Timeline for How would one calculate the entropy change for an adiabatic irreversible process?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 10, 2018 at 16:02 | comment | added | Archer | Ah, sorry! I was misinterpreting it. | |
| Mar 10, 2018 at 15:57 | comment | added | Chet Miller | I really don't understand what you have written here. In my equation, the expression on the left hand side is the change in internal energy of the ideal gas, and the term on the right hand side is the irreversible work done by the gas on its surroundings. Do you somehow disagree with the expressions on both sides of the equation? | |
| Mar 10, 2018 at 15:53 | comment | added | Archer | @ChesterMiller How did you obtain it? $nC_vT \ne -P_fV_f$ | |
| Mar 10, 2018 at 15:51 | comment | added | Chet Miller | Which part do you think is not true? (Incidentally, you're missing a minus sign on the right hand side of the equation). | |
| Mar 10, 2018 at 15:49 | comment | added | Chet Miller | @GaurangTandon By a one-step path, I meant a solely adiabatic path of infinitesimal increments or a solely isothermal path of infinitesimal increments. My point was that the overall reversible process can consist partly of an adiabatic path and partly of an isothermal path. | |
| Mar 10, 2018 at 15:42 | comment | added | Archer | In Step 1 of Physics Formums insights, it is written: "$\Delta U = nC_v(T_f- T_0)= P_f(V_f-V_o)$" Can you please explain how this is true? | |
| Mar 10, 2018 at 15:41 | comment | added | Gaurang Tandon | "the reversible process does not have to be done using a one-step path" don't all reversible processes always have infinite steps? | |
| Mar 10, 2018 at 15:29 | history | answered | Chet Miller | CC BY-SA 3.0 |