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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