Equation for chemical potential of pure substance. Unsure why this derives as it does

Dec 2018
2
0
Hello, I am new to this. I have attached a screenshot from a journal article which talks about deriving Fick's first law.

My calculus is a little rusty, and I don't understand how equation 5 is derived from equation 4.

I'd have a go, but I have little to offer. All I have to offer is that I thought that d(mu)/d(C) would be RT/C. I have no idea how equation 5 comes to be, so if anyone can help me out that would be great :)

Happy new year
 

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topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!
Hello, I am new to this. I have attached a screenshot from a journal article which talks about deriving Fick's first law.

My calculus is a little rusty, and I don't understand how equation 5 is derived from equation 4.

I'd have a go, but I have little to offer. All I have to offer is that I thought that d(mu)/d(C) would be RT/C. I have no idea how equation 5 comes to be, so if anyone can help me out that would be great :)

Happy new year
I'm assuming that the \(\displaystyle \mu\) in equation 4 is really \(\displaystyle \mu _i\)? And that R, T, \(\displaystyle \mu _i^0\) are constants wrt \(\displaystyle C_i\). If so, then:
\(\displaystyle \mu _i = \mu _i^0 + RT~ln(\gamma _i C_i )\)

Rewrite this a bit:
\(\displaystyle \mu _i = \mu _i^0 + RT~ln(\gamma _i ) + RT~ ln(C_i )\)

\(\displaystyle \dfrac{d \mu _i}{d C_i} = \dfrac{d \mu_i^0}{d C_i} + RT~\dfrac{d(ln( \gamma _i))}{d C_i} + RT \dfrac{d(ln(C_i))}{d C_i}\)

\(\displaystyle = 0 + RT~\dfrac{d(ln( \gamma _i))}{d C_i} + RT \dfrac{1}{C_i}\)

That's really all we can do, but we need to put it in the form the text wants. So I'm going to multiply the 1st non-zero term by \(\displaystyle C_i~/ C_i\)

\(\displaystyle \mu _i = \dfrac{C_i}{C_i} \cdot RT~\dfrac{d(ln( \gamma _i))}{d C_i} + RT \dfrac{1}{C_i}\)

Now factor the \(\displaystyle C_i\) and RT from both terms:
\(\displaystyle \mu _i = \dfrac{RT}{C_i} \left ( C_i \dfrac{d(ln( \gamma _i ))}{dC_i} + 1 \right )\)

-Dan
 
Dec 2018
2
0
Thanks very much Dan, this is very helpful :)