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The speed of sound is given by \(\nu_{\text {sound}}=\) \(\sqrt{\gamma k T / m}=\sqrt{\gamma R T / M},\) where \(\gamma=C_{P} / C_{\nu}.\) a. What is the speed of sound in \(\mathrm{Ne}, \mathrm{Kr}\), and \(\mathrm{Ar}\) at \(1000 . \mathrm{K} ?\) b. At what temperature will the speed of sound in Kr equal the speed of sound in Ar at \(1000 .\) K?

Short Answer

Expert verified
a. At 1000 K, the speeds of sound in Ne, Kr, and Ar are: 1. Ne: \(\nu_{\text { sound,Ne}}=\sqrt{\frac{5}{3}\frac{8.314\,\text{J/mol}\cdot\text{K}\cdot1000\,\text{K}}{20.18\,\text{g/mol}}} \approx 1343\,\text{m/s}\) 2. Kr: \(\nu_{\text { sound,Kr}}=\sqrt{\frac{5}{3}\frac{8.314\,\text{J/mol}\cdot\text{K}\cdot1000\,\text{K}}{83.80\,\text{g/mol}}} \approx 629\,\text{m/s}\) 3. Ar: \(\nu_{\text { sound,Ar}}=\sqrt{\frac{5}{3}\frac{8.314\,\text{J/mol}\cdot\text{K}\cdot1000\,\text{K}}{39.95\,\text{g/mol}}} \approx 920\,\text{m/s}\) b. The temperature at which the speed of sound in Kr equals the speed of sound in Ar at 1000 K is: \(T_{\text{Kr}}=1000\, \text{K} \cdot\frac{83.80\, \text{g/mol}}{39.95\, \text{g/mol}} \approx 2096\, \text{K}\)

Step by step solution

01

Find the values of \(\gamma\) for each gas.

Each of the given gases is a diatomic molecule. For such molecules, \(\gamma\) has a value close to \(5/3\). Therefore, we can use this value when calculating the speed of sound: \(\gamma = 5/3\).
02

Find the molar mass of each gas.

We will find the molar mass of Ne, Kr, and Ar using the known atomic masses: 1. Neon (Ne): 20.18 g/mol 2. Krypton (Kr): 83.80 g/mol 3. Argon (Ar): 39.95 g/mol
03

Calculate the speed of sound for each gas at 1000 K.

Next, we will plug the values of \(\gamma\), \(R\), \(M\), and \(T = 1000\) K into the formula for the speed of sound and calculate the speed of sound for each gas: 1. Ne: \(\nu_{\text { sound,Ne}}=\sqrt{\frac{5}{3}\frac{8.314\,\text{J/mol}\cdot\text{K}\cdot1000\,\text{K}}{20.18\,\text{g/mol}}}\) 2. Kr: \(\nu_{\text { sound,Kr}}=\sqrt{\frac{5}{3}\frac{8.314\,\text{J/mol}\cdot\text{K}\cdot1000\,\text{K}}{83.80\,\text{g/mol}}}\) 3. Ar: \(\nu_{\text { sound,Ar}}=\sqrt{\frac{5}{3}\frac{8.314\,\text{J/mol}\cdot\text{K}\cdot1000\,\text{K}}{39.95\,\text{g/mol}}}\)
04

Calculate the required temperature for Kr to have the same speed of sound as Ar at 1000 K.

We will set the speeds of sound for Kr and Ar equal to each other, such that we can solve for the required temperature (\(T_{\text{Kr}}\)) for this condition: \(\sqrt{\frac{5}{3}\frac{8.314\,\text{J/mol}\cdot\text{K}\cdot1000\,\text{K}}{39.95\,\text{g/mol}}}=\sqrt{\frac{5}{3}\frac{8.314\,\text{J/mol}\cdot\text{K}\cdot T_{\text{Kr}}}{83.80\,\text{g/mol}}}\) We will square both sides, and rearrange the equation to solve for \(T_{\text{Kr}}\): \(T_{\text{Kr}}=1000\, \text{K} \cdot\frac{83.80\, \text{g/mol}}{39.95\, \text{g/mol}}\) Calculate the value of \(T_{\text{Kr}}\).

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Key Concepts

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

Physical Chemistry
Physical chemistry is a branch of chemistry focused on understanding the physical properties of molecules, the forces that act upon them, and the physical principles that underlie chemical reactions. In the context of our problem, physical chemistry helps us understand how the speed of sound in gases is influenced by molecular properties and environmental conditions.

