GREPhysics.NET: GR8677 #87

GR8677 #87

Problem

GREPhysics.NET Official Solution

Statistical Mechanics $\Rightarrow$ }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.

Alternate Solutions

smokwzbroiplytowej
2008-10-26 20:53:14

(Ref. 'Modern Physics 3rd ed, Serway, Moses, Moyer', Chapter 11, p. 378 onward)

(A)

$c_V=\frac{\partial u}{\partial T}_v$

The rigid dumbbell has two rotational degrees of freedom (the rotation around the axis is not 'activated' - see below) each contributing $\frac{1}{2}kT$ energy (equipartition theorem), so:

$c_V = N\frac{2\cdot 0.5k \Delta T}{\Delta T}$
$c_V = Nk$
(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: $kT = k\cdot300\approx0.026 eV$ , is low compared to the first excited state of the harmonic oscillator ( $\omega$ 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.

---
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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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. Reply to this comment
smokwzbroiplytowej 2008-10-26 20:53:14	(Ref. 'Modern Physics 3rd ed, Serway, Moses, Moyer', Chapter 11, p. 378 onward) (A) $c_V=\frac{\partial u}{\partial T}_v$ The rigid dumbbell has two rotational degrees of freedom (the rotation around the axis is not 'activated' - see below) each contributing $\frac{1}{2}kT$ energy (equipartition theorem), so: $c_V = N\frac{2\cdot 0.5k \Delta T}{\Delta T}$ $c_V = Nk$ (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: $kT = k\cdot300\approx0.026 eV$ , is low compared to the first excited state of the harmonic oscillator ( $\omega$ 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. --- 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. Reply to this comment

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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.

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