How to write Kelvin equation in a different way

Question

My question is about the Kelvin equation which is as follows:

$$\ln(e/e_s) = \frac{2\cdot\sigma}{n\cdot k\cdot T\cdot r} $$

Keep in mind that the $e$ in $\ln(e/e_s)$ is not the constant $\mathrm{e}$.

I know that:

$$\frac{e}{e_s} = \mathrm{e}^{(\frac{2\cdot\sigma}{n\cdot k \cdot T \cdot r})}$$

I need to show that this is equal to:

$$\frac{e}{e_s} = 1 + \frac{a}{r}$$

Where $a = \frac{2\cdot \sigma}{n\cdot k\cdot T}$

How do I go about doing this?

You are aware that $\ln(\mathrm{e}^x)=x$, right? (Some people like to write $\mathrm{e}=2.71828…$ with an upright e. When you have other $e$s in the same context, this becomes essential.) — Harald Hanche-Olsen, Feb 25 '16 at 20:23
Sorry, this is my first time posting here. Thanks for the edit! — Supercell, Feb 25 '16 at 20:31

score 0 · Accepted Answer · answered Feb 25 '16 at 20:27

0

You want to show that $e^{a/r} = 1 + a/r$. But that isn't true. What is true is that $$ e^{a/r} = 1 + a/r + \frac{a^2}{2!r^2} + \frac{a^3}{3!r^3} + \cdots$$ So if $a/r$ were small, so that $a^2/r^2$ and all the following terms on the right were very small, then $1+a/r$ would be a good approximation for $e^{a/r}$.

answered Feb 25 '16 at 20:27

Matthew Towers

10,854

So it's a taylor expansion, right?
That would make sense since this questions is asked in the context of a numerical problem set.
– Supercell Feb 25 '16 at 20:35
Yes, that's the Taylor expansion for $e^x$ at zero. – Matthew Towers Feb 25 '16 at 20:41

How to write Kelvin equation in a different way

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