Fill in the algebraic steps to derive the Sackur-Tetrode equation(2.49).

Short Answer

Expert verified
  • The Sackur -Tetrode equation is

S=NklnVN4h232mπU3N32+52

Step by step solution

01

The entropy of substance 

  • The entropy of a substance is given as:

S=kln(Ω) (1)

  • where Ωis the number of microstates accessible to the substance. For a 3-dideal gas, this is given by Schroeder's equation 2.40:

Ω=VNN!h3N(2mπU)3N23N2! (2)

  • where Vis the volume, Uis the energy, Nis the number of molecules, mis the mass of a single molecule and his Planck's constant. We can further approximate this formula by using Stirling's approximation for the factorials:

n!2πnnnen

we get,

role="math" localid="1650298531903" N!2πNNNeN3N2!2π3N23N23N2e3N2 (3)&(4)

02

The entropy of a substance expression

3N2!N!2π3N23N23N2e3N22πNNNeN3N2!N!2π3N23N+12NN+12e5N2

  • When Nis large, we can throw away a couple of factors:

3N2!N!3N23N2NNe5N23N2!N!(3)3N2(2)3N2N5N2C25N (5)

substitute from (5)into (2), we get:

ΩVNh3N(2mπU)3N2(3)3N2(2)3N2N5N26N2ΩVN(mπU)3N2h3N(2)3N2(3)3N2(2)3N2N5N2e5N2ΩVN(mπU)3N2h3N(2)3Ne5N2(3)3N2N5N2

03

The entropy of a substance expression

ΩVN2h3NmπU33N2eN5N2ΩVN2h3mπU3N32Ne5N2ΩVN2h232mπU3N32Ne5N2ΩVN4h232mπU3N32Ne5N2ΩVN4mπU3h2N32Ne5N2

take the natural logarithm for both sides, and take into account role="math" localid="1650302000554" ln(ab)=ln(a)+ln(b)andlnab=ln(a)ln(b), so:

ln(Ω)NlnVN4h232mπU3N32+lne5N2ln(Ω)NlnVN4h232mπU3N32+5N2ln(Ω)NlnVN4h232mπU3N32+52

  • This gives the entropy of an ideal gas (from equation (1)) as:

S=NklnVN4h232mπU3N32+52

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Most popular questions from this chapter

Use a computer to plot formula 2.22directly, as follows. Define z=q_{A}/q, so that (1-z)=q_{B}/q. Then, aside from an overall constant that we'll ignore, the multiplicity function is [4z(1-z)]N, where zranges from 0to1and the factor of 4ensures that the height of the peak is equal to 1for any N. Plot this function forN=1,10,100,1000, and 10,000. Observe how the width of the peak decreases asNincreases.

Use Stirling's approximation to find an approximate formula for the multiplicity of a two-state paramagnet. Simplify this formula in the limit NNto obtain ΩNe/NN. This result should look very similar to your answer to Problem 2.17; explain why these two systems, in the limits considered, are essentially the same.

The mixing entropy formula derived in the previous problem actually applies to any ideal gas, and to some dense gases, liquids, and solids as well. For the denser systems, we have to assume that the two types of molecules are the same size and that molecules of different types interact with each other in the same way as molecules of the same type (same forces, etc.). Such a system is called an ideal mixture. Explain why, for an ideal mixture, the mixing entropy is given by

ΔSmixing=klnNNA

where Nis the total number of molecules and NAis the number of molecules of type A. Use Stirling's approximation to show that this expression is the same as the result of the previous problem when both Nand NAare large.

Use a computer to reproduce the table and graph in Figure2.4: two Einstein solids, each containing three harmonic oscillators, with a total of six units of energy. Then modify the table and graph to show the case where one Einstein solid contains six harmonic oscillators and the other contains four harmonic oscillators (with the total number of energy units still equal to six). Assuming that all microstates are equally likely, what is the most probable macrostate, and what is its probability? What is the least probable macrostate, and what is its probability?

Consider again the system of two large, identical Einstein solids treated in Problem 2.22.

(a) For the case N=1023, compute the entropy of this system (in terms of Boltzmann's constant), assuming that all of the microstates are allowed. (This is the system's entropy over long time scales.)

(b) Compute the entropy again, assuming that the system is in its most likely macro state. (This is the system's entropy over short time scales, except when there is a large and unlikely fluctuation away from the most likely macro state.)

(c) Is the issue of time scales really relevant to the entropy of this system?

(d) Suppose that, at a moment when the system is near its most likely macro state, you suddenly insert a partition between the solids so that they can no longer exchange energy. Now, even over long time scales, the entropy is given by your answer to part (b). Since this number is less than your answer to part (a), you have, in a sense, caused a violation of the second law of thermodynamics. Is this violation significant? Should we lose any sleep over it?

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