Change in entropy of reversible isothermal process

[Aurelius120]

TL;DR

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 $\Delta S_{sys.}$

 

[Chestermiller]

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 $$-T\Delta S=-nRT\ln{V_f/V_i}$$

 

[Aurelius120]

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?

 

[Chestermiller]

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?

$$dG=dU+d(PV)-d(TS)$$
$$dU=TdS-PdV$$

 

[Aurelius120]

$$dG=dU+d(PV)-d(TS)$$
$$dU=TdS-PdV$$

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 $$\Delta G=\Delta U+\Delta(PV) -T\Delta S \text{ for big change}$$

 

[Chestermiller]

For big change, $$\Delta G=\Delta U+\Delta (PV)-\Delta (TS)$$

 

[Aurelius120]

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.

 

[Chestermiller]

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.

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.

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?

 

[Aurelius120]

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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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.

 

[Chestermiller]

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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View attachment 342051

View attachment 342052

What is your understanding of what $\Delta G^0$ or $\Delta G$ represents in these problems?

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.

 

[Aurelius120]

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.$[-\Delta(TS)]$

 

[Chestermiller]

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.