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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: riboswitch on January 22, 2019, 10:51:58 AM

I'm trying to decipher a step undertaken to demonstrate the formula that:
[tex]A =  \frac{ln(Q_{NVT})}{ \beta } =  k_{B}Tln(Q_{NVT})[/tex]
One of the steps undertaken is:
[tex]A = E  TS[/tex]
[tex]\frac{dA}{dT}_{N,V} = S[/tex]
Since my professor's notes defined the Boltzmann factor as [tex] \beta = \frac{1}{k_{B}T} [/tex]
He then wrote that:
[tex]\frac{dA}{dT}_{N,V} = \frac{dA}{d\beta}_{N,V} \frac{ \beta }{T} = S[/tex]
I just don't understand how he obtained that last one... I don't think he just substituted T with 1/βk_{B}. As someone who lacks basic knowledge of mathematics, I have no idea how we arrived at that last part....

dA/dT = dA/dβ*dβ/dT
dβ/dT = 1/kT^{2} = β/T
I think you've missed out a minus sign.

dA/dT = dA/dβ*dβ/dT
dβ/dT = 1/kT^{2} = β/T
I think you've missed out a minus sign.
Now it makes sense. Thanks. Can I ask another question? I have another question related to the derivation of the average energy of a canonical ensemble and eventually proving that the Helmholtz free energy is related to the partition function Q_{NVT}:
[tex]<E> = \frac{\sum_v E_{v}e^{ \beta E_{v} } }{\sum_v e^{ \beta E_{v} }} [/tex]
[tex] =  \big(\frac{dln\big(\sum_v e^{ \beta E_{v} }\big) }{d \beta}\big) [/tex]
The transition from the first formula to the second formula is still not understood by me. Why are we using the derivative all of the sudden here? Am I missing something?

Try expressing d(lnΣ)/dβ as (1/Σ)*dΣ/dβ

Ok, I think I get it. First I have to derive the Q function (also known as the partition function):
[tex]Q_{NVT} = \sum_v e^{ \beta E_{v} } [/tex]
So, knowing that the derivative of e^{ax} is equal to ae^{ax}, then I am finally able to derive the function Q:
[tex]\frac{ \delta Q}{ \delta \beta } = \frac{ \delta \big(\sum_v e^{ \beta E_{v} }\big) }{ \delta \beta }[/tex]
[tex]\frac{ \delta Q}{ \delta \beta } =  \sum_v E_{v}e^{ \beta E_{v} } [/tex]
Then going back to the average energy in the canonical ensemble:
[tex]<E> = \sum_v P_{v} E_{v} = \frac{1}{Q} \sum_v E_{v} e^{ \beta E_{v} }[/tex]
[tex]<E> \ = \ \frac{1}{Q} \frac{ \delta Q}{ \delta \beta }[/tex]
The last equation can be rewritten in another way knowing the chain rule from my calculus class for derivative of a function inside a function:
[tex] \frac{ \delta }{ \delta x} f \big(g \big(x\big) \big) = f' \big(g \big(x\big) \big) g' \big(x\big) [/tex]
So now I can write the following, knowing that Q is also a function of β:
[tex] \frac{ \delta }{ \delta \beta} ln \big(Q \big( \beta \big) \big) = \frac{ \delta ln \big(Q\big) }{ \delta \beta } \frac{ \delta Q}{ \delta \beta } = \frac{1}{Q} \frac{ \delta Q}{ \delta \beta } [/tex]
So now I can finally define the average energy in the canonical ensemble as:
[tex]<E> \ =  \frac{ \delta ln \big(Q_{NVT} \big) }{ \delta \beta } [/tex]
Please correct me if there are mistakes in the procedures I have written.

So now I can write the following, knowing that Q is also a function of β:
δ/δβ ln(Q(β))=δln(Q)/δβ δQ/δβ =1/Q δQ/δβ
The middle expression ought to be dlnQ/dQ*dQ/dβ
Otherwise correct