Timeline for Issue with the work of a reversible process
Current License: CC BY-SA 3.0
19 events
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| S Feb 24, 2018 at 10:11 | history | bounty ended | Landau | ||
| S Feb 24, 2018 at 10:11 | history | notice removed | Landau | ||
| Feb 24, 2018 at 10:10 | vote | accept | Landau | ||
| Feb 20, 2018 at 11:17 | answer | added | knzhou | timeline score: 2 | |
| S Feb 20, 2018 at 10:53 | history | bounty started | Landau | ||
| S Feb 20, 2018 at 10:53 | history | notice added | Landau | Draw attention | |
| Feb 20, 2018 at 10:51 | history | edited | Landau | CC BY-SA 3.0 |
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| Feb 17, 2018 at 15:52 | history | edited | Landau | CC BY-SA 3.0 |
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| Feb 15, 2018 at 14:06 | history | edited | Landau | CC BY-SA 3.0 |
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| Feb 14, 2018 at 13:49 | history | edited | Landau | CC BY-SA 3.0 |
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| Feb 14, 2018 at 9:30 | history | edited | Landau | CC BY-SA 3.0 |
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| Feb 13, 2018 at 19:49 | answer | added | hyportnex | timeline score: 0 | |
| Feb 13, 2018 at 19:46 | comment | added | hyportnex | In general the 1st law would say $dU=\delta Q +\delta L$ but you have postulated a purely mechanical energy exchange, that is one with $dU=\delta L$. This is called an adiabatic process. If you further assume that the gas is internally frictionless then the partition will oscillate back and forth just as an ideal spring would do. If the gas has internal friction then the oscillation of the partition will be dissipated and the partition will stop at a point where the pressures will be equal, i.e, $\delta p = 0$. Read the subject called "the adiabatic piston". | |
| Feb 12, 2018 at 19:37 | history | edited | Landau | CC BY-SA 3.0 |
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| Feb 12, 2018 at 16:50 | comment | added | hyportnex | It is not "in contrast"; given the situation you defined by the constraint $dL_1 +dL_2=0$ you derived that the only virtual displacement compatible with your constraint and equilibrium is $dp=0$ (see d'Alembert principle). | |
| Feb 12, 2018 at 15:19 | comment | added | Landau | The result $dP=0$ is in contrast with my hypothesis. In this case, the pressure doesn't increase during the process, and so you haven't a compression/expansion of the gas: the system doesn't evolve in time. @hyportnex | |
| Feb 12, 2018 at 14:30 | comment | added | By Symmetry | What exactly do $L$ and $k$ mean here? | |
| Feb 12, 2018 at 14:00 | comment | added | hyportnex | while I do not quite understand what you wrote but it seems to me that you have deduced that in mechanical equilibrium $p_0=p_1$, which is correct. | |
| Feb 12, 2018 at 13:44 | history | asked | Landau | CC BY-SA 3.0 |