Is ΔG always equal to 0 when ΔSsystem ≠ 0 and qP = 0?

[sgstudent]

ΔG=ΔH-TΔS_system and ΔS_system=q_p/T and isn't q_p=ΔH? Which means ΔG=0 always which is obviously wrong.

But the derivation for Gibbs Free Energy is based on q_surrounding=-q_system so shouldn't the term be equals to 0?

 

[Ygggdrasil]

ΔG=ΔH-TΔS_system and ΔS_system=q_p/T and isn't q_p=ΔH? Which means ΔG=0 always which is obviously wrong.

ΔS_system=q_p/T is true only for reversible processes. What do you know about the ΔG of a reversible process?

 

[sgstudent]

ΔS_system=q_p/T is true only for reversible processes. What do you know about the ΔG of a reversible process?

It is 0. So for a reversible process q_p=ΔH and so ΔG=0?

For an irreversible process how would the ΔS look like?

Thanks

 

[Chestermiller]

It is 0. So for a reversible process q_p=ΔH and so ΔG=0?

For an irreversible process how would the ΔS look like?

Thanks

Are you aware that ΔG and ΔS are functions only of the initial and final thermodynamic equilibrium states of a system, and are independent of path between the two states, whether reversible or irreversible? From what you have learned, do you know how to determine the change in S between two thermodynamic equilibrium states of a system? Are you familiar with the equation dG = -SdT + VdP, which applies to a differential change in temperature and pressure between two closely neighboring thermodynamic equilibrium states of a system?

I think your real question is "how do one determine ΔG and ΔS between two thermodynamic equilibrium states of a system."

Chet

 

[Radhakrishnam]

It is possible for ΔS_system ≠ 0, even when q_P = 0. Therefore, q_P = 0, does not necessarily lead to ΔG = 0.

P. Radhakrishnamurty