Question

1. Ideal gas with internal degrees of freedom. Consider a free gas of diatomic molecules at temperature 7. Diatomic molecules have internal rotational excitations. The rotational energy levels of a single molecule are given by J(J+1) 2/2 J = 0,1,23 where J is the angular momentum and I is the moment of inertia. The degeneracy of the level J is 2J +1. Neglect any interaction between the molecules in the gas. The temperature is high enough so that the statistic of the molecules does not matter (classical regime). a). For gas of density n write down expressions for chemical potential, internal energy and specific heat Cy. b). Calculate the above quantities in two limits: low temperature where only the excitations with J = 1 contribute, and high temperature, where one can treat the rotational spectrum as continuous and thus approximate the sum over J by the integral.

Answer 1

free gas d. Consider a of diatomic molecule at temperature t. The rotational energy level is given by G - Jeft! ZIV Where I is angular momentum and I is the moment of Inertia The degeneracy is (2J+1). Now the partition for for single par. - ticle is > 20713883 TIME free energy f can be written as F = - KBT ln Lost for gas of densily in the free energy is fr=- nKBT ln Zoot then the chemical potential is As Of = - KB7km Font internal energy <b>= ko likte) =4n I (2341) TO:n Le - {RITU ER TIL ŞQI41) I(32) Toys 2[* Jestved!) ė Taher specific hear partich GECES =40 en content - [<E">-<EXY geran [ - ] 278) SATEN TCI+1) 2.7792 I Ile pole enam (6) Now in low temp J = 0, 1 stales only applicable M = - KBT ln [2.01116 J +8.1+1 El - = - KST ln [143 et CE) = 10+ 3.2.0 oflar / 1+ 3 PRATI + nie Leo Bar influen pri teploty llar 8 for high temperatur We can integrale instead of sum -BJ (Jelly Zoot = ((25110e at er = 22 2 IKT Chemical potential - M = - Katla REKATI <ED- - Anz Content KE>= ₂ = Kot] cv- @CE) - 240 Tu = KB Last modified: 8:54 pm