Timeline for Concerning the definitions of the enthalpy, the Helmholtz free energy and the Gibbs free energy of a system
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 4, 2019 at 20:04 | vote | accept | CommunityBot | ||
| Apr 4, 2019 at 18:26 | comment | added | BioPhysicist | @PiKindOfGuy Let us continue this discussion in chat. | |
| Apr 4, 2019 at 18:25 | comment | added | user113773 | My bad, I wasn't being literal, I meant that the pressures being well-defined are a consequence of the process being quasistatic. I agree that "if the changes occur too quickly then different parts of the system could have different values for these state variables." The conclusion "so the pressure of the system will be equal to the pressure of the environment" seems questionable to me. Can you back this up with an argument or an authoritative reference? Or at least tell me that you could provide an argument and have me take your word for it? | |
| Apr 4, 2019 at 18:12 | comment | added | BioPhysicist | @PiKindOfGuy No. By quasi-static I mean that changes to the system happen on time scales slower than the time it takes for the system to reach equilibrium after an infinitesimal change. If the changes occur too quickly then different parts of the system could have different values for these state variables. We assume the changes happen slowly (or at least slow enough), and so the pressure of the system will be equal to the pressure of the environment | |
| Apr 4, 2019 at 18:04 | comment | added | user113773 | By "quasistatic" you mean that the pressures of the system and environment are well-defined. This doesn't (seem to) imply that the pressures of the system and the environment are the same. | |
| Apr 4, 2019 at 17:54 | comment | added | BioPhysicist | @PiKindOfGuy Then some of these values don't have well defined values for the entire system, so the thermodynamic identity and definitions cannot be applied | |
| Apr 4, 2019 at 17:53 | comment | added | user113773 | Yes, what if the process in question isn't quasistatic? (See your second-to-last comment.) | |
| Apr 4, 2019 at 17:10 | comment | added | BioPhysicist | @PiKindOfGuy Is there anything else I can do to make things more clear here? | |
| Apr 3, 2019 at 9:24 | comment | added | BioPhysicist | @PiKindOfGuy We usually assume quasi-static processes where at any point in time the system is at equilibrium, so the pressure the system exerts on its boundary is the same as the pressure exerted on the system boundary by the environment | |
| Apr 3, 2019 at 3:13 | comment | added | user113773 | The pressure in those equations refers to the pressure of the environment, not that of the system, no? (See your comment.) | |
| Apr 3, 2019 at 3:06 | comment | added | BioPhysicist | @PiKindOfGuy Yes that is exactly right. As long as the system has well defined Pressure, Volume, Entropy, etc. then those definitions are fine. Why wouldn't they be? They are just definitions in terms of state variables. They don't refer to any processes. It is just how we define work to be $W=\int\mathbf F\cdot\text d\mathbf x$. As long as we have a force acting on an object over a distance the definition applies without any further assumptions of the system. | |
| Apr 3, 2019 at 0:15 | comment | added | user113773 | Sorry, I'm still confused. Are you claiming that if I don't assume anything about the environment those definitions still hold? | |
| Apr 2, 2019 at 12:58 | history | edited | BioPhysicist | CC BY-SA 4.0 |
added 71 characters in body
|
| Apr 2, 2019 at 11:39 | history | edited | BioPhysicist | CC BY-SA 4.0 |
added 127 characters in body
|
| Apr 2, 2019 at 10:29 | history | edited | BioPhysicist | CC BY-SA 4.0 |
added 10 characters in body
|
| Apr 2, 2019 at 10:23 | history | edited | BioPhysicist | CC BY-SA 4.0 |
added 423 characters in body
|
| Apr 2, 2019 at 10:11 | history | answered | BioPhysicist | CC BY-SA 4.0 |