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Topic: Helmholtz Energy (A) question  (Read 4187 times)

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Offline G O D I V A

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Helmholtz Energy (A) question
« on: November 10, 2009, 04:47:07 AM »
My book says:

A is a state function of the system such that its change provides a criterion to judge the direction of spontaneous change under constant T and V.

I don't get that.  Doesn't A ---> dA = -pdV -SdT.

Doesnt the dV/dT signify changes (infinitesmal/measureable) in V and T i.e. not constant; and so instead p and S should be the constant ones?

Offline Yggdrasil

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Re: Helmholtz Energy (A) question
« Reply #1 on: November 10, 2009, 02:36:04 PM »
In chemical reactions, you also have to consider the effect of changing the numbers of each chemical species in the reaction.  In these conditions, you have:



Where Nj represents the number of moles of chemical species j and µj represents the chemical potential of chemical species j.  At constant volume an temperature, the dV and dT terms drop out, so the change in free energy tells you directly about the changes in the numbers of moles of each chemical species in the system.


Offline G O D I V A

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Re: Helmholtz Energy (A) question
« Reply #2 on: November 10, 2009, 06:17:42 PM »
In chemical reactions, you also have to consider the effect of changing the numbers of each chemical species in the reaction.  In these conditions, you have:



Where Nj represents the number of moles of chemical species j and µj represents the chemical potential of chemical species j.  At constant volume an temperature, the dV and dT terms drop out, so the change in free energy tells you directly about the changes in the numbers of moles of each chemical species in the system.



How does the dV dT drop out? 

If it is constant then dV = 0 meaning -p*0 = 0, same for SdT meaning dA = 0?

Offline Yggdrasil

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Re: Helmholtz Energy (A) question
« Reply #3 on: November 11, 2009, 01:54:40 AM »
How does the dV dT drop out? 

If it is constant then dV = 0 meaning -p*0 = 0, same for SdT meaning dA = 0?

That's correct.

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