# Definition of activity

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?

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

##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.

BvU
Homework Helper
Yes they certainly can.

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:
$$\mu _{i}-\mu _{i}^{0}=RT\cdot \ln a_{i}$$

BvU
Homework Helper
What do you obtain ?

In that case I have an integral:

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

BvU
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

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

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

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).

BvU
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.

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:

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

BvU
Homework Helper
Why not ?

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

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

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

DrDu
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!

DrDu
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
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:
$$RT' \cdot \ln \left( a_i(T',p',T_{std}',p_{std}') \right) = RT \cdot \ln \left( a_i(T,p,T_{std},p_{std}) \right) - \int_{T}^{T'} \overline{S}_i\, dT + \int_{T_{std}}^{T_{std}'} \overline{S}_{std,i} \, dT + \int_{p}^{p'} \overline{V}_i\, dp - \int_{p_{std}}^{p_{std}'} \overline{V}_{std,i}\, dp$$

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