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Frostbite
 one year ago
Thermodynamics, statistical thermodynamics and spectroscopy COMBINED!
Frostbite
 one year ago
Thermodynamics, statistical thermodynamics and spectroscopy COMBINED!

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Frostbite
 one year ago
Best ResponseYou've already chosen the best response.2The electronic ground state of the Cl atom is fourfold degenerate, while the first electronically excited state is doubly degenerate and lies 881 cm–1 above the ground state. Calculate the contribution to the molar heat capacity at constant volume, \(C_{V,m}(T)\) of Cl atoms from electronic states at 500 Kelvin. There are two alternative ways solving this problem :)

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.2(answer should include some theory why) ;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Off the top of my head the only thing I think I am looking at is the energy difference given in wave numbers, so I have \[E = h c \tilde\nu\] And then I believe, not entirely sure: \[\left( \frac{\partial U}{\partial T} \right)_V = C_V\]

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.2Yup, just need to evaluate the derivative and you are almost there :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It looks like from your notes you gave me that \[C_v = \frac{N \sigma^2}{kT}\] However it's at about this time during the day that it's my bedtime. I've been awake for a while so I'll have to come back and finish this tomorrow. :)

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, I can see the notes somewhat is lacking in terms of the Boltzmann distribution to make it clear HOW to find the variance. So I make this an worked example: The average energy must be the sum of the energy multiplied with probability of finding the specie with the energy: \[\Large \langle \epsilon \rangle=\sum_{i}^{}p_i \epsilon_i\] The Boltzmann distribution allows us to calculate the probability of finding a specie with the energy \(\large \epsilon_i\): \[\Large p_i=\frac{ e^{ \beta \epsilon_i} }{ q }\] Where \(q\) is the partition function: \[\Large q=\sum_{\sf states~i}^{}e ^{ \beta \epsilon_i}=\sum_{\sf levels~i}^{}g_ie ^{ \beta \epsilon_i}\] Where \(g_i\) is the degeneracy of the ith energy level. The average energy is can therefor be written as: \[\Large \langle \epsilon \rangle=\sum_{i}^{}p_i \epsilon_i=\sum_{i}^{}\frac{ e ^{ \beta \epsilon_i } }{ q}\epsilon_i\].

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \left( \frac{ \partial U_m }{ \partial T } \right)_V \approx 1.95 \sf \frac{ J}{ mol \times K }\]

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.2And my handwritten notes (the reason I don't study math)
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