Gibbs and Helmholtz equations for thermodynamic processes

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SUMMARY

The Gibbs and Helmholtz equations are essential for analyzing thermodynamic processes, specifically ΔG=ΔH-TΔS and ΔA=ΔU-TΔS. For adiabatic, isothermic, constant volume, and constant pressure processes, the differential forms dG = -SdT + VdP and dA = -SdT - PdV are applicable. Each process requires setting specific terms to zero: dT=0 for isothermic, dP=0 for constant pressure, and dV=0 for constant volume. The discussion emphasizes the importance of understanding whether chemical reactions are involved, as this influences the applicability of the equations.

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tag16
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For a thermodynamic process, what equations would be used to find the change in Gibbs and Helmholtz free energy when:
a.)The process is adiabatic
b.)The process is isothermic
c.)The process is at constant volume
d.)The process is at constant pressure

I know ΔG=ΔH-TΔS and ΔA=ΔU-TΔS but do these equations apply to all four processes?
 
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I think you need to get into the thermodynamic partial derivative definition of these processes, i.e.

Gibbs Energy:
dG = -SdT + VdP
Helmholtz Energy:
dA = - SdT - PdV

Now you set the appropriate terms to zero.

i.e.
adiabatic: is it a reversible process?
isothermic: dT=0
isobaric (const pressure): dP=0
const vol: dV=0
 
Does the process include chemical reactions?
 
Basically, I'm asking if you were given a PV graph similar to this:

http://www.websters-online-dictionary.org/images/wiki/wikipedia/commons/thumb/d/dc/Stirling_Cycle.png/200px-Stirling_Cycle.png


How would you find ΔG and ΔA for the processes 1 to 2, 2 to 3, 3 to 4 and 4 to 1?
 
Last edited by a moderator:
For 1 to 2:
ΔG = ΔH - TΔS
ΔA = ΔU - TΔS

ΔU=NCvΔT=0
ΔH=NCpΔT=0

so ΔA = ΔG=-TΔS; you know T but need to find ΔS. There are two derivations; either will work for you. I'm going fast so may have mixed up negative signs & numerators/denominators, you need to double check the math when you do it yourself to make sure nothing is wrong.

Derivation 1
ΔU = Q + W = 0 b/c ΔU = 0.
Q = -W
dW = -PdV
dW= -(NRT/V)dV (plugged in ideal gas law)
W = -NRTln(V2/V1)
so Q = NRTln(V2/V1)
ΔS = Q/T = -NRln(V2/V1)Derivation 2 (my preference)
Use Maxwell's relations.
(dS/dV) at const T = (dP/dT) at const V. Plug in ideal gas law.
So dS/dV = NR/V; integrate to get ΔS = NR ln (V2/V1).
 
Last edited:
For 2 to 3:
Isochoric Cooling - Const Volume.

ΔU = NCvΔT = Q + W.
W = PΔV so W = 0.

ΔU = Q = NCvΔT.
ΔH=ΔU+Δ(PV)=ΔU+VΔP
 
Last edited:
tag16 said:
I know ΔG=ΔH-TΔS and ΔA=ΔU-TΔS but do these equations apply to all four processes?

I was asking whether chemical reactions are taking place (which, as I learned from your answer, is not the case) because the formulas above refer to the change of G or A due to a chemical reaction taking place at constant T and p. The "Delta" is not simply an abbreviation for a difference here but
\Delta X=\sum_i \nu_i \partial X/\partial n_i |_{T, P} which the nu_i being the stochiometric coefficients of the reaction taking place. So this formula is little helpful when you consider a process which does not involve chemical reactions.
 

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