Uranium Hexafluoride Isotopes

solved RmsSpeed

Q. Uranium has two isotopes of masses 235 g/mol $M_{U235} = 235$ g/mol and 238 g/mol $M_{U238} = 238$ g/mol. Both are present in uranium hexafluoride $M_{m235} = M_{U235} + 6 \cdot M_{f}$ , $M_{m238} = M_{U238} + 6 \cdot M_{f}$ gas. If the atomic mass of fluorine is 19 g/mol $M_{f} = 19$ g/mol, by what factor is the lighter molecule faster than the heavier at any temperature $T_{m235} = T_{m238}$ ?

Inputs

light

M_{U235}

235 g/mol

heavy

M_{U238}

238 g/mol

fluorine

M_{f}

19 g/mol

Outputs

ratio?

How it's solved

Solved with RmsSpeed

Given

$M_{U235} = 235$ g/mol
$M_{U238} = 238$ g/mol
$M_{f} = 19$ g/mol
$M_{m235} = M_{U235} + 6 \cdot M_{f}$ from “uranium hexafluoride”
$M_{m238} = M_{U238} + 6 \cdot M_{f}$ from “uranium hexafluoride”
$T_{m235} = T_{m238}$ from “at any temperature”

Find $\frac{v_{rms,m235}}{v_{rms,m238}}$ .

Working

rms speed of a gas(heavier) $v_{rms,m238} = {(\frac{3 \cdot R \cdot T_{m238}}{M_{m238}})}^{0.5}$
$v_{rms,m238} = {(\frac{24.9433879 \cdot T_{m238}}{0.352})}^{0.5}$
rms speed of a gas(lighter) $v_{rms,m235} = {(\frac{3 \cdot R \cdot T_{m235}}{M_{m235}})}^{0.5}$
$v_{rms,m235} = {(\frac{24.9433879 \cdot T_{m238}}{0.349})}^{0.5}$
$\frac{v_{rms,m235}}{v_{rms,m238}} = 1.00429$
$M_{U235} = 235$ g/mol $= 0.235$ kg/mol
$M_{U238} = 238$ g/mol $= 0.238$ kg/mol
$M_{f} = 19$ g/mol $= 0.019$ kg/mol

Conclusion

The two molecules differ by three parts in three hundred and fifty, so their speeds differ by half that; slight, but enough to separate the isotopes by diffusion.

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 :uranium_hexafluoride_isotopes do4  kind :numerical5  describes "Uranium has two isotopes of masses %{light} and %{heavy}. Both are present in uranium hexafluoride gas. If " \6            "the atomic mass of fluorine is %{fluorine}, %{ratio} at any temperature?"7  given light: "235 g/mol"8  given heavy: "238 g/mol"9  given fluorine: "19 g/mol"10  asks ratio: "by what factor is the lighter molecule faster than the heavier"11end

solution.rb

1require_relative "../questions/uranium_hexafluoride_isotopes"2require_relative "../science/kinetic_theory/rms_speed"34solve :uranium_hexafluoride_isotopes, using: RmsSpeed do5  objects :fluorine, lighter: :m235, heavier: :m238, light_uranium: :U235, heavy_uranium: :U23867  take light: light_uranium.molar_mass8  take heavy: heavy_uranium.molar_mass9  take fluorine: fluorine.molar_mass1011  deduce_from "uranium hexafluoride", lighter.molar_mass => light_uranium.molar_mass + 6 * fluorine.molar_mass,12                                      heavier.molar_mass => heavy_uranium.molar_mass + 6 * fluorine.molar_mass13  deduce_from "at any temperature", lighter.temperature => heavier.temperature1415  answer find_relation(lighter.rms_speed / heavier.rms_speed), figures: 41617  conclusion "The two molecules differ by three parts in three hundred and fifty, so their speeds differ by " \18             "half that; slight, but enough to separate the isotopes by diffusion."19end

playground.rb

1require_relative "../solutions/uranium_hexafluoride_isotopes"23play_with :uranium_hexafluoride_isotopes do4  title "Uranium Hexafluoride Isotopes"5end

