Timeline for A thermodynamic paradox regarding the first law
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2017 at 18:30 | vote | accept | AHB | ||
| Jan 18, 2017 at 4:18 | comment | added | AHB | @AlbertAspect . Yes. He meant our formula doesn't work here. I misunderstood. | |
| Jan 18, 2017 at 1:48 | comment | added | user126422 | @AHB I would not say that either. I am not really sure why Zemansky would say that. My guess is that he means that dW=PdV is undefined because P is undefined, so you should rather go back and use the definition: dW=Fdx=F/AdV instead. | |
| Jan 17, 2017 at 19:50 | answer | added | hyportnex | timeline score: 2 | |
| Jan 17, 2017 at 19:35 | answer | added | ZachMcDargh | timeline score: 2 | |
| Jan 17, 2017 at 19:18 | comment | added | AHB | @AlbertAspect So would it be correct to say that the infinitesimal work in undefined? This is what Zemansky is stating. That during a very short time, P is not defined, to the infinitesimal work is also undefined. | |
| Jan 17, 2017 at 19:17 | comment | added | AHB | @ZachMcDargh so they are both undefined? | |
| Jan 17, 2017 at 19:03 | comment | added | user126422 | They are both defined, even if P has not a defined global value the gas still makes a defined force on the piston (even if this force is unpredictable). Q is also defined, why do you think it is not? | |
| Jan 17, 2017 at 18:50 | comment | added | ZachMcDargh | Why would the sum of two undefined quantities be undefined? "Undefined" here is not like when you divide by zero, it just means you literally haven't defined it. | |
| Jan 17, 2017 at 18:41 | history | asked | AHB | CC BY-SA 3.0 |