A gas of neon atoms \(\left({ }^{20} \mathrm{Ne}_{10}\right)\), at temperature \(T=330 \mathrm{~K}\) and pressure \(P=10^{5} \mathrm{~Pa}\), is confined to a cubic box of volume \(V=8 \times 10^{3} \mathrm{~cm}^{3}\). Assume the radius of the atoms is approximately \(a=1.5 \mathrm{~A} .(\mathrm{a})\) What is the average speed \((V)\) of the neon atoms? (b) What is the mean free path \(\lambda\) of the neon atoms? (c) Compute the coefficient of thermal conductivity using the kinetic theory result, \(\left(K_{\text {lth }}=1 / 2 n(v) A k_{B}\right)\), where \(n=N / V\) is the particle number density, and the result obtained from Boltzmann's equation \(\left(K_{\text {Beq }}=75 k_{\mathrm{B}} /\left(256 a^{2}\right) \sqrt{k_{\mathrm{B}} T /(m \pi)}\right)\). How do they compare with the experimentally observed value of \(K_{\text {exp }} \approx 5 \times 10^{-2} \mathrm{~W} /(\mathrm{m} \mathrm{K}) ?(\mathrm{~d})\) If a temperature difference of \(\Delta T=\) \(3.0 \mathrm{~K}\) is created between two opposite walls of this system, what is the heat current flowing through the gas (in units of \(\left.\mathrm{J} /\left(\mathrm{m}^{2} s\right)\right)\) ?

Step by step solution

01

Calculate the Average Speed (V) of Neon Atoms

The average speed of gas atoms can be determined using the equation for the root mean square speed: \[ V = \sqrt{ \frac{3k_B T}{m} } \]where \( k_B = 1.38 \times 10^{-23} \mathrm{~J/K} \) (Boltzmann constant), \( T = 330 \mathrm{~K} \) (temperature), and \( m = \frac{20 \times 1.66 \times 10^{-27}}{10} \mathrm{~kg} \) (mass of neon atoms).First, compute the mass of a neon atom: \[ m = \frac{20 \times 1.66 \times 10^{-27}}{10} = 3.32 \times 10^{-26} \mathrm{~kg} \]Next, substitute the values into the equation for average speed: \[ V = \sqrt{ \frac{3 \times 1.38 \times 10^{-23} \times 330}{3.32 \times 10^{-26}} } = \sqrt{ \frac{1.3644 \times 10^{-20}}{3.32 \times 10^{-26}} } = \sqrt{ 4.11 \times 10^{5} } = 640.95 \mathrm{~m/s} \]

02

Calculate the Mean Free Path (λ)

The mean free path \(\lambda\) for neon atoms can be determined using the formula: \[ \lambda = \frac{k_B T}{ \sqrt{2} \pi a^2 P } \] where \( a = 1.5 \times 10^{-10} \mathrm{~m} \) (radius of neon atom), \( P = 10^{5} \mathrm{~Pa} \).Substitute values and solve for \(\lambda\): \[ \lambda = \frac{1.38 \times 10^{-23} \times 330}{ \sqrt{2} \pi \times (1.5 \times 10^{-10})^2 \times 10^5 } = \frac{4.554 \times 10^{-21}}{ \sqrt{2} \pi \times 2.25 \times 10^{-20} } = \frac{4.554 \times 10^{-21}}{ 7.95 \times 10^{-20} } = 5.73 \times 10^{-2} \mathrm{~m} \]

03

Compute the Coefficient of Thermal Conductivity Using Kinetic Theory (Klth)

The kinetic theory formula is given by: \[ K_{\text{lth}} = \frac{1}{2} n V a k_B \] where \( n = \frac{N}{V} \) (particle number density), \( V = 8 \times 10^{-3} \mathrm{~m^3} \) (volume), \( k_B = 1.38 \times 10^{-23} \mathrm{~J/K} \), \( N = \frac{P V}{k_B T} \), \( T = 330 \mathrm{~K} \), and \( P = 10^5 \mathrm{~Pa} \).First, compute \(N\) and then \(n\): \[ N = \frac{10^5 \times 8 \times 10^{-3}}{1.38 \times 10^{-23} \times 330} = \frac{8 \times 10^2}{4.554 \times 10^{-21}} = 1.756 \times 10^{23} \] \[ n = \frac{1.756 \times 10^{23}}{8 \times 10^{-3}} = 2.195 \times 10^{25} \mathrm{~m^{-3}} \]Substitute \(n\) into the \(K_{\text{lth}}\) entropy: \[ K_{\text{lth}} = \frac{1}{2} \times 2.195 \times 10^{25} \times 640.95 \times 1.5 \times 10^{-10} \times 1.38 \times 10^{-23} \]= \frac{3.05 \times 10^{-2}} \mathrm{ W/(Km)}

04

Compute the Coefficient of Thermal Conductivity Using Boltzmann's Equation (KBeq)

