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At S.T.P., the speed of sound in air is \(331 \mathrm{~m} / \mathrm{s}\). Determine the speed of sound in hydrogen at S.T.P. if the specific gravity of hydrogen relative to air is \(0.0690\) and if \(\gamma=1.40\) for both gases.

Short Answer

Expert verified
The speed of sound in hydrogen at S.T.P. is approximately 1258 m/s.

Step by step solution

01

Understanding the Problem

We need to find the speed of sound in hydrogen at Standard Temperature and Pressure (S.T.P.). We know the speed of sound in air, the specific gravity of hydrogen relative to air, and the adiabatic index, \( \gamma \), for both gases.
02

Using the Formula for Speed of Sound

The speed of sound \( v \) in a gas is given by the formula \( v = \sqrt{\frac{\gamma R T}{M}} \), where \( \gamma \) is the adiabatic index, \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas. However, we can use a simplified ratio form since we are comparing the speed of sound in two different gases.
03

Relating Speeds in Different Gases

For two gases at the same \( T \) and \( \gamma \), we can write the ratio of their speeds of sound as \( \frac{v_{1}}{v_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \). Here \( v_{1} \) is the speed in air, \( v_{2} \) in hydrogen, \( M_{1} \) the molar mass of air, and \( M_{2} \) that of hydrogen.
04

Using Specific Gravity to Find Molar Mass

Specific gravity is related to density, meaning \( \frac{\rho_{\text{hydrogen}}}{\rho_{\text{air}}} = 0.0690 \). Given the constant temperature and pressure, the molar mass ratio is the same as the density ratio. Therefore, \( \frac{M_{\text{hydrogen}}}{M_{\text{air}}} = 0.0690 \).
05

Substituting Values into the Ratio

Substitute the known quantities into the ratio formula: \( \frac{v_{\text{hydrogen}}}{v_{\text{air}}} = \sqrt{\frac{M_{\text{air}}}{M_{\text{hydrogen}}}} = \sqrt{\frac{1}{0.0690}} \).
06

Calculating the Speed of Sound in Hydrogen

\( v_{\text{hydrogen}} = v_{\text{air}} \times \sqrt{\frac{1}{0.0690}} = 331 \times \sqrt{\frac{1}{0.0690}} \approx 331 \times 3.803 \approx 1258 \mathrm{~m/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Adiabatic Index
The adiabatic index, often represented as \( \gamma \), is an important concept in thermodynamics. It is the ratio of specific heats of a gas at constant pressure \( C_p \) and constant volume \( C_v \). Mathematically, it is expressed as:\[ \gamma = \frac{C_p}{C_v}\]This ratio indicates how a gas will behave under adiabatic processes, where no heat is exchanged with the surroundings. When the adiabatic index has a higher value, it means the gas expands more easily when heated. This is crucial for calculations involving sound through gases because it affects the speed at which pressure waves move.
In this problem, we work with the same \( \gamma = 1.40 \) for both air and hydrogen, simplifying our calculations. Using a constant \( \gamma \), along with the known speeds of sound, allows us to relate the properties of different gases under the same conditions, especially when temperature and pressure do not vary.
  • The knowledge of \( \gamma \) helps in deriving relationships between temperature, pressure, and the speed of sound among different gases.
  • Because \( \gamma \) values differ for various gases, it affects how compressible gases are under pressure.
Specific Gravity
Specific gravity is a measure that indicates the ratio of the density of a substance to a reference substance. Commonly, the density of water is used as a reference for liquids and solids, while for gases, air is typically the reference. In our case, specific gravity of hydrogen relative to air is given as 0.0690. This means hydrogen is significantly lighter than air. Specific gravity is useful because it allows us to understand differences in behavior between different gases.
  • For gases, the specific gravity can directly correlate to molar mass at standard conditions of temperature and pressure, due to the Ideal Gas Law.
  • For example, a specific gravity of 0.0690 implies that hydrogen is much less dense, which is crucial for calculating sound speed, as less dense means faster propagation of sound.
By understanding the specific gravity, we were able to substitute it into the formula that relates the speed of sound in hydrogen with air. This provided a straightforward route to compute the desired speed in hydrogen without needing comprehensive other properties like actual gas Density or Molar masses.
Universal Gas Constant
The universal gas constant \( R \) is a fundamental constant in physical chemistry that appears in various important equations in thermodynamics, most notably the Ideal Gas Law:\[ PV = nRT\]Where \( P \) is pressure, \( V \) is volume, \( n \) is moles of gas, and \( T \) is temperature in Kelvin. The gas constant \( R \) links the amount of gas and its temperature to pressure and volume changes.In problems involving the speed of sound like in this exercise, though \( R \) does not explicitly appear in the final ratio of speeds, its presence is implicit in understanding how temperature and molar properties affect sound speed.
  • The constant allows us to generalize behaviors of gases under different conditions, establishing that certain variables remain constant across different gases (e.g., temperature and pressure).
  • While \( R \) remains a constant for all gases, its calculation alongside molar mass and temperature gives a foundation that influences how the adiabatic process calculates sound speed.
An inherent understanding of \( R \) is necessary to comprehend the full scope of gas dynamics, especially when working with gas properties not explicitly defined by the exercise but deeply influential in thermodynamic calculations.

