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[PoM] Rotational and vibrational heat capacity

  1. Jan 15, 2017 #1

    BRN

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    Hi guis, i need your help...

    1. The problem statement, all variables and given/known data

    Evaluate the rotational and vibrational contributions to the heat capacity of a gas of DBr (D=deuterium, Br=mixture at 50% of 79Br and 81Br) at 380 K temperature, knowing that the bond distance is 1.41 Å and the vibration frequency of 1H79Br is ##\nu_0=2650cm^{-1}##

    3. The attempt at a solution
    ##R_M##=1.41 Å=##1.41*10^{-10}[m]##
    ## \nu_0=2650[cm^{-1}]= \nu_0=265000[m^{-1}]=7.9235*10^{13}[Hz] ##

    Two isotopes have the same binding distance with inertia momentum:
    $$ I_1= \mu_1R_M^2= \frac{79}{80} \frac{10^{-3}}{N_A}R_M^2=3.2598*10^{-47}[Kgm^2] $$
    $$ I_2= \mu_2R_M^2= \frac{81}{82} \frac{10^{-3}}{N_A}R_M^2=3.2608*10^{-47}[Kgm^2] $$
    $$ I_{tot}=I_1+I_2=6.5206*10^{-47}[Kgm^2] $$

    The characteristic rotational temperature is:
    $$ \Theta_{rot}= \frac{ \hbar^2}{2I_{tot}k_B}=6.1760[K] $$
    I'm in ##T >> \Theta_{rot}## case, then:
    $$ C_{v,rot}=k_B=1.3806-10^{-23}[J/K] $$
    and
    $$ C_{v,vib}= \frac{k_B( \beta \hbar \omega)^2e^{- \beta \hbar \omega}}{(1-e^{- \beta \hbar \omega})^2} $$
    with
    ## \beta= \frac{1}{k_BT}## and ## \omega=2 \pi \nu_0=4.9784*10^{14}[rad/s] ##
    $$ \Rightarrow C_{v,vib}=6.2365*10^{-26}[J/K] $$

    ##C_{v,vib}## is wrong, why???

    SOLUTIONS:##C_{v,rot}=k_B=1.3806-10^{-23}[J/K]; C_{v,vib}=5.597*10^{-25}[J/K]##

    Thanks at all!
     
    Last edited: Jan 15, 2017
  2. jcsd
  3. Jan 15, 2017 #2

    TSny

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    Homework Helper
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    Have you taken into account that the vibrational frequencies depend on the reduced masses? Note that the given frequency is for 1H79Br, not D79Br.
     
  4. Jan 16, 2017 #3

    BRN

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    Hi TSny and Happy New Year!

    Yes, I know... but I don't know how to convert that frequency for D79Br.

    The only relationship I know that involves frequency and reduced mass is this:

    ##\omega=\sqrt{\frac{k}{\mu}}=2\pi\nu##
     
  5. Jan 16, 2017 #4

    TSny

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    Happy New Year!
    Use this equation to express the ratio of two frequencies in terms of the ratios of the masses.
     
  6. Jan 16, 2017 #5

    BRN

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    OK, then:
    for 1H79Br ##\omega=\sqrt{\frac{k}{\mu_1}}=2\pi\nu_0##
    for 2H79Br ##\omega=\sqrt{\frac{k}{\mu_2}}=2\pi\nu_2##
    and making the ratio, I get:

    ##\nu_2=\sqrt{\frac{\mu_1}{\mu_2}}\nu_0=5.6376*10^{13}[Hz]##

    Now, ##\omega=2\pi\nu_2=3.5422*10^{14}[rad/s]## and ##C_{v,vib}=5.6708*10^{-25}[J/K]##

    But the DBr gas is composed by 50% of 79Br and 50% of 81Br. I repeated as above for 2H81Br, getting ##C_{v,vib}=5.6799*10^{-25}[J/K]##

    I think serves the average of the two results. it's correct?
     
  7. Jan 16, 2017 #6

    TSny

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    Yes, I think that's the correct method. The accuracy of your answer will depend, of course, on the accuracy of the numbers you use for h, c, kB, and the atomic masses.
     
  8. Jan 16, 2017 #7

    BRN

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    Thanks a lot! As usual you've been patient, timely and accurate.:wink:
     
  9. Jan 17, 2017 #8
    For the rotation, you also calculated the moments of inertia for the 1H isotopomers instead of D. For some reason you added these together to give Itot, which is wrong. However in this case it didn't matter, as you say, T >> Θrot so Crot = kB.
     
  10. Jan 17, 2017 #9

    BRN

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    Oh yes! I know! I forgot to correct. Sorry...:sorry:

    The exact moment of inertia are:

    ##I_1= \mu_1R_M^2= \frac{158}{81} \frac{10^{-3}}{N_A}R_M^2=6.4396*10^{-47}[Kgm^2]##
    ##I_2= \mu_2R_M^2= \frac{162}{83} \frac{10^{-3}}{N_A}R_M^2=6.4396*10^{-47}[Kgm^2]##

    then, ## I_{tot}=1.2883*10^{-46}[kgm^2]##

    Thanks!
     
  11. Jan 17, 2017 #10
    Why do you add them together? That's meaningless. Do (if it was necessary) what you did for vibration - calculate C for both isotopomers, and take the average.
     
  12. Jan 18, 2017 #11

    BRN

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    Hi mjc123,
    I add them together because I think to particles system case where the inertia moments are added. It's wrong?
     
  13. Jan 18, 2017 #12
    I don't know what "particles system case" you are referring to, but here you are not looking for a total moment of inertia of a system of particles (in which case just adding two would be inadequate!) but for the molecular moment of inertia. Assuming your calculations are correct, I(D79Br) = 6.4396 x 10-47 kg m2. (Why are they both the same? I get 6.4419 and 6.4458 for the two isotopomers.) From that you work out the rotational constant and energy levels, and hence Cv, for that species. Do the same for D81Br. Take an average for Cv of the mixture.

    As T >> Θrot so Crot = kB, it doesn't matter in this case, but it's as well to get the principles right.
     
  14. Jan 18, 2017 #13

    BRN

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    Ops! I'm sorry! I made a copy / paste and did not correct the second value.:sorry:

    Ok, then I consider wrong the system of particles approach and I do the average of the two values.

    Thanks!
     
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