Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Use Boltzmann factors to derive the exponential formula for the density of an isothermal atmosphere, already derived in Problems 1.16 and 3.37. (Hint: Let the system be a single air molecule, let s1 be a state with the molecule at sea level, and let s2 be a state with the molecule at height z.)

Short Answer

Expert verified

Therefore, the exponential formula for the density of an isothermal atmosphere is:ρ(z)=ρ(0)e-mgz/kT

Step by step solution

01

Given information

Let the system be a single air molecule, let S1be a state with the molecule at sea level, and let S2 be a state with the molecule at height z.

02

Explanation

Consider a system with a single air molecule, where S1 is the state when the molecule is at sea level and S2is the state when the molecule is at a height of 2. Assume that the energy is only potential energy, so the difference in energy between the states S1and S2is the potential energy, which is ΔE=mgz and the ratio of S2state probability to state s1 probability is:

role="math" localid="1647369610229" Ps2Ps1=e-E2/kTe-E1/kT=e-ΔE/kTPs2Ps1=e-mgz/kT

This means that the air molecule is less likely to be at height of z than at the see level by a factor of e-mgz/kT, and that the number of molecules per unit volume at height of z is also smaller than the see level by the same ratio in the isothermal atmosphere, so:

ρ(z)=ρ(0)e-mgz/kT

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For a CO molecule, the constant ϵis approximately 0.00024eV. (This number is measured using microwave spectroscopy, that is, by measuring the microwave frequencies needed to excite the molecules into higher rotational states.) Calculate the rotational partition function for a COmolecule at room temperature (300K), first using the exact formula 6.30 and then using the approximate formula 6.31.

The most common measure of the fluctuations of a set of numbers away from the average is the standard deviation, defined as follows.

(a) For each atom in the five-atom toy model of Figure 6.5, compute the deviation of the energy from the average energy, that is, Ei-E¯,fori=1to5. Call these deviations ΔEi.

(b) Compute the average of the squares of the five deviations, that is, ΔEi2¯. Then compute the square root of this quantity, which is the root-mean- square (rms) deviation, or standard deviation. Call this number σE. Does σEgive a reasonable measure of how far the individual values tend to stray from the average?

(c) Prove in general that

σE2=E2¯-(E¯)2

that is, the standard deviation squared is the average of the squares minus the square of the average. This formula usually gives the easier way of computing a standard deviation.

(d) Check the preceding formula for the five-atom toy model of Figure 6.5.

Show explicitly from the results of this section thatG=Nμfor an ideal gas.

Although an ordinary H2 molecule consists of two identical atoms, this is not the case for the molecule HD, with one atom of deuterium (i.e., heavy hydrogen, 2H). Because of its small moment of inertia, the HD molecule has a relatively large value of ϵ:0.0057eV At approximately what temperature would you expect the rotational heat capacity of a gas of HD molecules to "freeze out," that is, to fall significantly below the constant value predicted by the equipartition theorem?

Consider a classical particle moving in a one-dimensional potential well ux, as shown. The particle is in thermal equilibrium with a reservoir at temperature T, so the probabilities of its various states are determined by Boltzmann statistics.

{a) Show that the average position of the particle is given by

x=xe-βuxdxe-βuxdx

where each integral is over the entire xaxis.

A one-dimensional potential well. The higher the temperature, the farther the particle will stray from the equilibrium point.

(b) If the temperature is reasonably low (but still high enough for classical mechanics to apply), the particle will spend most of its time near the bottom of the potential well. In that case we can expand u(z) in a Taylor series about the equilibrium point

ux=ux0+x-x0dudxx0+12x-x02d2udx2x0+13!x-x03d3udx3x0+........

Show that the linear term must be zero, and that truncating the series after the quadratic term results in the trivial prediction x=x0.

(c) If we keep the cubic term in the Taylor series as well, the integrals in the formula for xbecome difficult. To simplify them, assume that the cubic term is small, so its exponential can be expanded in a Taylor series (leaving the quadratic term in the exponent). Keeping only the smallest temperature-dependent term, show that in this limit x differs from by a term proportional to kT. Express the coefficient of this term in terms of the coefficients of the Taylor series ux

(d) The interaction of noble gas atoms can be modeled using the Lennard Jones potential,

ux=u0x0x12-2x0x6

Sketch this function, and show that the minimum of the potential well is at x=x0, with depth u0. For argon, x0=3.9Aand u0=0.010eV. Expand the Lennard-Jones potential in a Taylor series about the equilibrium point, and use the result of part ( c) to predict the linear thermal expansion coefficient of a noble gas crystal in terms of u0. Evaluate the result numerically for argon, and compare to the measured value α=0.0007K-1

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free