Question

At $STP$ conditions, the speed of sound (according to Newton's formula) in air is: (Take the mass of $1$ mole of air to be $29\times 10^{-3}kg$ )

• A. $250m{s}^{-1}$
• B. $260m{s}^{-1}$
• C. $270m{s}^{-1}$
• D. $280m{s}^{-1}$

Answer 1

Answer: (D)

$1$ mole of any gas occupies $22.4$ litres at $STP$ .
Therefore, density of air at $STP$ is
$\rho =\frac{Massofonemoleofair}{VolumeofonemoleofairatSTP}$
$=\frac{29\times 10^{-3}kg}{22.4\times 10^{-3}{m}^3}=1.29kg{m}^{-3}$
At $STP$ , $P=1atm=1.01\times 10^{5}N{m}^{-2}$
According to Newton’s formula, the speed of sound in air at $STP$ is
$v=\sqrt{\left(\frac{P}{\rho }\right)}=\sqrt{\frac{1.01\times {\left(10\right)}^{5}N{m}^{-2}}{1.29\times kg{m}^{-3}}}=280m{s}^{-1}$