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Question

Velocity of sound will be

The correct answer is

Greater in moist air than in dry air

Understanding the Velocity of Sound in Air

The question asks how the velocity of sound compares in moist air versus dry air.

The velocity of sound in a gas is determined by the properties of the gas, primarily its elasticity and density. A common formula for the velocity of sound \(v\) in a gas is given by:

\[v = \sqrt{\frac{\gamma P}{\rho}}\]

where:

  • \(\gamma\) (gamma) is the adiabatic index of the gas
  • \(P\) is the pressure of the gas
  • \(\rho\) (rho) is the density of the gas

Alternatively, considering the molar mass \(M\) of the gas at temperature \(T\), the velocity can also be expressed as:

\[v = \sqrt{\frac{\gamma RT}{M}}\]

where \(R\) is the universal gas constant.

Comparing Moist Air and Dry Air

Moist air is essentially a mixture of dry air and water vapor. Dry air is mainly composed of nitrogen (\(N_2\)), oxygen (\(O_2\)), argon (\(Ar\)), etc. Water vapor is \(H_2O\).

Let's compare the molar masses:

  • Average molar mass of dry air is approximately \(29 \text{ g/mol}\).
  • Molar mass of water (\(H_2O\)) is approximately \(18 \text{ g/mol}\).

When water vapor is added to dry air at a constant temperature and pressure, some of the heavier dry air molecules are replaced by lighter water molecules. This means that for the same volume, moist air has a lower mass than dry air. Therefore, the density of moist air (\(\rho_{moist}\)) is less than the density of dry air (\(\rho_{dry}\)) at the same temperature and pressure.

\[\rho_{moist} < \rho_{dry}\]

From the formula \(v = \sqrt{\frac{\gamma P}{\rho}}\), if pressure \(P\) is constant and \(\gamma\) is considered relatively similar or the density change is dominant, the velocity \(v\) is inversely proportional to the square root of the density (\(\rho\)).

Since moist air has lower density than dry air at the same temperature and pressure, the velocity of sound will be greater in moist air.

\[v_{moist} = \sqrt{\frac{\gamma P}{\rho_{moist}}}\] \[v_{dry} = \sqrt{\frac{\gamma P}{\rho_{dry}}}\]

Since \(\rho_{moist} < \rho_{dry}\), it follows that \(v_{moist} > v_{dry}\).

Another way to look at it is using the molar mass formula \(v = \sqrt{\frac{\gamma RT}{M}}\). Adding lighter water molecules reduces the average molar mass of the air mixture. A lower average molar mass \(M\) at constant temperature \(T\) and pressure \(P\) (assuming \(\gamma\) doesn't change significantly or its effect is less dominant) results in a higher velocity of sound.

Conclusion

Based on the relationship between the velocity of sound, density, and molar mass, the velocity of sound is greater in moist air compared to dry air when temperature and pressure are held constant. This is primarily because moist air is less dense due to the presence of lighter water vapor molecules.

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Important Questions from Wave

  1. Which of the following is correct?

    I. Sound is a mechanical wave

    II. Sound wave does not need any medium to propagate

  2. At a particular temperature, sound propagates in ______ at the fastest speed .

  3. The atmospheric green house effect is produced mainly by the absorption and re-emission of:

  4. Which of the following are examples of electromagnetic waves?

    a. Television waves

    b. Ultraviolet rays

    c. X-rays

    d. Sun rays

  5. When light passes from one medium into another medium, then the physical property which does not change is

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