Helmholtz and Gibbs free energy for an adiabatic process

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1. Apr 16, 2017

wwildlifee

1. The problem statement, all variables and given/known data
Calculate changes in A and G of one mole of an ideal gas that undergoes the following processes respectively.
1. adiabatic expansion from (T1, P1) to (T2, P2)
2. isobaric expansion from (P, V1, T1) to (P, V2, T2) (if it is not isothermal)
3. isochoric expansion from (V, P1, T1) to (V, P2, T) (if it is not isothermal)

2. The attempt at a solution
I can get a solution for isothermal processes, but for cases that temperature changes, I cannot calculate changes of A and G because I cannot even find a clue for handling d(TS) = TdS + SdT.
I tried to solve these by using the Gibbs-Helmholtz equation, but I failed.
And also I separated the path into two steps, but I failed again.

How can I calculate the change in A and G for processes in which temperature changes?

Thank you.

2. Apr 16, 2017

stevendaryl

Staff Emeritus
3. Apr 16, 2017

Staff: Mentor

In case 1, if $S=S_1$ at (T1, P1), what is the value of S at (T2,P2)?

4. Apr 16, 2017

wwildlifee

Thanks
However this is not what I do not know.
What I want to know is how to handle the integration of (-SdT) in dA or dG for the processes above(adiabatic, isobaric, isochoric).

5. Apr 16, 2017

wwildlifee

I can get the changes in S for the processes above, so if I already know S1, S2 = delta_S + S1 respectively.
Is it a hint for calculating the delta_A and delta_G for processes in which temperature changes?

6. Apr 16, 2017

Staff: Mentor

Sure.

Is this for a class in classical thermodynamics or a class in statistical thermodynamics?

What is the equation for $\Delta G$ in terms of $\Delta H$ and $\Delta(TS)$?

7. Apr 16, 2017

wwildlifee

This course is the classical thermodynamics.
So all I've got are
dU = Tds - pdV
dH = Tds + Vdp
dA = d(U-TS) = dU - TdS - SdT = -pdV - SdT
dG = d(H-TS) = dH - TdS - SdT = Vdp - SdT
Maxwell relations, and the Gibbs-Helmholtz equation.

I I tried to solve case 1 by separating the adiabatic process into two steps: an isothermal process and then an isochoric process.
In other words,
A: (T1, P1, V1)
B: (T1, P, V)
C: (T2, P2, V2)
A -> C (adiabatic process) = A -> B -> C
But in this case, I still need to calculate ΔA and ΔG for another isochoric process...

In order to solve this problem, the statistical thermodynamics is needed?

8. Apr 16, 2017

Staff: Mentor

No. Just checking.

You still haven't answered my question regarding the definition of the Gibbs Free Energy G: What is the equation for $\Delta G$ in terms of $\Delta H$ and $\Delta(TS)$?

Another question: If you know the initial and final states of an ideal gas, do the details of the process that took you from the initial to the final state matter in determining the changes in U, H, S, A, and G between these states?

9. Apr 16, 2017

wwildlifee

The definition of G is G = H - TS, so ΔG is ΔG = ΔH - Δ(TS) = ΔH - TΔS - SΔT

I'm not sure that I understand your question correctly, but U, H, S, A and G are state functions, so if I know the initial and final state, I think the details of the process do not matter.

10. Apr 16, 2017

Staff: Mentor

ΔH - Δ(TS) = ΔH - TΔS - SΔT??? This is not correct mathematically.
So, since you are told the initial and final states, why are you worried about "adiabatic," "isochoric," and "isobaric?"

For an ideal gas, what is the change in entropy between the two end states of processes 1, 2, and 3? What is the change in internal energy? What is the change in enthalpy?

11. Apr 16, 2017

wwildlifee

So, is it enough to express ΔG as ΔH - Δ(TS) ?

I assume the processes are all reversible. And I think it is okay to use Cv and Cp here.
1) ΔU = Cv (T2 - T1), ΔH = Cp (T2 - T1), ΔS = 0 (because dq = 0)
2) ΔU = Cv (T2 - T1), ΔH = Cp (T2 - T1), ΔS = Cp ln(T2/T1) - R ln(P2/P1) (from dH = TdS + Vdp)
3) ΔU = Cv (T2 - T1), ΔH = Cp (T2 - T1), ΔS = Cv ln(T2/T1) (from dU = TdS - pdV)

I'm getting confused more and more

12. Apr 16, 2017

Staff: Mentor

You are very close now. I'll be back tomorrow.

13. Apr 17, 2017

Staff: Mentor

yes.
If the actual process was irreversible, then, even if it was adiabatic, $\Delta S$ would not be zero. It is impossible to find an adiabatic reversible path that gives the same change in entropy as an irreversible adiabatic path; the reversible path between the same two end states must always involve some heat transfer. For the adiabatic path to be reversible, the temperatures and pressures at the two end points must be related by Cp ln(T2/T1) - R ln(P2/P1)=0. This is not necessarily the case here. For this case, the most we can say is that ΔS = Cp ln(T2/T1) - R ln(P2/P1)

The problem statements says that $P_2=P_1=P$. Therefore, for this case, ΔS = Cp ln(T2/T1), without the pressure term.

Correct!!!

Now for the changes in A and G. Considering Case 2, $$\Delta G=\Delta H-\Delta (TS)=\Delta H-T_2(S_1+\Delta S)+T_1S_1$$So,
$$\Delta G=C_p(T_2-T_1)-S_1(T_2-T_1)-C_pT_2\ln(T_2/T_1)$$
This can also be written as:
$$\Delta G=C_p(T_2-T_1)-\left(S_1+\frac{C_p\ln(T_2/T_1)}{2}\right)(T_2-T_1)-\frac{(T_1+T_2)}{2}C_p\ln(T_2/T_1)$$
Note that, in neither version is it possible to eliminate the entropy in the initial state $S_1$ from the equation for the change in G.

14. Apr 17, 2017

wwildlifee

I also thought that the entropy of the initial state S1 (and final state) might be needed.
However, I thought that there is a way to express ΔA and ΔG only in terms of the variables which are given in the problems.
So I tried, but it is an impossible work (at this point)!

I think you are right, that $\Delta A$ involves the quantity $S_1 \Delta T$ (for an adiabatic change). So it seems to me that you need to know the original entropy in order to solve for $\Delta A$.