A Helium Atom on the Sun

solved RmsSpeed

Q. Estimate the average thermal energy of a helium atom at the temperature on the surface of the Sun (6000 K $T = 6000$ K), in joules?

Inputs

temperature

T

6000 K

Outputs

energy

E_{avg}

How it's solved

Solved with RmsSpeed

Given

$T = 6000$ K

Find $E_{avg}$ .

Working

three halves k t for each molecule(helium) $E_{avg} = \frac{3 \cdot R \cdot T}{2 \cdot N_{A}}$
$E_{avg} = \frac{3 \cdot 8.31446262 \cdot 6000}{2 \cdot 6.02214076 \times 10^{23}}$
$E_{avg} = 1.243 \times 10^{−19}$ J
$E_{avg} \approx 1.24 \times 10^{−19}$ Jto 3 significant figures

Conclusion

The average energy of motion grows in step with the absolute temperature, so the Sun's surface gives each atom some twenty times what a room does.

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_helium_atom_on_the_sun do4  kind :numerical5  describes "%{energy} at the temperature on the surface of the Sun (%{temperature}), in joules?"6  given temperature: "6000 K"7  asks energy: "Estimate the average thermal energy of a helium atom"8end

solution.rb

1require_relative "../questions/a_helium_atom_on_the_sun"2require_relative "../science/kinetic_theory/rms_speed"34solve :a_helium_atom_on_the_sun, using: RmsSpeed do5  objects :helium67  take temperature: helium.temperature89  energy = helium.average_thermal_energy10  answer energy:, in: "J", figures: 31112  conclusion "The average energy of motion grows in step with the absolute temperature, so the Sun's surface " \13             "gives each atom some twenty times what a room does."14end

playground.rb

1require_relative "../solutions/a_helium_atom_on_the_sun"23play_with :a_helium_atom_on_the_sun do4  title "A Helium Atom on the Sun"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 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)