A Helium Atom at Room Temperature

Q. Estimate the average thermal energy of a helium atom at room temperature (27 °C), in joules? expects 6.2 × 10⁻²¹ J

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The web of concepts

given read from the words equation asked

The equations it used

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Check an answer

Q. Estimate the average thermal energy of a helium atom at room temperature (27 °C) $T = 300$ K, in joules?

The playground

Outputs

energy

E_{avg}

The solution

Solved with RmsSpeed

Given

$T = 300$ K from “room temperature (27 °C)”

Find $E_{avg}$ .

Working

three halves k t for each molecule(helium) $E_{avg} = \frac{3 \cdot R \cdot T}{2 \cdot N_{A}}$
$E_{avg} = \frac{3 \cdot 8.31446262 \cdot 300}{2 \cdot 6.02214076 \times 10^{23}}$
$E_{avg} = 6.213 \times 10^{−21}$ J
$E_{avg} \approx 6.2 \times 10^{−21}$ Jto 2 significant figures

Conclusion

A helium atom only moves, three ways, and each way holds half of Boltzmann's constant times the temperature on average.

The Ruby

1require_relative "../physics"23question :a_helium_atom_at_room_temperature do4  kind :numerical5  describes "%{energy} at room temperature (27 °C), in joules?"6  asks energy: "Estimate the average thermal energy of a helium atom"7end

1require_relative "../questions/a_helium_atom_at_room_temperature"2require_relative "../science/kinetic_theory/rms_speed"34solve :a_helium_atom_at_room_temperature, using: RmsSpeed do5  objects :helium67  deduce_from "room temperature (27 °C)", helium.temperature => "300 K"89  energy = helium.average_thermal_energy10  answer energy:, in: "J", figures: 21112  conclusion "A helium atom only moves, three ways, and each way holds half of Boltzmann's constant times the " \13             "temperature on average."14end

1require_relative "../solutions/a_helium_atom_at_room_temperature"23play_with :a_helium_atom_at_room_temperature do4  title "A Helium Atom at Room Temperature"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 helium	$P$	Pa	0 … ∞
volume of helium	$V$	m³	0 … ∞
amount of helium	$n$	mol	—
temperature of helium	$T$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of helium	$ρ$	kg/m³	0 … ∞
molar mass of helium	$M$	kg/mol	—
rms speed of ground	$v_{rms,g}$	m/s	—
rms speed of helium	$v_{rms}$	m/s	—
average thermal energy of ground	$E_{avg,g}$	J	—
average thermal energy of helium	$E_{avg}$	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 \cdot V = n \cdot R \cdot T

ideal gas law(helium)

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(helium)

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

three halves k t for each molecule(helium)