Stuck proving the relation $$\mu_i=\mu_i^\circ + RT\ln a_i,\tag{1}$$ I looked up the definiton of activity. It is


but the the intuition/motivation for such a definition is unclear. Is it just because we wish for a relation like $(1)$ to exist by analogy with $(3)$? $$G_i=G_i^\circ+n_iRT\ln{\frac{p_i}{p^\circ}}\tag{3}$$

That is, if we define activity as given in $(2)$, we can take the natural logarithm of both sides, multiply with $RT$, add $\mu_i^\circ$ to both sides and arrive at $(1)$. And hence $(1)$ has a similar form to $(3)$ which holds for ideal gases.

Or am I mistaken and there are other reasons at play?

  • $\mu_i$ is chemical potential
  • The superscript $^\circ$ denotes the standard state
  • $a_i$ is activity
  • The subscript $_i$ denotes the $i$-th component of a mixture

So I'm taking Pchem right now so I'm no experts, but here is my two cents.

I'll be honest I essentially followed my books lead on the determination of $a_{i}$, and $\mu_{i}$. When looking at problems like this I try to go backwards. I see that $a_{i}$ is derived from $\mu_{i}$, and I try to find where $\mu_{i}$ comes from then see where that comes from until I get a definition or a well known relation like Gibbs free energy. I'm not savy enough to be like "Oh yea $\mu_{i}$, that's basically $\int dG(T, P_{2})$ I'll just show all that". But, I can slowly go backward. So, I followed the following path. $$a_{i} \Rightarrow \mu_{i} \Rightarrow (\mu(T,P_{2})-\mu(T,P_{1}) ) \Rightarrow d \mu \Rightarrow d G$$ Then when you have this general path you reverse it to get the derivation.

Also remember that you have to keep track of all the assumptions you make. Its important to remember that. My answer is only applicable for a perfect gas at constant temperature, and a mixture of gas at low pressure.

From deferential of Gibbs free energy the chemical potential of a pure gas is, $$dG = -SdT +VdP = d\mu$$ at constant temperature, $$d \mu = V_{m}dP = \frac {RT}{P}dP$$ To get $\mu$ we integrate with constant temperature essentially expanding the gas on an isotherm for $P_{1}$ to $P_{2}$. $$\int _{1}^{2} d \mu = RT \int _{P_{1}}^{P_{2}} \frac{1}{P}dP$$ $$\mu(T,P_{2})-\mu(T,P_{1})=RT \ln \Big{(}\frac {P_{2}}{P_{1}}\Big{)}$$ Let $\mu(T,P_{1})=\mu^{\circ} $, $P_{1}=P^{\circ}=1bar$, $P_{2}=P$, and $\mu(T,P_{2})=\mu_{i} $,

$$\mu = \mu_{i}+ RT \ln \Big{(}\frac {P}{P^{\circ}}\Big{)}$$

Side note your equation 3 comes from this equation. $\mu_{i}=G_{i}$.

For a mix of gas at low pressure, the activity is equal to the ratio of the partial pressure of the gas over the standard pressure, $$a_{i}=\Big{(}\frac{P}{P^{\circ}}\Big{)}$$

$$\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i}$$

I think that's how you prove that relationship. I hope that helped. $$\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i} \Rightarrow a_{i}= e^{\frac{\mu_{i}-\mu^{\circ}}{RT}} $$ This relation is trivial and just some algebra. When you are trying prove something its important to start as far back as possible.


Chemical Activity

Levine, Ira N. Physical Chemistry. 6th ed. pg 175-176 . New York: McGraw-Hill, 1978. Print.

Notes from my professor (sorry).

| improve this answer | |
  • $\begingroup$ OP asked for the motivation/intuition behind the definition, perhaps you should try and address that? :) $\endgroup$ – getafix Oct 26 '16 at 2:01
  • $\begingroup$ Is that paragraph a good comment on motivation, or should I actually show the intoution for each step of the derivation, and do the derivation backwards? $\endgroup$ – stephen zilch Oct 26 '16 at 2:54
  • 1
    $\begingroup$ I have to agree with @getafix. Note that IUPAC defines activity as $$a_i=e^{\frac{\mu_i-\mu_i^\circ}{RT}}.\tag{1}$$ It is not possible to say that (1) holds because the following $\mu_{i} = \mu_{i}^{\circ} +RT \ln a_{i}$ is true. How do you then know that $$a_{i}=\Big{(}\frac{P}{P^{\circ}}\Big{)}?$$ To keep your argument from being circular, you must then define activity in an alternative manner. How would you go about it? In essence you are still saying that hey, since we have a pretty relationship for ideal gases, let us not mess that up! $\endgroup$ – Linear Christmas Oct 26 '16 at 16:09
  • $\begingroup$ I'm confused by your question then? In my opinion the relationship between your equation 1 and 2 is trivial, and my teacher wouldn't except that as an answer. Your equation 1 has to come from somewhere I showed one way of finding that for a pure gas at near standard pressure. $a_{i}$ is equal to a lot of things if you have it in terms of activity coefficient and molar fraction $\gamma-{i} x_{i}$ you can prove it that way, but I did in terms of partial pressure so I found a relationship between partial pressure and activity. Correct me if I'm wrong but that's the way I play the derivation game $\endgroup$ – stephen zilch Oct 26 '16 at 20:33
  • $\begingroup$ My point was that your argument is circular in a mathematical sense. You cannot say activity is equal to anything before defining it first. Right now you have the mathematical relationship $$\text{definition of activity} \implies a_{i}=\Big{(}\frac{P}{P^{\circ}}\Big{)} \implies a_i=e^{\frac{\mu_i-\mu_i^\circ}{RT}}.$$ Do you see the problem? $\endgroup$ – Linear Christmas Oct 27 '16 at 8:26

As wikipedia reports on the first lines: " In chemical thermodynamics, activity (symbol a) is a measure of the “effective concentration” of a species in a mixture, in the sense that the species' chemical potential depends on the activity of a real solution in the same way that it would depend on concentration for an ideal solution. " In other words, you should and have to find the same equations of an ideal solution. It is just that X and Y might have different activities/concentrations from Xideal and Yideal.

