Timeline for What happens to energy changes in non-reversible processes?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 19, 2015 at 8:25 | vote | accept | RobChem | ||
| Jan 12, 2015 at 18:09 | comment | added | RobChem | Thank you. How does the system lose the energy? As heat? | |
| Jan 12, 2015 at 17:15 | comment | added | DavePhD | yes, I added to my answer | |
| Jan 12, 2015 at 15:48 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Jan 12, 2015 at 14:15 | comment | added | RobChem | Okay. May you please address the second point in my question. | |
| Jan 12, 2015 at 14:12 | comment | added | DavePhD | when the process in irreversible, S can increase without any change in U. | |
| Jan 12, 2015 at 14:10 | comment | added | RobChem | Why? It has units of energy. | |
| Jan 12, 2015 at 14:09 | comment | added | DavePhD | (response to second to last comment) the right hand side no longer represents energy put into the system, because TdS no longer represents q | |
| Jan 12, 2015 at 14:03 | comment | added | RobChem | Perhaps my question is better put in another way. When at a constant P and T $dG\leq w_e$ indicating that the change in gibbs free energy at a constant P and T is the electrical work done. In a non-reversible process, where the loss of Gibbs free energy is greater in magnitude (the electrical work is also negative) then where does the rest of the energy go? | |
| Jan 12, 2015 at 14:01 | comment | added | RobChem | I understand that but the right hand side still represents energy put into the system and the left is the change in energy of the system right? So where does the energy go if he two sides of the equation are not equal (i.e. when the process is irreversible) | |
| Jan 12, 2015 at 14:00 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Jan 12, 2015 at 13:59 | comment | added | DavePhD | as an inequality, $dU\leq TdS-PdV$ is valid even without assuming equilibrium, but assuming only PV work. As an equality, it assumes quasistatic (near equilibrium) and only PV work. | |
| Jan 12, 2015 at 13:57 | comment | added | RobChem | I have edited my question now - I see my error. | |
| Jan 12, 2015 at 13:56 | comment | added | RobChem | Sorry, I see now I think. However, $dU \leq TdS-PdV$ is a valid inequality isn't it? Assuming equality at equilibrium? | |
| Jan 12, 2015 at 13:56 | comment | added | DavePhD | If you write $dU\leq TdS-PdV$ as an equality, that assumes quasistatic (near equilibrium) and only PV work. | |
| Jan 12, 2015 at 13:53 | comment | added | DavePhD | $\Delta U = q + w$ and $dU = dq + dw$ do not assume equilibrium. | |
| Jan 12, 2015 at 13:49 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Jan 12, 2015 at 13:46 | comment | added | RobChem | The equality assumes equilibrium though doesn't it? This is certainly the case for G: $dG\leq VdP-SdT$. What happens in that instance? | |
| Jan 12, 2015 at 13:44 | history | answered | DavePhD | CC BY-SA 3.0 |