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

Answer 1

Answer: (D)

$1 \, $ mole of any gas occupies $22.4 \, $ litres at $STP$ .
Therefore, density of air at $STP \, $ is
$\rho =\frac{M a s s \, o f \, o n e \, m o l e \, o f \, a i r}{V o l u m e \, o f \, o n e \, m o l e \, o f \, a i r \, a t \, S T P}$
$=\frac{29 \times 10^{- 3} \, k g}{22.4 \times 10^{- 3} m^{3}}=1.29 \, kg \, m^{- 3}$
At $STP$ , $P=1 \, atm=1.01\times 10^{5} \, N \, m^{- 2}$
According to Newton’s formula, the speed of sound in air at $STP \, $ is
$v=\sqrt{\left(\frac{P}{\rho }\right)}=\sqrt{\frac{1.01 \times \left(10\right)^{5} \, N m^{- 2}}{1.29 \times k g \, m^{- 3}}}=280 \, m \, s^{- 1}$