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A scientifically untrained but curious friend asks, "When I walk into a room, is there a chance that all the air will be on the other side?" How do you answer this question?

Short Answer

Expert verified

One present only one microstate, mathematically it is one of infinity.

Step by step solution

01

microstates

Microstate is a term that describes the microscopic properties of a thermodynamic system

02

show, thatis there a chance that all the air will be on the other side?

Although it is theoretically possible, it is borderline non-realistic, so it is ought to be dismissed.

Of the infinitive microstates that air molecules can have in that room, this one present only one microstate, mathematically it is one of infinity.

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Most popular questions from this chapter

When a star has nearly bumped up its intimal fuel, it may become a white dwarf. It is crushed under its own enormous 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 energies of all the electrons (by shortening their wavelengths). If gravitation and the electron exclusion principle are the only factors, there is minimum 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
Uclocimns=310(3Ï€2h3me32V)23N53

Note that as the volume Vis decreased, the energy does increase. For a neutral star. the number of electrons, N, equals the number of protons. If 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

Uelectrons=9h280me(3Ï€2M5mp5)131R2

Where, R is the star's radius?

(b) The gravitational potential energy of a sphere of mass Mand radius Ris given by

Ugray=-35GM2R

Taking both factors into account, show that the minimum total energy occurs when

R=3h28G(3Ï€2me3mp5M)13

(c) Evaluate this radius for a star whose mass is equal to that of our Sun 2x1030kg.

(d) White dwarfs are comparable to the size of Earth. Does the value in part (c) agree?

What is special about a metastable Stale, and why is it so useful ina laser? Why wouldn't a non-metastable state at the same energy work?

This problem investigates what fraction of the available charge must be transferred from one conductor to another to produce a typical contact potential. (a) As a rough approximation treat the conductors as10cmx 10 cm square plates2cm apart-a parallel-plate capacitors so thatq=CV , where C=σ0(0.01m2/0.02m). How much charge must be transferred from one plate to the other to produce a potential difference of 2V?(b) Approximately what fraction would this be of the total number of conduction electrons in a 100gpiece of copper. which has one conduction electron per atom?

A particle subject to a planet's gravitational pull has a total mechanical energy given by Emechanical=12mv2-GMmr, whereis the particle's mass.M the planet's mass, and Gthe gravitational constant6.67×10-11N·m3/kg2. It may escape if its energy is zero that is, if its positive KE is equal in magnitude to the negative PE holding if to the surface. Suppose the particle is a gas molecule in an atmosphere.

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We claim that the famous exponential decrease of probability with energy is natural, the vastly most probable and disordered state given the constraints on total energy and number of particles. It should be a state of maximum entropy ! The proof involves mathematical techniques beyond the scope of the text, but finding support is good exercise and not difficult. Consider a system of11oscillators sharing a total energy of just5³óÓ¬0 . In the symbols of Section 9.3. N=11andM=5 .

  1. Using equation(9-9) , calculate the probabilities ofn , being0,1,2, and3 .
  2. How many particlesNn , would be expected in each level? Round each to the nearest integer. (Happily. the number is still 11. and the energy still5³óÓ¬0 .) What you have is a distribution of the energy that is as close to expectations is possible. given that numbers at each level in a real case are integers.
  3. Entropy is related to the number of microscopic ways the macro state can be obtained. and the number of ways of permuting particle labels withN0 ,N1,N2 , and N3fixed and totaling11 is11!(N0!N1!N2!N3!) . (See Appendix J for the proof.) Calculate the number of ways for your distribution.
  4. Calculate the number of ways if there were6 particles inn=0.5 inn=1 and none higher. Note that this also has the same total energy.
  5. Find at least one other distribution in which the 11 oscillators share the same energy, and calculate the number of ways.
  6. What do your finding suggests?

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