If v_m is the velocity of sound in moist air | vedclass

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(A) The velocity of sound in a gas is given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$,where $\gamma$ is the adiabatic index,$P$ is the pressur

Vedclass · Accepted Answer

$v_m > v_d$

Vedclass · Answer

(A) The velocity of sound in a gas is given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$,where $\gamma$ is the adiabatic index,$P$ is the pressure,and $\rho$ is the density of the gas.Under identical conditions of pressure and temperature,the density of moist air (air mixed with water vapor) is less than the density of dry air because the molecular weight of water vapor $(18 \ g/mol)$ is less than the average molecular weight of dry air (approximately $29 \ g/mol$).Since $v \propto \frac{1}{\sqrt{\rho}}$,a lower density results in a higher velocity of sound.Therefore,$v_m > v_d$.

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

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