Chemical potential and fugacity

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  • Thread starter Kaguro
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Summary:
My textbook defined chemical potential in two different ways. How can same thing have 2 definitions?
I am trying to learn statistical physics. While learning MB statistics, my textbook defined chemical potential as ##\mu = (\frac{\partial F}{\partial N})_{V,T}##. That's nice.

Later when I started on Quantum statistics, my textbook described all three distribution functions via:
##n_i = \frac{g_i}{e^{\alpha + \beta E_i} + \kappa}##
We had already found out the value of beta from MB statistics (using MB distr. function. Why would that apply here is another mystery altogether)

Then suddenly book said:
##n_i = \frac{g_i}{e^{\frac{E_i - \mu}{K_B T}} + \kappa}##

Where we define chemical potential via the relation ##\alpha = -\mu/kT## (and its exponential is called fugacity)

How and why did the book define the same thing twice!?
 

Answers and Replies

  • #2
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To all future people who want to know: I found out how can we show equivalence.

We need to define mu only once, but neither in these places. We need to define mu in the fundamental thermodynamic equation:

First consider Grand Canonical Ensemble, that is, allow particle number to change.

Generalise first and second laws into this:

##dU = TdS - PdV + \mu dN##
We define ##\mu## as the rate of change of energy per unit change in particle number.

Then from this relation, ##\mu = (\frac{\partial U}{\partial N})_{S,V}##
F = U-TS
So, ##dF=-PdV - SdT + \mu dN##
##\Rightarrow \mu = (\frac{\partial F}{\partial N})_{T,V}## (Yaay!)

Now while deriving the distribution function by maximizing the log of number of microstates and using method of Lagrange multipliers we got:

##d(lnW) = \alpha dN + \beta dE##
So, ##\alpha = (\frac{\partial ln(W)}{\partial N})_{E,V}##

But we know ##S=k_B ln(W)## (separate derivation for that. But it is standalone)
##\Rightarrow lnW = \frac{S}{k_B}##
##\Rightarrow \alpha = \frac{1}{k_B} (\frac{\partial S}{\partial N})_{E,V}##

Now, ##TdS=dU+PdV-\mu dN##
##(\frac{\partial S}{\partial N})_{U,V} = \frac{-\mu}{T}##

Therefore, ##\alpha = \frac{-\mu}{kT}## (Yaaaay!)

Similarly we can show that ##\beta = \frac{1}{kT}##

Hence Proved.
 

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