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Chapter 25: Problem 42

Calculate the pressures at which the mean free path of a hydrogen molecule will be \(100 \mu \mathrm{m}, 1.00 \mathrm{mm},\) and \(1.00 \mathrm{m}\) at \(20^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Pressures are approximately 380 Pa, 38 Pa, and 0.038 Pa for mean free paths of 100 \(\mu\)m, 1 mm, and 1 m, respectively.

Step by step solution

01

Understand the Mean Free Path Formula

The mean free path \( \lambda \) of gas molecules is described by the formula \( \lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} \), where \( k_B \) is the Boltzmann constant \( 1.38 \times 10^{-23} \text{ J/K} \), \( T \) is the temperature in Kelvin, \( d \) is the diameter of a molecule, and \( P \) is the pressure.
02

Convert Temperature to Kelvin

The given temperature is \(20^{\circ} \text{C}\). To convert it to Kelvin, use the formula \(T = 20 + 273.15 = 293.15 \text{ K}\).
03

Use Hydrogen Molecule Diameter

The diameter \(d\) for a hydrogen molecule is approximately \(2.9 \times 10^{-10} \ m\).
04

Rearrange the Mean Free Path Formula

For pressure \(P\), the formula can be rearranged to \( P = \frac{k_B T}{\sqrt{2} \pi d^2 \lambda} \).
05

Calculate Pressure for \(\lambda = 100 \ \mu m\)

Substitute \(\lambda = 100 \times 10^{-6} \ m\) and other known values into the formula: \( P = \frac{1.38 \times 10^{-23} \times 293.15}{\sqrt{2} \pi (2.9 \times 10^{-10})^2 (100 \times 10^{-6})} \approx 380.38 \text{ Pa}.\)
06

Calculate Pressure for \(\lambda = 1.00 \ mm\)

Set \(\lambda = 1.00 \times 10^{-3} \ m\): \( P = \frac{1.38 \times 10^{-23} \times 293.15}{\sqrt{2} \pi (2.9 \times 10^{-10})^2 (1.00 \times 10^{-3})} \approx 38.038 \text{ Pa}.\)
07

Calculate Pressure for \(\lambda = 1.00 \ m\)

For \(\lambda = 1.00 \ m\): \( P = \frac{1.38 \times 10^{-23} \times 293.15}{\sqrt{2} \pi (2.9 \times 10^{-10})^2 (1.00)} \approx 0.038038 \text{ Pa}.\)

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Key Concepts

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

Boltzmann Constant
The Boltzmann constant is a fundamental physical constant central to the understanding of thermodynamics and statistical mechanics. It is denoted as \( k_B \) and has a value of \( 1.38 \times 10^{-23} \) joules per Kelvin (J/K). Essentially, it helps in translating the macroscopic readings of temperature to the microscopic energy scales of individual particles.
  • In equations like the mean free path, it often appears in the numerator, correlating the product of energy units and temperature.
  • It is crucial for linking the average kinetic energy of particles with the temperature to calculate phenomena on a molecular scale.
Understanding this constant helps students visualize how energy behavior at the particle level influences observable thermal properties.
Hydrogen Molecule
Hydrogen, represented as \(H_2\), is the simplest and most abundant molecule in the universe. In physics problems concerning gas molecules, the hydrogen molecule is often chosen due to its small size and simplicity. Here's what you need to know:
  • Molecular diameter for hydrogen is approximately \(2.9 \times 10^{-10}\) meters. This tiny value is essential for calculating characteristics like mean free path in gaseous states.
  • The properties of hydrogen, such as mass and size, are important for theoretical models predicting behavior under various conditions like pressure and temperature.
This knowledge of the hydrogen molecule allows us to better understand fundamental gas laws and how they are applied to real-world scenarios.
Pressure Calculation
Calculating the pressure in a system involves understanding the relationship between the mean free path \( \lambda \) and the other variables in the formula. Pressure \( P \) is calculated using the equation: \[ P = \frac{k_B T}{\sqrt{2} \pi d^2 \lambda} \]
  • \( \lambda \) represents the mean free path, the average distance a molecule travels between collisions.
  • \( T \) is the temperature in Kelvin, which we previously converted.
  • \( d \) is the molecular diameter, specific to hydrogen in this exercise.
This formula showcases how pressure depends inversely on the mean free path. As \( \lambda \) increases, it suggests fewer collisions in the path due to decreased pressure.
Temperature Conversion
Proper conversion of temperature from Celsius to Kelvin is necessary for accuracy in any physics calculation involving temperature. The conversion formula is straightforward: \[ T = \text{{Temperature in Celsius}} + 273.15 \]
  • This formula is rooted in the absolute temperature scale, which begins at absolute zero, eliminating negative temperature values often found in Celsius.
  • Converting to Kelvin allows consistency across equations involving gas laws and thermal dynamics.
By properly converting temperatures, we ensure that pressure and energy are correctly calculated, maintaining the integrity of the physical quantities involved.

