Question

The relation between speed of sound in moist air $v_m$ and dry air $v_d$ at the same temperature and pressure will be

• A. $v_m<v_d$
• B. $v_m>v_d$
• C. $v_m=v_d$
• D. $v_m\le v_d$

Answer 1

The correct option is B $v_m>v_d$

We know that the speed of sound in air is given by,

$v=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{\gamma RT}{M}}$

At the same temperature and pressure of air,

$v\propto \frac{1}{\sqrt{M}}.....\left(1\right)$

Dry air mostly contains nitrogen $\left(78\%\right)$ and oxygen $(\simeq 21\%)$ and other gases like $C{O}_2$ in traces.

Moist air will contain a higher volume fraction of moisture. The moisture will replace some of the volume occupied by gases like $C{O}_2,N_2,O_2,\text{etc}$.

Molar mass of $H_{2}O$ is $18\text{g/mol}$, which is lesser than the constituent components of dry air.

This will lead to reduction in the effective molar mass of the moist air$M_{\text{moist air}}<M_{\text{dry air}}$.

$\Rightarrow M_d>M_m.....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$ we get,

$v_m>v_d$