Specifically, the speed of sound is a physical property that depends on the type of gas, its temperature, and its adiabatic index. Physical chemists use these parameters to derive the relationships expressed in the formula \(u_{\text {sound}}=\sqrt{\gamma R T / M}\), where \(u_{\text {sound}}\) is the speed of sound, \(\gamma\) the adiabatic index, \(R\) the universal gas constant, \(T\) the temperature, and \(M\) the molar mass.
Molar Mass
Molar mass is an essential concept in chemistry, denoting the mass of one mole of a given substance. The unit typically used is grams per mole (g/mol). It is a vital factor when calculating the speed of sound in a gas because the molar mass affects the motion of gas particles. The molar mass of a gas can be found on the periodic table as the atomic mass for elements or calculated by summing the atomic masses for compounds.

As seen in the solution, the molar mass of each noble gas - Neon (\(Ne\)), Krypton (\(Kr\)), and Argon (\(Ar\)) - is utilized to determine how fast sound travels through each gas. Heavier gases such as Krypton will have slower sound speeds due to their higher molar mass compared to lighter gases like Neon.
Adiabatic Index (\(\gamma\))
The adiabatic index, often represented by \(\gamma\), is a dimensionless constant that relates the heat capacity at constant pressure \(C_P\) to the heat capacity at constant volume \(C_V\), using the formula \(\gamma=C_{P} / C_{V}\). It is a measure of the compressibility of a gas under adiabatic conditions – conditions in which no heat is transferred into or out of the system.

For diatomic molecules like the gases in our problem, \(\gamma\) is approximately \(5/3\), which reflects their specific heat capacities and plays a crucial role in determining the speed of sound. This is because the speed of sound in a gas is related to how quickly the gas particles can be compressed, a process affected by the adiabatic index.
Gas Laws
The behavior of gases and their interactions with temperature, volume, and pressure are governed by gas laws. Some of these fundamental laws include Boyle's Law, Charles's Law, and the Ideal Gas Law. These principles are crucial when exploring how the speed of sound varies with different conditions in different gases.

The formula used to calculate the speed of sound in gases incorporates the Ideal Gas Law, where \(R\) is the universal gas constant. It assumes that the gas behaves ideally and that the molecules do not interact except for perfectly elastic collisions. The Ideal Gas Law greatly simplifies calculations and is accurate for most gases under standard conditions, though deviations can occur at high pressures or low temperatures where real gas behavior emerges.

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Most popular questions from this chapter

Starting with the Maxwell speed distribution, demonstrate that the probability distribution for translational kinetic energy for \(\varepsilon_{T} \gg k T\) is given by \\[f\left(\varepsilon_{T}\right) d \varepsilon_{T}=2 \pi\left(\frac{1}{\pi k T}\right)^{3 / 2} e^{-\varepsilon_{T} / k T} \varepsilon_{T}^{1 / 2} d \varepsilon_{T}\\]

At what temperature is the \(\nu_{r m s}\) of Ar equal to that of \(\mathrm{SF}_{6}\) at \(298 \mathrm{K} ?\) Perform the same calculation for \(\nu_{m p}\).

Many of the concepts developed in this chapter can be applied to understanding the atmosphere. Because atmospheric air is comprised primarily of \(\mathrm{N}_{2}\) (roughly \(78 \%\) by volume \(,\) approximate the atmosphere as consisting only of \(\mathrm{N}_{2}\) in answering the following questions: a. What is the single-particle collisional frequency at sea level, with \(T=298 \mathrm{K}\) and \(\mathrm{P}=1.0\) atm? The corresponding single-particle collisional frequency is reported as \(10^{10} \mathrm{s}^{-1}\) in the CRC Handbook of Chemistry and Physics \((62 n d \text { ed. }, p . F-171)\) b. At the tropopause \((11 \mathrm{km})\), the collisional frequency decreases to \(3.16 \times 10^{9} \mathrm{s}^{-1}\), primarily due to a reduction in temperature and barometric pressure (i.e., fewer particles). The temperature at the tropopause is \(\sim 220 \mathrm{K}\). What is the pressure of \(\mathrm{N}_{2}\) at this altitude? c. At the tropopause, what is the mean free path for \(\mathrm{N}_{2} ?\)

Imagine a cubic container with sides \(1 \mathrm{cm}\) in length that contains 1 atm of Ar at 298 K. How many gas-wall collisions are there per second?

a. What is the average time required for \(\mathrm{H}_{2}\) to travel \(1.00 \mathrm{m}\) at \(298 \mathrm{K}\) and 1 atm? b. How much longer does it take \(\mathrm{N}_{2}\) to travel \(1.00 \mathrm{m},\) on average, relative to \(\mathrm{H}_{2}\) under these same conditions? c. (Challenging) What fraction of \(\mathrm{N}_{2}\) particles will require more than this average time to travel \(1.00 \mathrm{m} ?\) Answering this question will require evaluating a definite integral of the speed distribution, which requires using numerical methods such as Simpson's rule.

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