Question

According to Newtons formula, the speed of sound in air at $STP$ is
(Take the mass of $1$ mole of air is $29\times 10^{-3}kg$)

• 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 $STP$.
bnm Therefore, density of air at $STP$ is
$\rho =\frac{\text{Mass of one mole of air}}{\text{Volume of one mole of air at }STP}$
$=\frac{29\times 10^{-3}kg}{22.4\times 10^{-3}{m}^3}=1.29kg{m}^3$
At $STP$, $P=1atm=1.01\times 10^{5}N{m}^{-2}$
According to Newtons formula, the speed of sound in air at $STP$ is
$v=\sqrt{\left(\frac{p}{\rho }\right)}=\sqrt{\frac{1.01\times 10^{5}N{m}^{-2}}{29\times kg{m}^{-3}}}$
$=280m{s}^{-3}$