Question

According to Newton’s formula, the speed of sound in air at $STP$ is:
(Take the mass of $1$ mole of are is $29 \times {10^{ - 3}}\,Kg$ )
(A) $250\,m{s^{ - 1}}$
(B) $260\,m{s^{ - 1}}$
(C) $270\,m{s^{ - 1}}$
(D) $280\,m{s^{ - 1}}$

Answer 1

Hint: Use the formula of the density and substitute the mass and the volume in it to find the value of the density of the air. Substitute the obtained density and the pressure in the formula of the velocity to find the speed of the sound in the air.
Formula used:
(1) The formula of the density of the air at $STP$ is given by
$\rho = \dfrac{m}{V}$
Where $\rho $ is the density of the air at $STP$ , $m$ is the mass of the one mole of the air and $V$ is the volume of one mole of air at $STP$ .
(2) The formula of the speed of sound air at $STP$ is given by
$v = \sqrt {\dfrac{P}{\rho }} $
Where $v$ is the speed of the sound in the air, $P$ is the pressure of the air and $\rho $ is its density.

Complete answer:
It is given that the mass of $1$ mole of are is $29 \times {10^{ - 3}}\,Kg$
It is known that the mole of any gas occupies $22.4$ litres at $STP$ and hence $V = 22.4 \times {10^{ - 3}}\,{m^3}$
Let us use the density of the air formula,
$\rho = \dfrac{m}{V}$
Substituting the known values in the above formula, we get
$\rho = \dfrac{{29 \times {{10}^{ - 3}}}}{{22.4 \times {{10}^{ - 3}}}}$
By performing various arithmetic operations, we get
$\rho = 1.29\,Kg{m^{ - 3}}$
Using the formula of the velocity of the speed,
$v = \sqrt {\dfrac{P}{\rho }} $
Substituting the pressure as the one atmospheric pressure since the condition is taken as standard temperature and pressure in the above step,
$v = \sqrt {\dfrac{{1.01 \times {{10}^5}}}{{1.29}}} $
By the simplification of the above solution, we get
$v = 280\,m{s^{ - 1}}$
Hence the velocity of the sound in the air is obtained as $280\,m{s^{ - 1}}$ .

Thus the option (D) is correct.

Note:
Remember that the $STP$ used in the above solution is the standard temperature and the pressure in which the temperature is maintained as $273.15\,K$ and the pressure is maintained as $1$ atmospheric pressure i.e. $1.01 \times {10^5}\,N{m^{ - 2}}$. This condition is maintained throughout the process.