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GR8677 #87
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Problem
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Statistical Mechanics }Diatomic Molecules
Vibrational energy of a diatomic molecule goes to 0 at low temperatures. Thus, the springy dumbbell would be approximately a rigid dumbbell at a low-enough temperature. The other choices are too specific, and thus (E), the most general, must be it.
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Alternate Solutions |
smokwzbroiplytowej
2008-10-26 20:53:14 | (Ref. 'Modern Physics 3rd ed, Serway, Moses, Moyer', Chapter 11, p. 378 onward)
(A)
The rigid dumbbell has two rotational degrees of freedom (the rotation around the axis is not 'activated' - see below) each contributing  energy (equipartition theorem), so:
(A) is incorrect
(B) Model II has an additional degree of freedom because of the 'springy' interaction, so by an analogous calculation as above, we see that it has a greater specific heat than Model I - (B) is incorrect as well.
(C) Model I is not always correct. In fact it will fail whenever the vibrations of the molecule become important. Answer (C) is incorrect.
(D) Model II is not always correct either - it fails anytime the potential between the molecules is not approximated correctly by a harmonic potential. Answer (D) is incorrect.
(E) The choice between Models I and II does depend on the temperature, so (E) is correct.
Consider the harmonic potential of Model II. At typical energies:  , is low compared to the first excited state of the harmonic oscillator ( is determined by the particle, for details see the Serway reference) and Model I is a good approximation. For higher temperatures, Model II will be better.
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Note, that classically the rigid dumbbell should have 3 rotational degrees of freedom, because equipartition of energy in the classical model does not need 'activation'. Again, see the Serway chapter. |  |
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Comments |
tau1777
2008-11-04 15:54:40 | elimination
a) model I has five degrees of freedom, 3 translational and 2 rotational. specific heat at constant volume should be degrees of freedom/2 (Nk) , getting this from specific heat is temperature derivative of U (internal energy) also only at constant volume.
b) model II has 2 extra vibrational degrees of freedom than model I because of spring, so specific heat is 7/2Nk.
c) and d) if this were true why have two models
e) there's you're answer. |  | smokwzbroiplytowej
2008-10-26 20:53:14 | (Ref. 'Modern Physics 3rd ed, Serway, Moses, Moyer', Chapter 11, p. 378 onward)
(A)
The rigid dumbbell has two rotational degrees of freedom (the rotation around the axis is not 'activated' - see below) each contributing energy (equipartition theorem), so:
(A) is incorrect
(B) Model II has an additional degree of freedom because of the 'springy' interaction, so by an analogous calculation as above, we see that it has a greater specific heat than Model I - (B) is incorrect as well.
(C) Model I is not always correct. In fact it will fail whenever the vibrations of the molecule become important. Answer (C) is incorrect.
(D) Model II is not always correct either - it fails anytime the potential between the molecules is not approximated correctly by a harmonic potential. Answer (D) is incorrect.
(E) The choice between Models I and II does depend on the temperature, so (E) is correct.
Consider the harmonic potential of Model II. At typical energies: , is low compared to the first excited state of the harmonic oscillator ( is determined by the particle, for details see the Serway reference) and Model I is a good approximation. For higher temperatures, Model II will be better.
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Note, that classically the rigid dumbbell should have 3 rotational degrees of freedom, because equipartition of energy in the classical model does not need 'activation'. Again, see the Serway chapter. | |
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