Definition of activity

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  • #1
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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?
 

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  • #2
BvU
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##T## is the current temperature. Lemma is pretty clear about that !
 
  • #3
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##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.
 
  • #4
BvU
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Yes they certainly can.
 
  • #5
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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:
$$\mu _{i}-\mu _{i}^{0}=RT\cdot \ln a_{i}$$
 
  • #6
BvU
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What do you obtain ?
 
  • #7
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In that case I have an integral:

$$\mu _{i}-\mu _{i}^{0}=R\cdot \int_{a_{i}^{0}}^{a_{i}} T(a_{i})\cdot d\ln a_{i}$$
 
  • #8
BvU
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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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  • #9
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dμi=d(μi−μ⊖i)
That is not true. It that case μi should be always zero.
 
  • #10
BvU
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No. It is a constant. So the ## \ d(\mu _{i}^\ominus) = 0 ##.
 
  • #11
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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).
 
  • #12
BvU
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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.
 
  • #13
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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:

$$d\mu_i = d\left ( RT\ln a_i \right ) $$
 
  • #14
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Why not ?
 
  • #15
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Sorry I meant:
$$d\mu_i = RT\cdot d\ln a_i $$
 
  • #16
BvU
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Is wrong. T is not a constant like R
 
  • #17
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Yes the total differential should be probably:
$$d\mu _{i}(dT,dp,dx_{1},...,dx_{N})=RT\cdot d\ln a_{i}(dp,dx_{1},...,dx_{N})+R\cdot \ln a_{i}\cdot dT$$
 
  • #18
DrDu
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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!
 
  • #19
DrDu
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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)##.
 
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  • #20
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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:
$$RT' \cdot \ln \left( a_i(T',p',T_{std}',p_{std}') \right) = RT \cdot \ln \left( a_i(T,p,T_{std},p_{std}) \right) - \int_{T}^{T'} \overline{S}_i\, dT + \int_{T_{std}}^{T_{std}'} \overline{S}_{std,i} \, dT + \int_{p}^{p'} \overline{V}_i\, dp - \int_{p_{std}}^{p_{std}'} \overline{V}_{std,i}\, dp$$
 
  • #21
DrDu
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Also the chemical activity can be defined for current state and standard state with arbitrary temperature and pressure:
$$\mu_i(T,p) \equiv \mu_i^0(T^0,p^0) + RT \cdot \ln \left( a_i(T,p,T^0,p^0) \right) $$
I think the activity is always defined for the same temperature and pressure as the corresponding standard state. So I won't assume a to depend on two temperatures and pressures.
And the correction of chemical activity with changing temperature and pressure of current resp. standard state is:
Take in mind that the two integrals over T and p aren't independent. Rather it is a line integral over both variables. Only for a special path, like ##(p_0, T_0) \to (p_0, T'_0) \to (p'_0, T'_0)## can this integral be written as two consecutive integrals over T (at fixed ##p_0##) and p (at fixed ##T'_0##).
 

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