Argon as Fast as Cold Helium

Q. The atomic mass of Ar is 39.9 g/mol and of He is 4.0 g/mol. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C, in kelvin? expects 2.52 × 10³ K

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Q. The atomic mass of Ar is 39.9 g/mol $M_{a} = 39.9$ g/mol and of He is 4.0 g/mol $M_{h} = 4.0$ g/mol. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed $v_{rms,h} = v_{rms,a}$ of a helium gas atom at – 20 °C $T_{h} = 253$ K, in kelvin?

The playground

Inputs

argon

M_{a}

39.9 g/mol

helium

M_{h}

4 g/mol

Outputs

temperature

T_{a}

The solution

Solved with RmsSpeed

Given

$M_{a} = 39.9$ g/mol
$M_{h} = 4.0$ g/mol
$T_{h} = 253$ K from “a helium gas atom at – 20 °C”
$v_{rms,h} = v_{rms,a}$ from “equal to the rms speed”

Find $T_{a}$ .

Working

$M_{a} = 39.9$ g/mol $= 0.0399$ kg/mol
$M_{h} = 4.0$ g/mol $= 0.004$ kg/mol
rms speed of a gas(helium) $v_{rms,h} = {(\frac{3 \cdot R \cdot T_{h}}{M_{h}})}^{0.5}$
$v_{rms,h} = {(\frac{3 \cdot 8.31446262 \cdot 253}{0.004})}^{0.5}$
$v_{rms,h} = 1256.05306$ m/s
equal to the rms speed $v_{rms,h} = v_{rms,a}$
$1256.05306 = v_{rms,a}$
$v_{rms,a} = 1256.05306$ m/s
rms speed of a gas(argon) $v_{rms,a} = {(\frac{3 \cdot R \cdot T_{a}}{M_{a}})}^{0.5}$
$1256.05306 = {(\frac{3 \cdot 8.31446262 \cdot T_{a}}{0.0399})}^{0.5}$
$T_{a} = 2524$ K
$T_{a} \approx 2.52 \times 10^{3}$ Kto 3 significant figures

Conclusion

The rms speed goes as the square root of the temperature over the molar mass, so the heavier argon must be hotter in the same proportion as it is heavier.

The Ruby

1require_relative "../physics"23question :argon_as_fast_as_cold_helium do4  kind :numerical5  describes "The atomic mass of Ar is %{argon} and of He is %{helium}. %{temperature} of an atom in an argon gas " \6            "cylinder equal to the rms speed of a helium gas atom at – 20 °C, in kelvin?"7  given argon: "39.9 g/mol"8  given helium: "4.0 g/mol"9  asks temperature: "At what temperature is the root mean square speed"10end

1require_relative "../questions/argon_as_fast_as_cold_helium"2require_relative "../science/kinetic_theory/rms_speed"34solve :argon_as_fast_as_cold_helium, using: RmsSpeed do5  objects :argon, :helium67  take argon: argon.molar_mass8  take helium: helium.molar_mass910  deduce_from "a helium gas atom at – 20 °C", helium.temperature => "253 K"11  deduce_from "equal to the rms speed", helium.rms_speed => argon.rms_speed1213  temperature = argon.temperature14  answer temperature:, in: "K", figures: 31516  conclusion "The rms speed goes as the square root of the temperature over the molar mass, so the heavier argon " \17             "must be hotter in the same proportion as it is heavier."18end

1require_relative "../solutions/argon_as_fast_as_cold_helium"23play_with :argon_as_fast_as_cold_helium do4  title "Argon as Fast as Cold Helium"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 argon	$P_{a}$	Pa	0 … ∞
volume of argon	$V_{a}$	m³	0 … ∞
amount of argon	$n_{a}$	mol	—
temperature of argon	$T_{a}$	K	0 … 1e+09
pressure of helium	$P_{h}$	Pa	0 … ∞
volume of helium	$V_{h}$	m³	0 … ∞
amount of helium	$n_{h}$	mol	—
temperature of helium	$T_{h}$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of argon	$ρ_{a}$	kg/m³	0 … ∞
molar mass of argon	$M_{a}$	kg/mol	—
density of helium	$ρ_{h}$	kg/m³	0 … ∞
molar mass of helium	$M_{h}$	kg/mol	—
rms speed of ground	$v_{rms,g}$	m/s	—
rms speed of argon	$v_{rms,a}$	m/s	—
rms speed of helium	$v_{rms,h}$	m/s	—
average thermal energy of ground	$E_{avg,g}$	J	—
average thermal energy of argon	$E_{avg,a}$	J	—
average thermal energy of helium	$E_{avg,h}$	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_{a} \cdot V_{a} = n_{a} \cdot R \cdot T_{a}

ideal gas law(argon)

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

ideal gas law(helium)

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

density of a gas(ground)

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

density of a gas(argon)

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

density of a gas(helium)

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

rms speed of a gas(ground)

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

rms speed of a gas(argon)

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

rms speed of a gas(helium)

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

three halves k t for each molecule(ground)

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

three halves k t for each molecule(argon)

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

three halves k t for each molecule(helium)

v_{rms,h} = v_{rms,a}

equal to the rms speed