly %{radius}. The molecular mass of N2 is " \6            "%{molar_mass}. %{frequency} of a nitrogen molecule in a cylinder containing nitrogen at %{pressure} and " \7            "temperature 17 °C, in collisions per second?"8  given radius: "1.0 Å"9  given molar_mass: "28.0 g/mol"10  given pressure: "2.0 atm"11  asks frequency: "Estimate the collision frequency"12end

solution.rb

1require_relative "../questions/a_nitrogen_molecule_s_collisions"2require_relative "../science/kinetic_theory/mean_free_path"34solve :a_nitrogen_molecule_s_collisions, using: MeanFreePath do5  objects :nitrogen67  take radius: nitrogen.molecular_radius8  take molar_mass: nitrogen.molar_mass9  take pressure: nitrogen.pressure1011  deduce_from "temperature 17 °C", nitrogen.temperature => "290 K"1213  frequency = nitrogen.collision_frequency14  answer frequency:, in: "s-1", figures: 21516  conclusion "Each collision comes after a run of one mean free path, so a molecule moving at its rms speed " \17             "collides as often as that speed covers the path."18end

playground.rb

1require_relative "../solutions/a_nitrogen_molecule_s_collisions"23play_with :a_nitrogen_molecule_s_collisions do4  title "A Nitrogen Molecule's Collisions"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)