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Most popular questions from this chapter

A tiny sound source emits sound uniformly in all directions. The intensity level at a distance of \(2.0 \mathrm{~m}\) is \(100 \mathrm{~dB}\). How much sound power is the source emitting? The energy emitted by a point source can be considered to flow out through a spherical surface which has the source at its center. Hence, if we find the rate of flow through such a surface, it will equal the flow from the source. Take a concentric sphere of radius \(2.0 \mathrm{~m}\). We know that the sound level on its surface is \(100 \mathrm{~dB}\). You can show that this corresponds to \(I=0.010 \mathrm{~W} / \mathrm{m}^{2}\). Thus, the energy flowing each second through each \(\mathrm{m}^{2}\) of surface is \(0.010 \mathrm{~W}\). The total energy flow through the spherical surface is then \(I\left(4 \pi r^{2}\right)\), where \(I=0.010 \mathrm{~W} / \mathrm{m}^{2}\) and \(r=2.0 \mathrm{~m}\) : $$ \text { Power from source }=\left(0.010 \mathrm{~W} / \mathrm{m}^{2}\right)(4 \pi)(2 \mathrm{~m})^{2}=0.50 \mathrm{~W} $$ Notice how little power issues as sound from even such an intense source.

A locomotive moving at \(30.0 \mathrm{~m} / \mathrm{s}\) approaches and passes a person standing beside the track. Its whistle is emitting a note of frequency \(2.00 \mathrm{kHz}\). What frequency will the person hear \((a)\) as the train approaches and (b) as it recedes? The speed of sound is \(340 \mathrm{~m} / \mathrm{s}\).

A whisper has an intensity level of about \(15 \mathrm{~dB}\). What is the corresponding intensity of the sound?

Find the speed of sound in a diatomic ideal gas that has a density of \(3.50 \mathrm{~kg} / \mathrm{m}^{3}\) and a pressure of \(215 \mathrm{kPa}\) We know that \(v=\sqrt{\gamma R T / M}\) and can find the temperature from the pressure. Using the gas law \(P V=(m / M) R T\), $$ \frac{R T}{M}=P \frac{V}{m} $$ Moreover, \(\rho=m / V\), and so the expression for the speed becomes $$ v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{(1.40)\left(215 \times 10^{3} \mathrm{~Pa}\right)}{3.50 \mathrm{~kg} / \mathrm{m}^{3}}}=293 \mathrm{~m} / \mathrm{s} $$ We used the fact that \(\gamma \approx 1.40\) for a diatomic ideal gas, as discussed in Chapter 20 .

Two cars are heading straight at each other with the same speed. The horn of one \((f=3.0 \mathrm{kHz})\) is blowing, and is heard to have a frequency of \(3.4 \mathrm{kHz}\) by the people in the other car. Find the speed at which each car is moving if the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\).

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