Timeline for Is reversible work a point function?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 14, 2017 at 9:33 | comment | added | Novice C | This could be taken one step further by invoking the fundamental thermodynamic relationship. Then for a reversible process we have $\delta W_{\textrm{rev}} = -p \textrm{d} V + f \textrm{d} X$. It is clear in general that pdV work is path dependent from valerio92 example, and you would not call $\delta W$ a state function, as it depends wholly on the path. Just because something can be expressed in terms of it's initial and final state does not make it a state function, note you are only able to do so after assuming a path. This is the entire point of using $\delta W$ vs $\textrm{d}W$. | |
| Jun 23, 2016 at 13:20 | comment | added | valerio | Actually for reversible isothermal and reversible isentropic it is a trivial result because you are choosing a single curve connecting the two points in the thermodynamic plane. It is a non-trivial result in the case of the irreversible adiabatic, because in that case it is impossible to draw a curve in the thermodynamic plane. | |
| Jun 23, 2016 at 12:34 | comment | added | valerio | It is true for a reversible isothermal, general adiabatic ($\delta Q=0$) and reversible isentropic ($dS=0$) transformation. You could add those cases to your answer ;-) | |
| Jun 23, 2016 at 11:35 | history | edited | Dimitri | CC BY-SA 3.0 |
added 55 characters in body
|
| Jun 23, 2016 at 11:34 | comment | added | Dimitri | You're right, at least it seems rigorous for an isotherm but it probably is a more complicated question in the general case. | |
| Jun 23, 2016 at 8:59 | comment | added | Auburn | Yes but Tds is path dependent. | |
| Jun 22, 2016 at 13:01 | comment | added | valerio | *I meant isobar+isochor | |
| Jun 22, 2016 at 12:48 | comment | added | valerio | This is only true if $T$ is constant, i.e. along an isotherm. For example, consider the reversible transformation of an ideal gas from point $A$ to point $B$ along an isotherm and then along an isobar. | |
| Jun 22, 2016 at 12:11 | history | answered | Dimitri | CC BY-SA 3.0 |