Uranium Hexafluoride Isotopes

Q. Uranium has two isotopes of masses 235 g/mol and 238 g/mol. Both are present in uranium hexafluoride gas. If the atomic mass of fluorine is 19 g/mol, by what factor is the lighter molecule faster than the heavier at any temperature? expects 1.004

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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 $T_{m235} = T_{m238}$ ?

The playground

Inputs

light

M_{U235}

235 g/mol

heavy

M_{U238}

238 g/mol

fluorine

M_{f}

19 g/mol

Outputs

ratio?

The solution

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 Ruby

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

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

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

The laws

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