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Thread: Gas degrees of freedom

  1. #1
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    How many degrees of freedom does an ideal diatomic gas possess? How about a tri-atomic gas?


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    Originally posted by yevgeny

    How many degrees of freedom does an ideal diatomic gas possess? How about a tri-atomic gas?

    Looks like you've been studying hard for this test; I m taking it too this saturday(Nov 8th), and after reading all the messages sent between you and ecm; looks like you two will do in the 90 percentile; For me I m trying to do above average since there are several topics I have not yet covered in class ( I m a first semester senior, in a US institution). But back to your question:


    Like the energies, you have three groups that contribute to the DOF(Degrees of Freedom):

    - Translational DOF : 3 (always);

    - Rotational DOF : 0 (if it is a single atom),1 (for diatomic compounds) and 3 (for compounds made of more than 2 atoms)

    - Vibrational DOF: (3N-5) for linear molecules, and (3N-6) for non linear
    molecules; N is the number of atoms in the molecules. remark that if N<2 there is no (0) Vibrational DOF.

    the total DOF is the addition of all three contributions. May I ask which question on the GRE sample this one is pertinent too?

    PS:You may want to memorize the corresponding contributions to the internal energy also for the GRE.







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    Hi papucisse!

    Welcome to our forum! Tell us about yourself!

    Don't worry about the messages between me and ecm. Many of the issues we discussed are either beyond the GRE level or have a small chance to appear on GRE.

    As to your answer:
    a) Why is there only 1 rotational DOF for diatomic compounds? There are two possible axes of rotation!
    b) Most of my questions are not related to any specific question of the sample test, but this one originated from question 79. What do you think?

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    Originally posted by yevgeny

    Hi papucisse!

    Welcome to our forum! Tell us about yourself!
    Thank you, I m a physics major (senior) and, like you, I m trying to work with professors in some very good graduate schools( sorry caltech is not (yet) one of them but stanford is one in california); So perhaps we will all see each other at the orientations of those schools we're applying to. Oh yeah, I m from an african country but I'm doing my undergrad in the US.


    Don't worry about the messages between me and ecm. Many of the issues we discussed are either beyond the GRE level or have a small chance to appear on GRE.

    As to your answer:
    a) Why is there only 1 rotational DOF for diatomic compounds? There are two possible axes of rotation!

    I am sorry, what was I thinking[|)]?; you're right! indeed even for molecules that have more than two atoms and that are linear in shape ( I believe CO2 is an example) there are 2 rotational DOF.

    so let me rephrase it in case someone else ( like me ) will be reading the post:

    Rotational DOF: 0 for Single atoms; 2 for linear molecules, and 3 for non linear (and non-single) molecules.

    b) Most of my questions are not related to any specific question of the sample test, but this one originated from question 79. What do you think?
    Thank you, Somehow the first time I took the practice test ( and did so badly) I couldn't understand what most questions were asking. Problem 79 was one of them. But after I read your reply, I went back to redo it and got the right answer (7/3); Perhaps I should Take the practice exam again tomorrow, now that I have memorized my basic fomulas, hopefully It will boost my confidence for the exam a bit before the test this saturday

    You guys have changed my perspectives on the exam, (you, ecm, romad, and physicsgre), thanks a lot!

    and i'll try to be here more often before this saturday!

    Ps: I enjoyed your question on physicsgre.com . It threw me away at first though cause you said "assume velocity and wavelength are not related".

    Ok I m going back to memorize some more formulas, read you soon.

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    Hi there!

    Nice meeting you, papucisse! (and until we all meet in person in California, eh? ). Im glad more people are joining the forum, it was somewhat empty.

    You and yevgeny have me a bit intrigued here. Can somebody please explain question 79? I must be very stupid, but I still dont get it

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    Originally posted by ecm

    Hi there!

    Nice meeting you, papucisse! (and until we all meet in person in California, eh? ). Im glad more people are joining the forum, it was somewhat empty.

    You and yevgeny have me a bit intrigued here. Can somebody please explain question 79? I must be very stupid, but I still dont get it
    thanks ecm,

    Here's question 79:

    It's based on the fact that at high tempereature you have all three groups: Translational, Rotational and vibrational energies that contribute to the internal energy; and at low temperature there is only the translational energy.

    so for high temperature:

    Translational: 3 Degrees of Freedom, each contributes (1/2)RT to energy

    Rotational: 2 DOF(beacuse diatomic=linear shape), each contributes (1/2)RT to the energy

    Vibrational3N-5)DOF (linear shape); since N=2 we get(3*(2)-5)=1 DOF each vibrational DOF contributes (1)RT to the energy.

    so at high T the internal energy is:
    U=U(trans) +U(rot) +U(vib)= (3/2)RT+(2/2)RT+(1)RT=(7/2)RT
    since heat capacity at constant V is Cv=dU/dT (derivative wrt to T)
    so for high temps Cv1=(7/2)R.

    And for low temp the internal energy is:
    U=U(trans)= (3/2)RT; and since Cv=dU/dT;
    Cv2=(3/2)R

    so the ratio (Cv1/Cv2)=[(7/2)R/(3/2)R]=7/3 which I believe is the answer in the practice test.

    So Remember: at Low temperature only Translational energy, if you boost the temperatur a bit, you get both Translational and Rotational; then only at high temperature do you have all three( Translational, Rotational, and Vibrational).

    I assume if they change the question to put it on the GRE again, they will tell you at exaclty what temperatures you start getting different energies, since it varies with the gaz.

    read you soon

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    Thanks so much papucisse!!

    I understand it now.

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    degrees of freedom of co2 molecule

    Quote Originally Posted by papucisse View Post
    thanks ecm,

    Here's question 79:

    It's based on the fact that at high tempereature you have all three groups: Translational, Rotational and vibrational energies that contribute to the internal energy; and at low temperature there is only the translational energy.

    so for high temperature:

    Translational: 3 Degrees of Freedom, each contributes (1/2)RT to energy

    Rotational: 2 DOF(beacuse diatomic=linear shape), each contributes (1/2)RT to the energy

    Vibrational3N-5)DOF (linear shape); since N=2 we get(3*(2)-5)=1 DOF each vibrational DOF contributes (1)RT to the energy.

    so at high T the internal energy is:
    U=U(trans) +U(rot) +U(vib)= (3/2)RT+(2/2)RT+(1)RT=(7/2)RT
    since heat capacity at constant V is Cv=dU/dT (derivative wrt to T)
    so for high temps Cv1=(7/2)R.

    And for low temp the internal energy is:
    U=U(trans)= (3/2)RT; and since Cv=dU/dT;
    Cv2=(3/2)R

    so the ratio (Cv1/Cv2)=[(7/2)R/(3/2)R]=7/3 which I believe is the answer in the practice test.

    So Remember: at Low temperature only Translational energy, if you boost the temperatur a bit, you get both Translational and Rotational; then only at high temperature do you have all three( Translational, Rotational, and Vibrational).

    I assume if they change the question to put it on the GRE again, they will tell you at exaclty what temperatures you start getting different energies, since it varies with the gaz.

    read you soon

    CO2 molecule although polyatomic has linear structure, so shouldnt it have 5 degrees of freedom at low temperature. 3 translation and 2 rotation.

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