/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Show that the rms speed of a gas... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that the rms speed of a gas molecule, defined as \(v_{\mathrm{mm}} \equiv \sqrt{\bar{v}^{2}},\) is given by \(\sqrt{3 k_{\mathrm{B}} T / m}\).

Short Answer

Expert verified
The root mean square speed of a gas molecule \(v_{\mathrm{rms}}\) is given by \(\sqrt{3 k_{\mathrm{B}} T / m}\)

Step by step solution

01

Understanding the Concept of rms Speed and Average of Squared Speed

The root mean square (rms) speed of a gas molecule is defined as the square root of the average of the squares of the speeds of the molecules. The kinetic theory of gases provides a relationship between the speed of molecules and the macroscopic properties of a gas, such as temperature and pressure. The rms speed of a gas molecule can be represented as \(v_{\mathrm{rms}} = \sqrt{{v}^{2}}\).
02

Relate You RMS Speed to the Kinetic Energy

The definition of the rm speed is comparative to the definition of kinetic energy. In kinetic theory of gases, each molecule of an ideal gas has average kinetic energy equivalent to \(\frac{3}{2}\) times Boltzmann’s constant, \(k_{\mathrm{B}}\) times the temperature of gas in Kelvins, \(T\). Hence the average kinetic energy can be mentioned as \( \frac{1}{2} m {v}^{2} = \frac{3}{2} k_{\mathrm{B}} T \). Where \(m\) is the mass of each molecule and \(v\) is the speed of the molecule.
03

Rearrange the Equation and Solve the rms Speed

Rearrange the equation obtained in step 2 which gives \(v^{2} = 3k_{\mathrm{B}} T / m\). Where \(v = \sqrt{{v}^{2}}\), so \(v_{\mathrm{rms}} = \sqrt{3 k_{\mathrm{B}} T / m}\)

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Key Concepts

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

Kinetic Theory of Gases
The kinetic theory of gases is a powerful tool in the field of physics that helps us understand the behavior of gases at the molecular level. According to this theory, gas molecules are considered to be in constant, random motion, colliding with each other and with the walls of their container. This microscopic motion is connected to macroscopic properties like temperature, pressure, and volume.

One of the key ideas of the kinetic theory is that the temperature of a gas is directly proportional to the average kinetic energy of its molecules. When the temperature increases, the molecules move faster, thus increasing their kinetic energy and, consequently, the gas pressure if the volume is kept constant. This relationship is crucial in deriving the root mean square (rms) speed of gas molecules, providing a link between the micro and macro worlds of physics.

The rms speed is particularly useful because it gives us an idea of the speed of a 'typical' molecule in the gas, as opposed to simply knowing the average or most probable speed. By combining this with the mass of a molecule, we can find out a great deal about its kinetic energy and how that relates to the observed properties of the gas.
Boltzmann's Constant
Boltzmann's constant (\(k_{\rm B}\)) is a fundamental constant in physics that appears in various equations relating to the kinetic theory of gases. It acts as a bridge between the micro-scale behavior of individual atoms or molecules and the macro-scale properties of the gas.

Numerically, it's valued at approximately \(1.38 \times 10^{-23} J/K\), where J stands for joules and K for kelvins. Boltzmann's constant essentially tells us how much kinetic energy there is per degree of temperature in each molecule. A key formula in this context is the expression for the average kinetic energy (\(E_{k}\)) of a gas molecule: \(E_{k} = \frac{3}{2} k_{\rm B} T\), where \(T\) is the temperature of the gas in kelvins. This pivotal equation demonstrates how the motions of individual molecules relate to the temperature of a gas, providing a fundamental understanding of thermodynamics.
Average Kinetic Energy
Average kinetic energy is a concept that ties together the microscopic and macroscopic views of gas. It represents the average amount of energy associated with the motion of the molecules in a gas. Kinetic energy is given by the formula \( \frac{1}{2} mv^{2} \), where \(m\) is the mass of a molecule and \(v\) is its velocity.

The average kinetic energy of gas molecules depends solely on the temperature of the gas, regardless of the type of gas or the number of molecules present. This is due to the fact that temperature is a measure of the average kinetic energy per particle in the system. Hence, at a given temperature, all gases have the same average kinetic energy, which is represented by the equation \( \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} k_{\rm B} T \). The bar above \(v^{2}\) signifies that it's the average of the squared velocities of the molecules. Understanding this concept is crucial when solving problems involving the rms speed of gas molecules.
Macroscopic Properties of Gas
Macroscopic properties of gas refer to the physical attributes of a gas that can be observed and measured without the need to look at individual molecules. These include temperature, pressure, volume, and moles of gas, commonly known as the variables in the ideal gas law: \(PV=nRT\).