The motivation is very simple: keep ideal equations, models but characterise the chemicals "just" by the activity. Or mathematically: fit on the activity as free parameter the ideal equations.

I hope it fits also the intuition part.

| improve this answer | |

The activity is as you have defined it and can easily be expressed in mathematically equivalent ways. It arises from the need to use equations developed for ideal gases to the behaviour of real gases (or solutions, or electrolytes etc.)

The variation of the free energy G of an ideal gas at pressure p (bar) from its standard state $G^0$ is expressed as $$G=G^o +RT\ln \left(\frac{p}{p_0}\right) $$ where $p_0$ is 1 bar pressure and is added only so that the log is dimensionless. (Many books do not include this but this means that the log has dimensions, which it cannot have. You can also use chemical potential as this is just free energy/mole. You can use atm. instead of bar if you want).
For clarity I remove the $p_0$ in the following equations thus

$$G=G^o +RT\ln \left(p\right) $$

When we want to measure free energy of a real gas then it would be useful to still to be able to use this equation. However, real gases deviate in their behaviour from an ideal gas, the more so as the pressure is increased.

To cope with this we define a new quantity, the activity a. Some authors use fugacity $f$ which is equal to activity for a gas. The activity is used to define a new standard state which is that at which the activity would be equal to $1$ bar assuming that the gas behaves ideally to this pressure. Thus an ideal gas always has an activity equal to the pressure but a real gas deviates from this and has, perhaps, an activity that is smaller than the pressure. This may be due to attractive intermolecular forces.

Although a real gas may never reach the standard state of unit activity (fugacity) it does not prevent us from using it since measurements are always relative to the standard value.
The standard states for liquids or solids is always taken to be 1 bar, whereas the standard state for a gas is not that of unit pressure but of unit activity.

It can be shown that as the activity and hence free energy at low pressure has ideal behaviour then$^*$ $$G=G^o+RT\ln(a) \tag 1$$

If we compare this to the equation for an ideal gas then we can write it as $$G=G^o+RT\ln (\gamma p) $$

where $\gamma = a/p $ is called the activity coefficient.

Similarly to low vs. high pressure gases, activity is used for higher concentrations in solution instead of concentration.

$^*$ As activity and pressure are equal at low pressure we can arrive at eqn (1) by subtracting the free energy at some pressure (represented by activity a ) from that at low pressure in which case its activity is equal to the pressure itself thus
$G-G_{low} = +RT\ln(a) - RT\ln(p_{low})$
and now do the same with the standard where $a=1$
$G^o - G_{low}= +RT\ln(1)- RT\ln(p_{low}) = - RT\ln(p_{low})$

then by subtracting, eqn. (1) follows.

| improve this answer | |
  • $\begingroup$ I suggested a few edits but the sentence the activity a [---] that defines a new standard state as that at which the activity would be equal to 1 bar if the gas behaved ideally up to 1 bar and the footnote are a little confusing. $\endgroup$ – Linear Christmas Oct 29 '16 at 18:33
  • $\begingroup$ have tried to make things clearer. $\endgroup$ – porphyrin Oct 30 '16 at 14:12
  • $\begingroup$ There are still a few problems, some of which were reintroduced after your edit. $(1)$ It is more correct to denote Gibbs energy per mole with a subscript $m$. $(2)$ For a gas, the standard state is the gas itself under standard pressure and ideal behavior. Thus, activity as such is not part of the definition of a standard state. $(3)$ Activity does not have dimensions, hence it does not make sense to say 'activity is equal to 1 bar'. $(4)$ Fugacity is not equal to activity, note that fugacity has the same dimensionality as pressure. $\endgroup$ – Linear Christmas Oct 30 '16 at 16:10
  • $\begingroup$ ok here goes! (1) don't care whether m is subscript you can define it how you want. The symbol $\mu$ is usually used for chemical potential /mole. (2) yes for an ideal gas, but we do not deal with ideal gases but with real gases when we use activity/fugacity. (3) for a gas activity = fugacity, $a=f$ , and $a/p=1$ as $p \rightarrow 0$. Often in books we see $\ln(f)$ but this really means $ln(f/1 atm)$ , yes its confusing ; same applies to $ln(p)$. The fugacity is equal to the vapour pressure when the vapour is an ideal gas, and in general it may be regarded as a corrected pressure. $\endgroup$ – porphyrin Oct 30 '16 at 18:06
  • $\begingroup$ $(1)$ The subscript $m$ is for the molar Gibbs energy, that is $G_m$. No, the symbol $\mu$ denotes chemical potential itself. It is just that since $$\mu = \frac{\partial G}{\partial n}$$ and $G = n\cdot G_m$, then $$\mu = \frac{\partial nGm}{\partial n}=G_m.$$ In other words, chemical potential is molar Gibbs free energy. $(2)$ See here for proper definition of a standard state. $(3)$ The limit you gave may equal $1$ but that hasn't got to do with anything :). And still, activity $\neq$ fugacity. The correct relationship is $$a=\frac{f}{p^\circ}.$$ $\endgroup$ – Linear Christmas Oct 30 '16 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.