B Change in entropy of reversible isothermal process

AI Thread Summary
The discussion centers on the change in entropy (ΔS) for a system undergoing a reversible isothermal process, highlighting a conflict between two calculations yielding different ΔS values. The first calculation, based on the equation TΔS_sys = Q, suggests ΔS_sys = nRln(V2/V1), while the second approach leads to ΔS_sys = 0 due to the assumption of equilibrium and zero change in internal energy (ΔU) during the isothermal process. Participants explore the implications of Gibbs energy (ΔG) and its relationship with enthalpy (ΔH) and entropy (ΔS), questioning the applicability of certain thermodynamic equations in isothermal conditions. The conversation emphasizes the need for understanding chemical potential terms in reactions and the thermodynamics of mixtures to fully grasp Gibbs energy in chemical processes. The conclusion reinforces that ΔG = ΔH - TΔS is primarily applicable to isothermal reactions, necessitating further study of thermodynamic principles for clarity.
Aurelius120
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Why is there a difference between changes in entropy calculated by these methods? One using Gibbs energy and the other using the equation of first law of thermodynamics
So I had to find change in entropy of system in reversible isothermal process.
$$T\Delta S_{sys.}=Q\implies \Delta S_{sys.}=nRln\left(\frac{V_2}{V_1}\right)$$
This was good because for isothermal process ##\Delta U=0\implies Q=W##

Then I read this
Throughout an entire reversible process, the system is in thermodynamic equilibrium, both physical and chemical, and nearly in pressure and temperature equilibrium with its surroundings. This prevents unbalanced forces and acceleration of moving system boundaries, which in turn avoids friction and other dissipation.

Everything went wrong from here.
For equilibrium, value of Gibb's energy is zero.
$$dG=0\implies dH-TdS_{sys.}=0$$
$$\implies dU+d(PV)-TdS_{sys.}=0$$
Change in internal energy and change in ##(PV)## iis zero for an isothermal process. Therefore,
$$TdS_{sys.}=0\implies dS_{sys.}=0\implies \Delta S_{sys.}=0$$

So why are there two different values of ##\Delta S_{sys.}##?

I thought it could be that one of the above equations involve ##\Delta S_{univ.}## but both involve ##\Delta S_{sys.}##
 
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Along the process path, $$dG=VdP-SdT$$So for a constant temperature path $$dG=VdP=\frac{nRT}{P}dP$$So, $$\Delta G=nRT\ln{P_f/P_i}=-nRT\ln{V_f/V_i}$$Also, for an isothermal change, ##\Delta H=0##, so $$\Delta G=0-T\Delta S$$So $$-T\Delta S=-nRT\ln{V_f/V_i}$$
 
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Chestermiller said:
Along the process path, $$dG=VdP-SdT$$So for a constant temperature path $$dG=VdP=\frac{nRT}{P}dP$$S
How do you get this??
Shouldn't it be:
If ##\Delta G=\Delta H-T\Delta S##
Then ##dG=dH-TdS## (Small Change)

Also if ##\Delta H=\Delta U+\Delta(PV)##
Then ##dH=dU+d(PV)=dU+VdP+PdV##
##dU=0##

Finally $$dG=VdP+PdV-TdS$$

Correct?
 
Aurelius120 said:
How do you get this??
Shouldn't it be:
If ##\Delta G=\Delta H-T\Delta S##
Then ##dG=dH-TdS## (Small Change)

Also if ##\Delta H=\Delta U+\Delta(PV)##
Then ##dH=dU+d(PV)=dU+VdP+PdV##
##dU=0##

Finally $$dG=VdP+PdV-TdS$$

Correct?
$$dG=dU+d(PV)-d(TS)$$
$$dU=TdS-PdV$$
 
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Chestermiller said:
$$dG=dU+d(PV)-d(TS)$$
$$dU=TdS-PdV$$
If ##d## represents a small change and ##\Delta## represents a large change
Why does the formula change from $$dG=dU+d(PV)-d(TS) \text{ for small change }$$
To $$\Delta G=\Delta U+\Delta(PV) -T\Delta S \text{ for big change}$$
 
For big change, $$\Delta G=\Delta U+\Delta (PV)-\Delta (TS)$$
 
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So the formula I learned in chemistry $$\Delta G=\Delta H-T\Delta S= \Delta U+\Delta(PV)-T\Delta S$$ is only for isothermal cases??
Since I have used it for reactions in chemistry, most chemical reactions are isothermal??

And since products and reactants are different compounds that would explain why ##\Delta H## is not zero despite it being isothermal.
 
Aurelius120 said:
So the formula I learned in chemistry $$\Delta G=\Delta H-T\Delta S= \Delta U+\Delta(PV)-T\Delta S$$ is only for isothermal cases??
Yes.
Aurelius120 said:
Since I have used it for reactions in chemistry, most chemical reactions are isothermal??
If there is a reaction taking place, you need additional chemical potential terms in your equation for dG.
Aurelius120 said:
And since products and reactants are different compounds that would explain why ##\Delta H## is not zero despite it being isothermal.
Have you learned about thermodynamics of mixtures yet?
 
Chestermiller said:
Yes.

If there is a reaction taking place, you need additional chemical potential terms in your equation for dG.
Most questions we have on Gibb's Energy are either on spontaneity, or application of that formula in isothermal case.##\Delta G=\Delta H-T\Delta S##
##\Delta H=\Delta U +\Delta n_gRT##
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Chestermiller said:
Have you learned about thermodynamics of mixtures yet?
Unless you are referring to calorimetry of mixtures,(calculating final temperature given the value specific heats, heat gain, amonut of substances mixed, initial temperature etc.), I am afraid not.
 
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Aurelius120 said:
Most questions we have on Gibb's Energy are either on spontaneity, or application of that formula in isothermal case.##\Delta G=\Delta H-T\Delta S##
##\Delta H=\Delta U +\Delta n_gRT##
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What is your understanding of what ##\Delta G^0## or ##\Delta G## represents in these problems?
Aurelius120 said:
Unless you are referring to calorimetry of mixtures,(calculating final temperature given the value specific heats, heat gain, amonut of substances mixed, initial temperature etc.), I am afraid not.
You really need to learn about the thermodynamics of mixtures (solutions) if you are going to understand the use of G with regard to chemical reactions and chemical equilibrium. See Chapters 10 and 11 in Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics.
 
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Chestermiller said:
What is your understanding of what ##\Delta G^0## or ##\Delta G## represents in these problems?
Since it represents the useful work, I take it to be the extra energy to be provided or the energy released after comtribution from the environment.##[-\Delta(TS)]##
 
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Aurelius120 said:
Since it represents the useful work, I take it to be the extra energy to be provided or the energy released after comtribution from the environment.##[-\Delta(TS)]##
It represents the change in G in going reversibly from pure reactants in separate containers at T and P to pure products in separate containers at T and P. The useful work and heat added depend on process path.
 
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