Question

If $V_m$ is the velocity of sound in moist air, and the velocity of sound in dry air, under identical conditions of pressure and temperature is $V_d$ then:

• A. $V_m<V_d$
• B. $V_m>V_d$
• C. $V_m{V}_d=1$
• D. $V_m=V_d$

Answer 1

The correct option is B

$V_m>V_d$

The velocity of sound in air:

1. The distance traveled by a sound wave in a medium per unit time is the velocity of sound in that medium. It is different for different mediums as it depends upon several factors.
2. The velocity of sound can be mathematically expressed as $v=\sqrt{\frac{\gamma P}{d}}$, where is $\gamma$ specific heat ratio, $P$ is the pressure of gas and $d$ is the density of the medium.
3. According to this, the velocity of sound in a medium is inversely proportional to the square of the density of that medium, $v\propto \frac{1}{\sqrt{d}}$. So, the velocity will be lower for denser medium and higher for rarer medium.

The velocity of sound in dry air $\left(v_d\right)$ and moist air $\left(v_m\right)$:

1. Molecular weight is the mass of a given molecule. It can be described as the sum of the atomic masses of all atoms present in the molecule.
2. The molecular weight of a water atom is $W\left(H_{2}O\right)=1\times 2+16=18amu$ . And the molecular weights of oxygen and nitrogen atoms are $W\left(O_2\right)=16\times 2=32amu$ and $W\left(N_2\right)=14\times 2=28amu$ respectively.
3. Dry air is comprised of mostly oxygen and nitrogen molecules, whereas, moist air comprises of water molecules mostly. Thus, dry air is denser than moist air.
4. Therefore, the velocity of sound in dry air will be less than that in moist air $v_d<v_m$.

Hence, option (B) is the correct answer.