Question

If the velocity of sound wave in humid air is \(\mathrm{v}_{\mathrm{m}}\) and that in dry air is \(\mathrm{v}_{\mathrm{d}}\), then \(\ldots \ldots\) (A) \(\mathrm{v}_{\mathrm{m}}>\mathrm{v}_{\mathrm{d}}\) (B) \(\mathrm{v}_{\mathrm{m}}<\mathrm{v}_{\mathrm{d}}\) (C) \(\mathrm{v}_{\mathrm{m}}=\mathrm{v}_{\mathrm{d}}\) \((\mathrm{D}) \mathrm{v}_{\mathrm{m}} \gg \mathrm{v}_{\mathrm{d}}\)

Short answer

The short answer based on the step-by-step solution is: The velocity of sound wave in humid air, \(v_m\), is greater than that in dry air, \(v_d\) (A) \(v_m > v_d\).

Step-by-step solution

Understand the speed of sound in air formula

The speed of sound in air depends on temperature, pressure, and humidity. One formula that takes these factors into account is: \[ v = \sqrt{\frac{\gamma R T}{M}} \] Where: - \(v\) is the speed of sound - \(\gamma\) is the adiabatic index (usually 1.4 for air) - \(R\) is the specific gas constant (287 J/(kg·K) for air) - \(T\) is the absolute temperature in Kelvin - \(M\) is the molar mass of the gas in kg/mol

Compare vm and vd using the formula

Since both vm and vd are velocities of sound in air, they will have the same values for \(\gamma\) and \(R\). The only difference between them is the molar mass, which will be affected by the humidity. Humid air has a lower molar mass than dry air because water vapor (molar mass = 18 g/mol) is lighter than the gases that make up dry air (mostly nitrogen and oxygen, with molar masses of 28 g/mol and 32 g/mol, respectively). We can rewrite the formula for vm and vd as follows: \[ v_m = \sqrt{\frac{\gamma R T}{M_m}} \] \[ v_d = \sqrt{\frac{\gamma R T}{M_d}} \]

Determine the relationship between vm and vd

Since humid air has a lower molar mass (Mm) than dry air (Md), the denominator in the formula for vm will be smaller than the denominator for vd. When you divide by a smaller number, the result is larger. Therefore: \[ v_m > v_d \] So the correct answer is: (A) \(v_m > v_d\)