Question

At what temperature, the speed of sound in air will become double of its value at 27⁰C.
A.54⁰C
B.627⁰C
C.327⁰C
D.927⁰C

Answer 1

Hint: The speed of a sound is directly proportional to the square root of temperature of its medium i.e., with the increase in temperature, speed of sound increases and vice-versa.

Step by step answer: The speed of sound v in a gas (air) increases with the increase in temperature of the gas. The reason is that with the increase of temperature, there is a decrease in density as the volume of the gas increases. From the relation,
$v = \sqrt {\dfrac{{\gamma P}}{\rho }} $
where γ is the ratio of specific heat at constant pressure to the specific heat at constant volume, P is the pressure of the gas and ρ is the density of the gas. So, v is inversely proportional to the square root of density of the gas i.e. v is proportional to$\dfrac{1}{{\sqrt \rho }}$ .
Initially let the speed of sound be v₁ , temperature be T₁ .
As v α√T. [v is speed of sound and T is temperature]
${v^2} = kT$ [ where k is proportionality constant]
Now, we put the value of T₁ and v in the above equation,
${v_1}^2 = k \times 27$
${v_1}^2 = k\left( {27 + 273} \right)$ [ changing the temperature from ⁰C to K]
$
  {v_1}^2 = 300k \\
  k = \dfrac{{{v_1}^2}}{{300}} \\
$
Now, when the speed of sound is 2 v₁ and temperature be T₂. On substituting the value in the same equation,
${\left( {2{v_1}} \right)^2} = k{T_2}$
$4{v_1}^2 = k{T_2}$
Substituting the value of k in the equation,
$4{v_1}^2 = \left( {\dfrac{{{v_1}^2}}{{300}}} \right){T_2}$
$
  {T_2} = 4{v_1}^2 \times \dfrac{{300}}{{{v_1}^2}} \\
  \therefore {T_2} = 1200K \\
$
Temperature in ⁰C = $1200 - 273 = 927$⁰C
Hence Option (D) is correct.


Note: The temperature in degree Celsius should be converted to Kelvin. To convert temperature in degree Celsius to Kelvin, we have to add 273 ⁰C to the given temperature and to convert Kelvin to degree Celsius, subtract 273 ⁰C from the temperature in Kelvin.