# Definition of activity

ussername
The definition of activity is:
$$\mu _{i}=\mu _{i}^{0}+RT\cdot \ln a_{i}$$
where μi is the chemical potential of i in current state and μi0 is the chemical potential of i in standard state.
The current and standard state have the same temperature or can their temperature differ?
If their temperature can differ, than T in the definition equation is the temperature of current state?

## Answers and Replies

Homework Helper
##T## is the current temperature. Lemma is pretty clear about that !

ussername
##T## is the current temperature. Lemma is pretty clear about that !
But I'm asking if the temperature of current state and standard state can differ. Probably yes but it is not mentioned within the definition on wikipedia.

Homework Helper
Yes they certainly can.

ussername
In such a case this definition is not equivalent to the definition of activity:
$$d\mu _{i}=RT\cdot d\ln a_{i}$$ $$a_{i}^{0}=1$$
because when integrating with changing temperature, I generally do not obtain this equation:
$$\mu _{i}-\mu _{i}^{0}=RT\cdot \ln a_{i}$$

Homework Helper
What do you obtain ?

ussername
In that case I have an integral:

$$\mu _{i}-\mu _{i}^{0}=R\cdot \int_{a_{i}^{0}}^{a_{i}} T(a_{i})\cdot d\ln a_{i}$$

Homework Helper
I don't recognize $$d\mu _{i}=RT\cdot d\ln a_{i}$$ as a definition. ##a## is not an independent variable, so ##T(a)## seems weird.

The expression is kind of a tautology when I substitute the definition : ##\ d\mu _{i} = d(\mu _{i} - \mu _{i}^\ominus)##

Check here under activity and activity coefficients

Alternatively we can consult @Chestermiller who might well have a didactically more responsible answer

Last edited:
ussername
dμi=d(μi−μ⊖i)
That is not true. It that case μi should be always zero.

Homework Helper
No. It is a constant. So the ## \ d(\mu _{i}^\ominus) = 0 ##.

ussername
No. It is a constant. So the ## \ d(\mu _{i}^\ominus) = 0 ##.
Now I don't know what you wanted to show, but i is the infinitesimal change of chemical potential of i during the infinitesimal process (eg. addition of dni).

Homework Helper
And here's me thinking $$\mu_i = \left ( \partial G\over \partial N_i \right )_{T,P,N_{j\ne i}}$$ (fortunately some others seem to think so too).

I don't know what you wanted to show
What I showed is that ##d\mu_i = d\left ( RT\ln a_i \right ) \ ## boils down to ##d\mu_i = d\mu_i \ ## if you insert ##\ a_i = e^{\mu_i-\mu^\ominus_i\over RT}\ ##, not very surprising and not very interesting.

If you define $$a_i \equiv e^{\mu_i-\mu^\ominus_i\over RT}$$ then trivially and without integrating, just taking logarithms: $$\mu_i=\mu^\ominus_i+ RT\ln a_i$$ And we could have ended this thread after post #2.

ussername
What I wanted to say:
if we take the definition:
$$\mu_i=\mu^\ominus_i+ RT\ln a_i$$
for arbitrary ##T,T^\ominus##, than this is not valid:

$$d\mu_i = d\left ( RT\ln a_i \right )$$

Homework Helper
Why not ?

ussername
Sorry I meant:
$$d\mu_i = RT\cdot d\ln a_i$$

Homework Helper
Is wrong. T is not a constant like R

ussername
Yes the total differential should be probably:
$$d\mu _{i}(dT,dp,dx_{1},...,dx_{N})=RT\cdot d\ln a_{i}(dp,dx_{1},...,dx_{N})+R\cdot \ln a_{i}\cdot dT$$

Yes the total differential should be probably:
$$d\mu _{i}(dT,dp,dx_{1},...,dx_{N})=RT\cdot d\ln a_{i}(dp,dx_{1},...,dx_{N})+R\cdot \ln a_{i}\cdot dT$$
Differentials aren't functions of differentials!

For fixed concentrations ##d \mu_i=-s_i dT +v_i dP## where ##s_i## and ##v_i## are molar entropy and volume, respectively.
A similar equation holds for the standard chemical potential.
##d \mu^\ominus_i=-s^\ominus_i dT +v^\ominus_i dP##, with the standard molar entropy and volume.
So if you change T, you get
##\mu(T)=\mu^\ominus(T) +RT_0 \ln a_i(T_0) -(\int_{T_0}^T (s_i(T')-s^\ominus(T')) dT')= \mu^\ominus(T) +RT \ln a_i(T)## which you may easily solve for ##a_i(T)##.

Greg Bernhardt
ussername
Also the change of chemical potential with temperature resp. pressure is:
$$\left( \frac{\partial \mu_i}{\partial T} \right)_{p,\vec{n}} = - \overline{S}_i$$ $$\left( \frac{\partial \mu_i}{\partial p} \right)_{T,\vec{n}} = \overline{V}_i$$
##\mu(T)=\mu^\ominus(T) +RT_0 \ln a_i(T_0) -(\int_{T_0}^T (s_i(T')-s^\ominus(T')) dT')= \mu^\ominus(T) +RT \ln a_i(T)##

Also the chemical activity can be defined for current state and standard state with arbitrary temperature and pressure:
$$\mu_i(T,p) \equiv \mu_i^0(T^0,p^0) + RT \cdot \ln \left( a_i(T,p,T^0,p^0) \right)$$

And the correction of chemical activity with changing temperature and pressure of current resp. standard state is:
$$RT' \cdot \ln \left( a_i(T',p',T_{std}',p_{std}') \right) = RT \cdot \ln \left( a_i(T,p,T_{std},p_{std}) \right) - \int_{T}^{T'} \overline{S}_i\, dT + \int_{T_{std}}^{T_{std}'} \overline{S}_{std,i} \, dT + \int_{p}^{p'} \overline{V}_i\, dp - \int_{p_{std}}^{p_{std}'} \overline{V}_{std,i}\, dp$$

$$\mu_i(T,p) \equiv \mu_i^0(T^0,p^0) + RT \cdot \ln \left( a_i(T,p,T^0,p^0) \right)$$