Timeline for What stops entropy from forming if you extract 100% of heat flow and use it for work?
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Jun 22, 2022 at 13:36 | vote | accept | bananenheld | ||
| May 14, 2022 at 21:41 | history | edited | Bob D | CC BY-SA 4.0 |
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| May 12, 2022 at 1:34 | comment | added | Bob D | @JánLalinský See my edit | |
| May 12, 2022 at 1:34 | history | edited | Bob D | CC BY-SA 4.0 |
additional information
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| May 12, 2022 at 1:30 | comment | added | Bob D | @JánLalinský OK, I will include it. Thanks. | |
| May 12, 2022 at 1:21 | comment | added | Ján Lalinský | I suppose if that is an assumption, it would be good to state it in your answer. Because the question does not seem to imply it. | |
| May 12, 2022 at 1:19 | comment | added | Bob D | @JánLalinský I explained it in an earlier answer to this post, but there has been so many iterations since then it got lost. I can restate the assumptions if that would satisfy you. | |
| May 12, 2022 at 1:13 | comment | added | Ján Lalinský | @BobD where does this irreversible process with a sudden removal of a weight come from? It is not stated in the question. Is it your own additional assumption? | |
| May 12, 2022 at 1:06 | comment | added | Bob D | Now after the weight is removed, the gas is allowed to expand (irreversibly, since the gas is not in equilibrium with the external pressure) until the gas comes into thermal equilibrium with the thermal reservoir of the surroundings, i.e., in equilibrium at state 2. We call this path isothermal because even though the gas is not in internal equilibrium, the temperature of the gas at the boundary with the surroundings equals the temperature of the thermal reservoir | |
| May 12, 2022 at 1:06 | comment | added | Bob D | Now suddenly the weight is removed. The external pressure suddenly drops to the final external pressure $P_2$, but the internal pressure of the gas is not in equilibrium with the external pressure . Also, since the reduction in external pressure happens so quickly (non quasi-statically) it is an irreversible process and there is not enough time for heat transfer to occur, i.,e in effect it happens adiabatically | |
| May 12, 2022 at 1:06 | comment | added | Bob D | @JánLalinský Because the path occurs as follows: At equilibrium state 1, visualize a vertically oriented piston/cylinder with a weight on top. That weight plus atmospheric pressure constitutes the initial external pressure on the system at state 1. It also equals the equilibrium pressure of the gas. | |
| May 12, 2022 at 0:43 | comment | added | Ján Lalinský | @BobD why do you assume the horizontal and vertical lines are irreversible, and why do you assume they are isothermal? | |
| May 10, 2022 at 18:46 | history | edited | Bob D | CC BY-SA 4.0 |
correction PV diagram
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| May 10, 2022 at 15:26 | comment | added | Bob D | @bananenheld After re reading my answer I realized it did not correctly respond to your original question. So I have completely revised it as shown above. Check it out and see if it helps. | |
| May 10, 2022 at 15:25 | history | undeleted | Bob D | ||
| May 10, 2022 at 15:24 | history | edited | Bob D | CC BY-SA 4.0 |
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| May 10, 2022 at 9:20 | history | edited | Bob D | CC BY-SA 4.0 |
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| May 10, 2022 at 9:19 | history | deleted | Bob D | via Vote | |
| May 9, 2022 at 20:39 | comment | added | bananenheld | Ahh I think I now understand it. For an isothermal reversible process: $Q_{in}=-W$ For an isothermal irreversible process $Q_{in}=T \Delta S-W$. Since $T$ is constant and not all heat is converted into work, this means $Q_{out} \neq 0 \implies T\Delta S \neq 0 \implies \Delta S \neq 0$. But this means taht an isothermal path can be irreversible if it doesn't reach max $W$ | |
| May 9, 2022 at 18:52 | comment | added | bananenheld | I don't understand how it is the 'other way around', isn't that kind of redundant? I am probably misunderstanding here. But an isothermal expansion is reversible because you transfer all heat into work so the work is 'captured' and no energy is 'lost'. That is why it stays isothermal. If it didn't do that, the temperature would change wouldn't it? | |
| May 9, 2022 at 17:45 | history | edited | Bob D | CC BY-SA 4.0 |
additional information
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| May 9, 2022 at 13:31 | history | answered | Bob D | CC BY-SA 4.0 |