# Meaning of small triangle in chemical potential

• etotheipi
It is given that the solution is ideal, i.e. that we can take ##\gamma_A = 1##.

I wondered what that small triangle signifies in the second definition? Thanks!

Homework Helper
Good question. I've never seen that before. But it looks like in the first equation the reference state is a solution with [A] = 1M, while in the second, the reference state is xA = 1, i.e. pure A. But I've never seen the little triangle symbol.

etotheipi
Thanks, that's probably what's intended. It's likely just outdated notation . If I stumble across the small-print sometime in the future, then I'll post an update

Mentor
It's likely just outdated notation

I am outdated and I have never seen it as well, so more like an obscure notation

berkeman and etotheipi
Mentor
I wondered what that small triangle signifies in the second definition?
Where is that screenshot from?

etotheipi
It's from the Oxford Chemistry Primers series, there are some really nice undergrad-level overviews of different subjects in Chemistry; if you like you can have a look at some of the titles here .

berkeman
Gold Member
I got my PhD in chemistry in 2018 so I believe my knowledge on recent textbooks is rather updated, but me too have never seen that kind of symbol.

That being said, considering that xA is a notation used for Mole fraction, I believe μA equals to μA* (chemical potential of a pure A solvent), and the only difference is that the former notation is not what the IUPAC recommends. This is the chemical potential of the solution based on vapor pressure.

Chemical potential (Gibbs energy) for mixed gas is:
$\mu=\mu^{\ominus}+RTln\frac{p}{p^{\ominus}}$
where $p^{\ominus}$ is the standard pressure (1 bar) and IUPAC recommends using this symbol. Now turning to solution case, the chemical potential of the vapor pressure of fully pure liquid A (which in equilibrium, this is identical to the chemical potential of the liquid) is:
$\mu _{A}^{*}=\mu _{A}^{\ominus}+RTln \frac{p _{A}^{*}}{p^{\ominus}}$
If the liquid is not pure A, then:
$\mu _{A}=\mu _{A}^{\ominus}+RTln \frac{p _{A}}{p^{\ominus}}$
So combining these two equation gives:
$\mu _{A}=\mu _{A}^{*}+RTln\frac{p _{A}}{p _{A}^{*}}$
We can write this in terms of chemical activity $a_{A}=\frac{p _{A}}{p _{A}^{*}}$, which gives:
$\mu _{A}=\mu _{A}^{*}+RTlna _{A}$
And using activity coefficient and mole fraction,
$\mu _{A}=\mu _{A}^{*}+RTln\gamma _{A}x_{A}$
Ideal solution with sufficiently low solute B and pure solvent A (Roult's Law) means that $\gamma _{A}=1$ so,
$\mu _{A}=\mu _{A}^{*}+RTlnx_{A}$
This is now identical to the second equation of the OP's question.

Last edited:
etotheipi