The laws it drew on

quantity	written	units	within	from
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	—	RmsSpeed
rms speed	$v_{rms}$	m/s	—	RmsSpeed
average thermal energy	$E_{avg}$	J	—	RmsSpeed
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 fluorine	$P_{f}$	Pa	0 … ∞
volume of fluorine	$V_{f}$	m³	0 … ∞
amount of fluorine	$n_{f}$	mol	—
temperature of fluorine	$T_{f}$	K	0 … 1e+09
pressure of lighter	$P_{m235}$	Pa	0 … ∞
volume of lighter	$V_{m235}$	m³	0 … ∞
amount of lighter	$n_{m235}$	mol	—
temperature of lighter	$T_{m235}$	K	0 … 1e+09
pressure of heavier	$P_{m238}$	Pa	0 … ∞
volume of heavier	$V_{m238}$	m³	0 … ∞
amount of heavier	$n_{m238}$	mol	—
temperature of heavier	$T_{m238}$	K	0 … 1e+09
pressure of light uranium	$P_{U235}$	Pa	0 … ∞
volume of light uranium	$V_{U235}$	m³	0 … ∞
amount of light uranium	$n_{U235}$	mol	—
temperature of light uranium	$T_{U235}$	K	0 … 1e+09
pressure of heavy uranium	$P_{U238}$	Pa	0 … ∞
volume of heavy uranium	$V_{U238}$	m³	0 … ∞
amount of heavy uranium	$n_{U238}$	mol	—
temperature of heavy uranium	$T_{U238}$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of fluorine	$ρ_{f}$	kg/m³	0 … ∞
molar mass of fluorine	$M_{f}$	kg/mol	—
density of lighter	$ρ_{m235}$	kg/m³	0 … ∞
molar mass of lighter	$M_{m235}$	kg/mol	—
density of heavier	$ρ_{m238}$	kg/m³	0 … ∞
molar mass of heavier	$M_{m238}$	kg/mol	—
density of light uranium	$ρ_{U235}$	kg/m³	0 … ∞
molar mass of light uranium	$M_{U235}$	kg/mol	—
density of heavy uranium	$ρ_{U238}$	kg/m³	0 … ∞
molar mass of heavy uranium	$M_{U238}$	kg/mol	—
rms speed of ground	$v_{rms,g}$	m/s	—
rms speed of fluorine	$v_{rms,f}$	m/s	—
rms speed of lighter	$v_{rms,m235}$	m/s	—
rms speed of heavier	$v_{rms,m238}$	m/s	—
rms speed of light uranium	$v_{rms,U235}$	m/s	—
rms speed of heavy uranium	$v_{rms,U238}$	m/s	—
average thermal energy of ground	$E_{avg,g}$	J	—
average thermal energy of fluorine	$E_{avg,f}$	J	—
average thermal energy of lighter	$E_{avg,m235}$	J	—
average thermal energy of heavier	$E_{avg,m238}$	J	—
average thermal energy of light uranium	$E_{avg,U235}$	J	—
average thermal energy of heavy uranium	$E_{avg,U238}$	J	—

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

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

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

ideal gas law(ground)

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

ideal gas law(fluorine)

P_{m235} \cdot V_{m235} = n_{m235} \cdot R \cdot T_{m235}

ideal gas law(lighter)

P_{m238} \cdot V_{m238} = n_{m238} \cdot R \cdot T_{m238}

ideal gas law(heavier)

P_{U235} \cdot V_{U235} = n_{U235} \cdot R \cdot T_{U235}

ideal gas law(light uranium)

P_{U238} \cdot V_{U238} = n_{U238} \cdot R \cdot T_{U238}

ideal gas law(heavy uranium)

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

density of a gas(ground)

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

density of a gas(fluorine)

ρ_{m235} = \frac{n_{m235} \cdot M_{m235}}{V_{m235}}

density of a gas(lighter)

ρ_{m238} = \frac{n_{m238} \cdot M_{m238}}{V_{m238}}

density of a gas(heavier)

ρ_{U235} = \frac{n_{U235} \cdot M_{U235}}{V_{U235}}

density of a gas(light uranium)

ρ_{U238} = \frac{n_{U238} \cdot M_{U238}}{V_{U238}}

density of a gas(heavy uranium)

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

rms speed of a gas(ground)

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

rms speed of a gas(fluorine)

v_{rms,m235} = {(\frac{3 \cdot R \cdot T_{m235}}{M_{m235}})}^{0.5}

rms speed of a gas(lighter)

v_{rms,m238} = {(\frac{3 \cdot R \cdot T_{m238}}{M_{m238}})}^{0.5}

rms speed of a gas(heavier)

v_{rms,U235} = {(\frac{3 \cdot R \cdot T_{U235}}{M_{U235}})}^{0.5}

rms speed of a gas(light uranium)

v_{rms,U238} = {(\frac{3 \cdot R \cdot T_{U238}}{M_{U238}})}^{0.5}

rms speed of a gas(heavy uranium)

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

three halves k t for each molecule(ground)

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

three halves k t for each molecule(fluorine)

E_{avg,m235} = \frac{3 \cdot R \cdot T_{m235}}{2 \cdot N_{A}}

three halves k t for each molecule(lighter)

E_{avg,m238} = \frac{3 \cdot R \cdot T_{m238}}{2 \cdot N_{A}}

three halves k t for each molecule(heavier)

E_{avg,U235} = \frac{3 \cdot R \cdot T_{U235}}{2 \cdot N_{A}}

three halves k t for each molecule(light uranium)

E_{avg,U238} = \frac{3 \cdot R \cdot T_{U238}}{2 \cdot N_{A}}

three halves k t for each molecule(heavy uranium)

M_{m235} = M_{U235} + 6 \cdot M_{f}

uranium hexafluoride (molar mass of lighter)

M_{m238} = M_{U238} + 6 \cdot M_{f}

uranium hexafluoride (molar mass of heavier)

T_{m235} = T_{m238}

at any temperature