[PoM] Rotational and vibrational heat capacity

[BRN]

Hi guis, i need your help...

1. Homework Statement
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 $\nu_0=2650cm^{-1}$

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!

 

[TSny]

Have you taken into account that the vibrational frequencies depend on the reduced masses? Note that the given frequency is for ¹H⁷⁹Br, not D⁷⁹Br.

 

[BRN]

Hi TSny and Happy New Year!

Note that the given frequency is for ¹H⁷⁹Br, not D⁷⁹Br.

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:

$\omega=\sqrt{\frac{k}{\mu}}=2\pi\nu$

 

[TSny]

Happy New Year!

$\omega=\sqrt{\frac{k}{\mu}}=2\pi\nu$

Use this equation to express the ratio of two frequencies in terms of the ratios of the masses.

 

[BRN]

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?

 

[TSny]

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, k_B, and the atomic masses.

 

[BRN]

Thanks a lot! As usual you've been patient, timely and accurate.

 

[mjc123]

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 I_tot, which is wrong. However in this case it didn't matter, as you say, T >> Θ_rot so C_rot = k_B.

 

[BRN]

Oh yes! I know! I forgot to correct. 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!

 

[mjc123]

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.

 

[BRN]

Hi mjc123,
I add them together because I think to particles system case where the inertia moments are added. It's wrong?

 

[mjc123]

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 C_v, for that species. Do the same for D⁸¹Br. Take an average for C_v of the mixture.

As T >> Θ_rot so C_rot = k_B, it doesn't matter in this case, but it's as well to get the principles right.

 

[BRN]

(Why are they both the same? I get 6.4419 and 6.4458 for the two isotopomers.)

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

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

Thanks!