The Speed of Sound in Air

solved SpeedOfSound

Q. The mass of 1 mole of air is 29.0 × 10⁻³ kg/mol $M = 29.0 \times 10^{−3}$ kg/mol. Taking the ratio of specific heats of air as 1.4 $γ = 1.4$ , what is the speed of sound in air at standard temperature and pressure $T = 273$ K, $P = 1.01 \times 10^{5}$ Pa, in m/s?

Inputs

molar mass

M

0.029 kg/mol

gamma

γ

1.4

Outputs

speed

v

How it's solved

Solved with SpeedOfSound

Given

$M = 29.0 \times 10^{−3}$ kg/mol
$γ = 1.4$
$T = 273$ K from “standard temperature and pressure”
$P = 1.01 \times 10^{5}$ Pa from “standard temperature and pressure”

Find $v$ .

Working

density of an ideal gas(air) $ρ = \frac{P \cdot M}{R \cdot T}$
$ρ = \frac{101000 \cdot 0.029}{8.31446262 \cdot 273}$
$ρ = 1.29039461$ kg/m³
laplaces speed of sound(air) $v = \sqrt{\frac{γ \cdot P}{ρ}}$
$v = \sqrt{\frac{1.4 \cdot 101000}{1.29039461}}$
$v = 331.0 m/s$
$v \approx 331$ m/sto 3 significant figures

Conclusion

Sound squeezes the air too fast for heat to flow, so the springiness is the adiabatic one, γ times the pressure; Newton's guess without γ came out fifteen percent slow.

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 :the_speed_of_sound_in_air do4  kind :numerical5  describes "The mass of 1 mole of air is %{molar_mass}. Taking the ratio of specific heats of air as %{gamma}, " \6            "%{speed}, in m/s?"7  given molar_mass: "29.0e-3 kg/mol"8  given gamma: "1.4"9  asks speed: "what is the speed of sound in air at standard temperature and pressure"10end

solution.rb

1require_relative "../questions/the_speed_of_sound_in_air"2require_relative "../science/waves/speed_of_sound"34solve :the_speed_of_sound_in_air, using: SpeedOfSound do5  objects :air67  take molar_mass: air.molar_mass8  take gamma: air.ratio_of_specific_heats910  deduce_from "standard temperature and pressure", air.temperature => "273 K", air.pressure => "1.01e5 Pa"1112  speed = air.speed_of_sound13  answer speed:, in: "m/s", figures: 31415  conclusion "Sound squeezes the air too fast for heat to flow, so the springiness is the adiabatic one, " \16             "γ times the pressure; Newton's guess without γ came out fifteen percent slow."17end

playground.rb

1require_relative "../solutions/the_speed_of_sound_in_air"23play_with :the_speed_of_sound_in_air do4  title "The Speed of Sound in Air"5end

The laws it drew on

quantity	written	units	within	from
gas constant	$R$	J mol⁻¹ K⁻¹	—	SpeedOfSound
pressure	$P$	Pa	0 … ∞	SpeedOfSound
temperature	$T$	K	0 … ∞	SpeedOfSound
molar mass	$M$	kg/mol	—	SpeedOfSound
density	$ρ$	kg/m³	0 … ∞	SpeedOfSound
ratio of specific heats	$γ$	—	—	SpeedOfSound
speed of sound	$v$	m/s	—	SpeedOfSound
density of ground	$ρ_{g}$	kg/m³	0 … ∞
pressure of ground	$P_{g}$	Pa	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
temperature of ground	$T_{g}$	K	0 … ∞
density of air	$ρ$	kg/m³	0 … ∞
pressure of air	$P$	Pa	0 … ∞
molar mass of air	$M$	kg/mol	—
temperature of air	$T$	K	0 … ∞
speed of sound of ground	$v_{g}$	m/s	—
ratio of specific heats of ground	$γ_{g}$	—	—
speed of sound of air	$v$	m/s	—
ratio of specific heats of air	$γ$	—	—

ρ_{g} = \frac{P_{g} \cdot M_{g}}{R \cdot T_{g}}

density of an ideal gas

v_{g} = \sqrt{\frac{γ_{g} \cdot P_{g}}{ρ_{g}}}

laplaces speed of sound

ρ_{g} = \frac{P_{g} \cdot M_{g}}{R \cdot T_{g}}

density of an ideal gas(ground)

ρ = \frac{P \cdot M}{R \cdot T}

density of an ideal gas(air)

v_{g} = \sqrt{\frac{γ_{g} \cdot P_{g}}{ρ_{g}}}

laplaces speed of sound(ground)

v = \sqrt{\frac{γ \cdot P}{ρ}}

laplaces speed of sound(air)