Question

The velocity of sound in the air is:
(A) 330 m/sec
(B) 220 m/sec
(C) 110 m/sec
(D) 232 m/sec

Answer 1

Hint The speed of sound in a particular fluid is directly proportional to the square root of the adiabatic Bulk Modulus and inversely proportional to the square root of the density of the fluid at a particular temperature. Using this relation, we can find the velocity of sound in air.

Formula used: $ v = \sqrt {\dfrac{K}{\rho }} $ where $ K $ is the adiabatic Bulk modulus and $ \rho $ is the density of the fluid at a particular temperature.

Complete step by step answer
Sound is a mechanical wave which means that they travel through a material medium and thus cannot travel through vacuum. Sound propagates by setting the molecules of the medium nearest to the source into vibration, which then set its own surrounding molecule into vibration, these newly vibrating molecules then also set other molecules into vibration, and so on.
To solve the problem, we shall note that the speed of sound is given by the equation
 $\Rightarrow v = \sqrt {\dfrac{K}{\rho }} $ where $ K $ is the adiabatic Bulk modulus and $ \rho $ is the density of the fluid at a particular temperature.
For air, the average adiabatic bulk modulus is regarded to be $ K = 142kPa = 1.42 \times {10^5}Pa $ while the density of air at $ 0^\circ C $ is given as $ \rho = 1.28kg/{m^3} $
Inserting the values into the equation above, we have that
$\Rightarrow v = \sqrt {\dfrac{{142000}}{{1.28}}} = \sqrt {110937.5} $
$\Rightarrow v = 333m/s $
The closest to the answer is A.
Hence, the correct option is A.

Note
Alternatively, in ideal gases, the speed of sound can be given as
 $\Rightarrow v = \sqrt {\dfrac{{\gamma RT}}{M}} $ where $ \gamma $ is the adiabatic index, $ R $ is the universal gas constant, $ T $ is the absolute temperature and $ M $ is the molecular mass of a gas.
For air, $ \gamma = 1.4 $ , $ M = 2.89 \times {10^{ - 2}}kg/mol $
Hence, Inserting the values into equation for $ T = 0^\circ C = 273K $ , we have
$\Rightarrow v = \sqrt {\dfrac{{1.4 \times 8.314 \times 273}}{{2.89 \times {{10}^{ - 2}}}}} = \sqrt {\dfrac{{3117.6108}}{{0.0289}}} $
$ \Rightarrow v = \sqrt {109951.93} = 332m/s$
Which is very similar to our answer in the solution step.