The Speed of Sound in Air

Q. The mass of 1 mole of air is 29.0 × 10⁻³ kg/mol. Taking the ratio of specific heats of air as 1.4, what is the speed of sound in air at standard temperature and pressure, in m/s? expects 331 m/s

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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?

The playground

Inputs

molar mass

M

0.029 kg/mol

gamma

γ

1.4

Outputs

speed

v

The solution

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 Ruby

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

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

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

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)