The formula is: \[ K_{\text{Beq}} = \frac{75 k_B}{256 a^2} \sqrt{\frac{k_B T}{m \pi}} \] where \( k_B = 1.38 \times 10^{-23} \mathrm{~J/K} \) (Boltzmann constant), \( T = 330 \mathrm{~K} \), \( a = 1.5 \times 10^{-10} \mathrm{~m} \) (radius), and \( m = 3.32 \times 10^{-26} \mathrm{~kg} \).Substitute values and solve for \(K_{\text{Beq}}\): \[ K_{\text{Beq}} = \frac{75 \times 1.38 \times 10^{-23}}{256 \times (1.5 \times 10^{-10})^2} \times \sqrt{\frac{1.38 \times 10^{-23} \times 330}{3.32 \times 10^{-26} \pi}} \] \[ K_{\text{Beq}} = \frac{103.5 \times 10^{-23}}{5.76 \times 10^{-20}} \times \sqrt{\frac{4.554 \times 10^{-21}}{1.04 \times 10^{-25}}} \] \[ K_{\text{Beq}} = 1.80 \times 10^{-3} \times \sqrt{4.38 \times 10^{4}} \] \[ K_{\text{Beq}} = 1.8 \times 2.09 \times 10^{-2} \mathrm{~W/(mK)} \]

05

Compare the Results With Experimental Value

Compare \(K_{\text{lth}}\), \(K_{\text{Beq}}\) with \(K_{\text{exp}}\), which is approximately \(5 \times 10^{-2} \mathrm{~W/(mK)} \)

06

Calculate the Heat Current

The thermal conductivity can be used to find the heat current, given by \[ Q = K_{\text{exp}} \frac{A \Delta T}{d} \] For the cubic box, surface area \(A\) and thickness \(d\) can be calculated as \[ A = (8 \times 10^{-3})^{2/3} = 0.2 \text{~m}^2 \]\[ d = (8 \times 10^{-3})^{1/3} = 0.02 \text{~m}\]Substituting values into the formula gives \[ Q= 5 \times 10^{-13} \mathrm{~W}\]

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Key Concepts

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

average speed of gas particles

The average speed of gas particles is an essential concept in the kinetic theory of gases. It helps us understand how fast the particles are moving on average under a given set of conditions. We typically calculate this using the root mean square speed formula: \( V = \sqrt{ \frac{3k_B T}{m} } \) Here,

• \( k_B \) is the Boltzmann constant and has a value of \( 1.38 \times 10^{-23} \mathrm{~J/K} \)
• \( T \) is the temperature in Kelvin
• \( m \) is the mass of one gas atom, which for neon is calculated as: \( m = \frac{20 \times 1.66 \times 10^{-27}}{10} = 3.32 \times 10^{-26} \mathrm{~kg} \).

By plugging the numbers into the formula, neon atoms at \( 330 \mathrm{~K} \) have an average speed of approximately \( 640.95 \mathrm{~m/s} \). This high speed highlights how energetic gas particles are due to thermal energy.

mean free path

The mean free path is another critical concept in the kinetic theory of gases. It represents the average distance a gas particle will travel before colliding with another particle. This concept is necessary for understanding transport properties in gases such as diffusion, viscosity, and thermal conductivity. We calculate the mean free path using the following formula: \( \lambda = \frac{k_B T}{ \sqrt{2} \pi a^2 P } \) Here,

• \( k_B \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \mathrm{~J/K} \))
• \( T \) is the temperature in Kelvin \( (330 \mathrm{~K}) \)
• \( a \) is the radius of a gas atom \( (1.5 \times 10^{-10} \mathrm{~m}) \)
• \{ P \} is the pressure \( (10^{5} \mathrm{~Pa}) \)

By substituting the values, we find that the mean free path for neon atoms in this scenario is approximately \( 5.73 \times 10^{-2} \mathrm{~m} \). This indicates that neon atoms can, on average, travel about a few centimeters before colliding with another atom.

thermal conductivity

Thermal conductivity measures a material’s ability to conduct heat. In gases, this property depends on how quickly and how far the gas particles can travel. We can compute thermal conductivity in two ways using the kinetic theory of gases and Boltzmann's equation.

Kinetic Theory

Thermal conductivity according to kinetic theory is given by: \( K_{\text{lth}} = \frac{1}{2} n \text{V} a k_B \) Where,

• \( n \) is the particle number density
• \( V \) is the average speed
• \( a \) is the particle radius
• \( k_B \) is the Boltzmann constant

First, determine the particle number density \( n = \frac{N}{V} \), where \( N \) is the number of particles derived from \( PV = Nk_B T \). Then, we can calculate: \( K_{\text{lth}} = \frac{3.05 \times 10^{-2}} \mathrm{~W/(mK)} \).

Boltzmann's Equation

The second method uses Boltzmann's result: \( K_{\text{Beq}} = \frac{75 k_B}{256 a^2} \sqrt{\frac{k_B T}{m \pi}} \)Here, substituting the values, we get: \( K_{\text{Beq}} = 2.09 \times 10^{-2} \mathrm{~W/(mK)} \). Comparing the results helps us check the accuracy. The kinetic theory result is closer to the experimentally observed value of \( K_{\text{exp}} \approx 5 \times 10^{-2} \mathrm{~W/(mK)} \), suggesting that our theoretical approaches give us useful approximations.