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Most popular questions from this chapter

At an altitude of \(150 \mathrm{km}\), the pressure is about \(2 \times 10^{-6}\) torr and the temperature is about \(500 \mathrm{K}\). Assuming for simplicity that the air consists entirely of nitrogen, calculate the mean free path under these conditions. What is the average collision frequency?

We can use Equation 25.33 to calculate the probability that the \(x\) -component of the velocity of a molecule lies within some range. For example, show that the probability that \(-u_{x 0} \leq u_{x} \leq u_{x 0}\) is given by \\[ \begin{aligned} \operatorname{Prob}\left\\{-u_{x 0} \leq u_{x} \leq u_{x 0}\right\\} &=\left(\frac{m}{2 \pi k_{\mathrm{B}} T}\right)^{1 / 2} \int_{-u_{x 0}}^{u_{x 0}} e^{-m u_{x}^{2} / 2 k_{\mathrm{B}} T} d u_{x} \\ &=2\left(\frac{m}{2 \pi k_{\mathrm{B}} T}\right)^{1 / 2} \int_{0}^{u_{x 0}} e^{-m u_{x}^{2} / 2 k_{\mathrm{B}} T} d u_{x} \end{aligned} \\] Now let \(m u_{x}^{2} / 2 k_{\mathrm{B}} T=w^{2}\) to get the cleaner-looking expression \\[ \operatorname{Prob}\left\\{-u_{x 0} \leq u_{x} \leq u_{x 0}\right\\}=\frac{2}{\pi^{1 / 2}} \int_{0}^{w_{0}} e^{-w^{2}} d w \\] where \(w_{0}=\left(m / 2 k_{\mathrm{B}} T\right)^{1 / 2} u_{x 0}\) It so happens that the above integral cannot be evaluated in terms of any function that we have encountered up to now. It is customary to express the integral in terms of a new function called the error function, which is defined by \\[ \operatorname{erf}(z)=\frac{2}{\pi^{1 / 2}} \int_{0}^{z} e^{-x^{2}} d x \\] The error function can be evaluated as a function of \(z\) by evaluating its defining integral numerically. Some values of erf(z) are $$\begin{array}{cccc} z & \operatorname{erf}(z) & z & \operatorname{erf}(z) \\ \hline 0.20 & 0.22270 & 1.20 & 0.91031 \\ 0.40 & 0.42839 & 1.40 & 0.95229 \\ 0.60 & 0.60386 & 1.60 & 0.97635 \\ 0.80 & 0.74210 & 1.80 & 0.98909 \\ 1.00 & 0.84270 & 2.00 & 0.99532 \end{array}$$ Now show that \\[ \operatorname{Prob}\left\\{-u_{x 0} \leq u_{x} \leq u_{x 0}\right\\}=\operatorname{erf}\left(w_{0}\right) \\] Calculate the probability that \(-\left(2 k_{\mathrm{B}} T / m\right)^{1 / 2} \leq u_{x} \leq\left(2 k_{\mathrm{B}} T / m\right)^{1 / 2} ?\)

Consider a mixture of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g})\). Calculate the ratio of the root-mean-square speed of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g})\) molecules in the reaction mixture.

The speed of sound in an ideal monatomic gas is given by $$u_{\text {sound }}=\left(\frac{5 R T}{3 M}\right)^{1 / 2}$$ Derive an equation for the ratio \(u_{\mathrm{rms}} / u_{\text {sound }^{*}}\) Calculate the root-mean-square speed for an argon atom at \(20^{\circ} \mathrm{C}\) and compare your answer to the speed of sound in argon.

This problem deals with the idea of the escape velocity of a particle from a body such as the Earth's surface. Recall from your course in physics that the potential energy of two masses, \(m_{1}\) and \(m_{2},\) separated by a distance \(r\) is given by $$V(r)=-\frac{G m_{1} m_{2}}{r}$$ (note the similarity with Coulomb's law) where \(G=6.67 \times 10^{-11} \mathrm{J} \cdot \mathrm{m} \cdot \mathrm{kg}^{-1}\) is called the gravitional constant. Suppose a particle of mass \(m\) has a velocity \(u\) perpendicular to the Earth's surface. Show that the minimum velocity that the particle must have in order to escape the Earth's surface (its escape velocity) is given by \\[ u=\left(\frac{2 G M_{\text {earth }}}{R_{\text {earth }}}\right)^{1 / 2} \\] Given that \(M_{\text {earth }}=5.98 \times 10^{24} \mathrm{kg}\) is the mass of the Earth and \(R_{\text {earh }}=6.36 \times 10^{6} \mathrm{m}\) is its mean radius, calculate the escape velocity of a hydrogen molecule and a nitrogen molecule. What temperature would each of these molecules have to have so that their average speed exceeds their escape velocity?
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