Activity Definition: Is T Temp of Current State?

In summary: I'm sorry, I misunderstood what you meant. I thought you meant that the activity should depend on both the current state and the standard state's temperature and pressure. Yes, I agree with you that the activity should only depend on the temperature and pressure of the current state and the standard state, not on the temperature and pressure of both states simultaneously.
  • #1
ussername
60
2
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
##T## is the current temperature. Lemma is pretty clear about that !
 
  • #3
BvU said:
##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
Yes they certainly can.
 
  • #5
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
What do you obtain ?
 
  • #7
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
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
BvU said:
dμi=d(μi−μ⊖i)
That is not true. It that case μi should be always zero.
 
  • #10
No. It is a constant. So the ## \ d(\mu _{i}^\ominus) = 0 ##.
 
  • #11
BvU said:
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
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).

ussername said:
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
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
Why not ?
 
  • #15
Sorry I meant:
$$d\mu_i = RT\cdot d\ln a_i $$
 
  • #16
Is wrong. T is not a constant like R
 
  • #17
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
ussername said:
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
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
Also the change of chemical potential with temperature resp. pressure is:
$$\left( \frac{\partial \mu_i}{\partial T} \right)_{p,\vec{n}} = - \overline{S}_i$$ $$\left( \frac{\partial \mu_i}{\partial p} \right)_{T,\vec{n}} = \overline{V}_i$$
DrDu said:
##\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
ussername said:
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##).
 

1. What is the purpose of activity definition in scientific research?

The purpose of activity definition is to clearly outline the steps and procedures necessary to conduct a scientific experiment or study. It helps to ensure that all aspects of the research are well-documented and organized, making it easier to replicate or build upon in the future.

2. How do you determine the temperature of the current state in an activity definition?

The temperature of the current state can be determined through various methods, depending on the specific experiment or study. Common techniques include using a thermometer, temperature sensors, or thermal imaging cameras. It is important to use reliable and accurate methods to obtain the most precise temperature measurement.

3. Why is the temperature of the current state important in scientific research?

The temperature of the current state can provide valuable information about the conditions and environment in which an experiment or study is being conducted. It can affect the results and outcomes of the research, and can also help to identify any potential sources of error or bias.

4. How does the temperature of the current state impact the validity of scientific results?

The temperature of the current state can have a significant impact on the validity of scientific results. Changes in temperature can affect the behavior of materials, chemical reactions, and biological processes, ultimately influencing the outcomes of the research. It is important to control and monitor the temperature to ensure reliable and accurate results.

5. What are some common factors that can affect the temperature of the current state in scientific research?

There are many factors that can affect the temperature of the current state in scientific research, including location, time of day, weather conditions, and human intervention. It is crucial to consider and control these factors to maintain consistency and accuracy in the research.

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