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

#### key524

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

#### Attachments

• 131.6 KB Views: 10

#### topsquark

Forum Staff
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

#### key524

Thanks very much Dan, this is very helpful 