Temperature, as we have learned, is a measure of the average kinetic energy of the particles in the gas. Pressure is the force exerted by the gas molecules when they collide with the walls of their container. Volume is the space occupied by the gas, and moles (n), a fundamental unit in chemistry, represent the amount of substance or the number of molecules present.

Understanding these macroscopic properties helps in connecting molecular-level phenomena, such as individual molecule speeds and energies, to larger-scale behaviors that we can observe and measure. Being able to calculate quantities like the rms speed of gas molecules - which is derived from kinetic theory and involves Boltzmann's constant - demonstrates the intimate relationship between the microscopic and macroscopic domains in the study of gases.

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Most popular questions from this chapter

Pruve that fur any sine function \(\sin (k x+\phi)\) of wavelength shonter than \(2 a\), where \(a\) is the atomic spacing. there is a sine function with a wavelength longer than \(2 a\) that has the same values at the points \(x=a .20,3 a\). and so on. Wotr: It is probably easier to work with wave number than with wavelength. We seck to show that for every wave number greater than \(\pi / a\) there is an equivalent une less than \(\pi / a\).)

Consider the two-sided room, (a) Which is more likely to have an imbalance of five particles (ie., \(N_{\mathrm{k}}=\frac{1}{2} N+5\) ): a room with \(N=20\) or a room with \(N=60\) ? (Note: The total number of ways of distributing particles, the sum of \(W_{N_{k}}^{N}\) from \(O\) to \(N,\) is \(2^{N}\).) (b) Which is more likely to have an imbalance of \(5 \%\) (i.e., \(\left.N_{R}=\frac{1}{2} N+0.05 N\right) ?\) (c) An average-size room is quite likely to have a trillion more air molecules on one side than on the other. Why may we say that precisely half will be on each side?

When a star has nearly bumed up its intemal fuel, it may become a white dwarf. It is crushed under its own enonnous gravitational forces to the point at which the exclusion principle for the electrons becomes a factor. A smaller size would decrease the gravitational potential energy, but assuming the electrons to be packed into the lowest energy states consistent with the exclusion principle. "squeezing" the potential well necessarily increases the ener gies of all the electrons (by shortening their wavelengths). If gravitation and the electron exclusion principle are the only factors, there is a mini. mum total energy and corresponding equilibrium radius. (a) Treat the electrons in a white dwarf as a quantum gas. The minimum energy allowed by the exclusion principle (see Exercise 67 ) is $$ U_{\text {elecimns }}=\frac{3}{10}\left(\frac{3 \pi^{2} \hbar^{3}}{m_{e}^{3 / 2} V}\right)^{2 / 3} N^{5 / 3} $$ Note that as the volume \(V\) is decreased, the energy does increase. For a neutral star. the number of electrons, \(N\), equals the number of protons. Assuming that protons account for half of the white dwarf's mass \(M\) (neutrons accounting for the other half). show that the minimum electron energy may be written $$ U_{\text {electrons }}=\frac{9 \hbar^{2}}{80 m_{e}}\left(\frac{3 \pi^{2} M^{5}}{m_{\mathrm{p}}^{5}}\right)^{1 / 3} \frac{1}{R^{2}} $$ where \(R\) is the star's radius. (b) The gravitational potential energy of a sphere of mass \(M\) and radius \(R\) is given by $$ U_{\operatorname{mav}}=-\frac{3}{5} \frac{G M^{2}}{R} $$ Taking both factors into account, show that the minimum total energy occurs when $$ R=\frac{3 h^{2}}{8 G}\left(\frac{3 \pi^{2}}{m^{3} m_{p}^{5} M}\right)^{1 / 3} $$ (c) Evaluate this radius for a star whose mass is equal to that of our Sun, \(\sim 2 \times 10^{30} \mathrm{~kg}\). (d) White dwarfs are comparable to the size of Eath. Does the value in part (c) agree?

A two-sided room contains six particles, \(a, b, c, d\). \(e .\) and \(f\), with two on the left and four on the right. (a) Describe the macrostate. (b) Identify the possible microstates. (Note: With only six particles, this isn't a thermodynamic system, but the general idea stitl applies. and the number of combinations is tractable.)

The entropy of an ideal monatomic gas is \((3 / 2) N k_{B} \ln E+\) \(N k_{\mathrm{B}} \ln V-N k_{\mathrm{B}} \ln N,\) to within an additive constant. Show that this implies the correct relationship between internal energy \(E\) and temperature.

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