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Old Dec 29th 2018, 04:51 PM   #1
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Equation for chemical potential of pure substance. Unsure why this derives as it does

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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Old Dec 29th 2018, 07:11 PM   #2
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Originally Posted by key524 View Post
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
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Old Jan 3rd 2019, 05:39 AM   #3
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Thanks very much Dan, this is very helpful
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