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. $250\, ms^{-1}$
• B. $260\, ms^{-1}$
• C. $270\, ms^{-1}$
• D. $280\, ms^{-1}$

Answer 1

Answer: (D)

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