at what temperature will the speed of sound be double its value at 273K - Physics - Waves

Question

at what temperature will the speed of sound be double its value
at 273K

Kranav Sharma · Accepted Answer

The speed of sound in air is given by the following relation

$v = \sqrt{(\frac{γ R T}{M})}$

here,

γ - adiabatic index

R - gas constant

T - temperature

M - molar mass (kg/mol)

now,

the speed of sound at T₀ (= 273 K) will be

$v_{0} = \sqrt{(\frac{γ R T_{0}}{M})}$ ......................(1)

and

the speed of sound at unknown temperature T' will be

$v' = \sqrt{(\frac{γ R T'}{M})}$

now, we know that v' = 2v₀ at temperature T'

so,

$2 v_{0} = \sqrt{(\frac{γ R T'}{M})}$

or

$v_{0} = \frac{1}{2} \sqrt{(\frac{γ R T'}{M})}$ ......................(2)

by equating (1) and (2), we get

$\sqrt{(\frac{γ R T_{0}}{M})} = \frac{1}{2} \sqrt{(\frac{γ R T'}{M})}$

or

by solving further we get..

T₀ = (1/4)T

or

T = 4T₀ = 4 x 273 K

thus, the temperature at which the speed of sound would be double is

T = 1092 K or T = 819 ^oC