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Chapter 9: 41E (page 405)

Show that the rms speed of a gas molecule, defined as $v_{rms} \equiv \sqrt{v^{2}}$ , is given by $\sqrt{\frac{3 k_{B} T}{m}}$ .

Short Answer

Expert verified

The rms speed of a gas molecule is $\sqrt{\frac{3 k_{B} T}{m}}$

Step by step solution

01

Maxwell Probability Distribution

$v_{rms} = \sqrt{\bar{v_{2}}}$

$v_{rms} = {(\int_{0}^{\infty} v^{2} P (v) dv)}^{\frac{1}{2}}$ ……(1)

Where,

P(v)is the Maxwell probability distribution, which is given by

$P (v) = {(\frac{m}{2 {πk}_{B} T})}^{\frac{3}{2}} 4 {πv}^{2} e^{\frac{- {mv}^{2}}{2 k_{B} T}}$ …..(2)

Where,

mis the mass of the particle.

vis velocity of particle.

k_Bis Boltzmann constant.

02

determine the speed of gas

$v_{rms} = {(\int_{0}^{\infty} v^{2} {(\frac{m}{2 {πk}_{B} T})}^{\frac{3}{2}} 4 {πv}^{2} e^{- \frac{m^{2}}{2 k_{B} T}} dv)}^{\frac{1}{2}}$

Let $b = \frac{1}{2 a^{2}} = \frac{m}{2 k_{B} T}$

$\begin{array}{rcl} v_{rms} & = & {(4 π {(\frac{b}{π})}^{\frac{3}{2}} \int_{0}^{\infty} v^{4} e^{- {bv}^{2}} dv)}^{\frac{1}{2}} \\ = & {(4 π {(\frac{b}{π})}^{\frac{3}{2}} \frac{3}{8} \sqrt{\frac{π}{b^{5}}})}^{\frac{1}{2}} \\ = & {(\frac{3}{2 b})}^{\frac{1}{2}} \\ = & \sqrt{\frac{3 k_{B} T}{m}} \end{array}$

Therefore, The rms speed of a gas molecule is $\sqrt{\frac{3 k_{B} T}{m}}$ .

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Most popular questions from this chapter

In Exercise 35, a simple two-state system is studied. Assume that the particles are distinguishable. Determine the molar specific heat $C_{v}$ of this material and plot it versus T. Explain qualitatively why it should behave as it does.

Discusses the energy balance in a white dwarf. The tendency to contract due to gravitational attraction is balanced by a kind of incompressibility of the electrons due to the exclusion principle.

(a) Matter contains protons and neutrons, which are also fanions. Why do the electrons become a hindrance to compression before the protons and neutrons do?

(b) Stars several times our Sun's mass has sufficient gravitational potential energy to collapse further than a white dwarf; they can force essentially all their matter to become neutrons (formed when electrons and protons combine). When they cool off, an energy balance is reached like that in the white dwarf but with the neutrons filling the role of the incompressible fermions. The result is a neutron star. Repeat the process of Exercise 89. but assume a body consisting solely of neutrons. Show that the equilibrium radius is given by

$R = \frac{3 ℏ^{2}}{2 G} {(\frac{3 π^{2}}{2 m_{n}^{8} M})}^{1 / 3}$

(c) Show that the radius of a neutron star whose mass is twice that of our Sun is only about $10 km$ .

By carrying out the integration suggested just before equation (9-28), show that the average energy of a one-dimensional oscillator in the limit $k_{B} T ≫ {hω}_{0}$ is $k_{B} T$ .

Classically, what would be the average energy of a particle in a system of particles fine to move in the xy-plane while rotating about the $i$ -axis?

Using density of states $D (E) = \frac{(2 s + 1)}{{hω}_{0}}$ , which generalizes equation (9-27) to account for multiple allowed spin states (see Exercise 52), the definition ${Nhω}_{0} / (2 s + 1) = ε$ and $N = \int_{0}^{\infty} N (E) D (E) dE$ . Solve for $B$ in distributions (9-32) and (9-33) careful use of $\pm$ will cut your work by about half. Then plug back in and show that for a system of simple harmonic oscillators, the distributions become $γ {(E)}_{BE} = \frac{1}{\frac{e^{E / k_{B} T}}{1 - e^{- δ / k_{B} T}} - 1} and N {(E)}_{FD} = \frac{1}{\frac{e^{E / k_{B} T}}{e^{+ δ / k_{B} T} - 1} + 1}$ .

You will need the following integral: $\int_{0}^{\infty} {({Be}^{2} \pm 1)}^{- 1} dz =$ $\pm \ln (1 \pm \frac{1}{B})$ .

See all solutions

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