Timeline for Compressing and cooling air in a Coolgardie Safe, then decompressing
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 23, 2020 at 6:44 | comment | added | El Flea | So basically if I just used $$Vfinal/V$$, I wouldn't have to bother using the 2/7, right? | |
| May 22, 2020 at 15:11 | comment | added | probably_someone | @ElFlea Using the adiabatic relation between $P$ and $V$ before and after expansion, you can find an expression for $V_{final}/V$ and substitute it into the last equation. | |
| May 22, 2020 at 9:16 | vote | accept | El Flea | ||
| May 22, 2020 at 9:14 | comment | added | El Flea | How did the final equation have an exponent of 2/7? I'm a 10th grader so it would help to explain how $$T_{cool}\left(\frac{V_{final}}{V}\right)\left(\frac{P_{ambient}}{P_{cool}}\right)$$ became $$T_{cool}\left(\frac{P_{ambient}}{P_{cool}}\right)^{2/7}$$ | |
| May 16, 2020 at 2:21 | vote | accept | El Flea | ||
| May 22, 2020 at 9:12 | |||||
| May 15, 2020 at 20:30 | comment | added | probably_someone | @ElFlea The decompressed air will be coldest if it decompresses quickly. Slowing the decompression will make the decompressed air warmer. If you do the decompression extremely slowly, then the air will be at room temperature once decompressed, since it has enough time to come to equilibrium with the (comparatively warm) surrounding air as it decompresses. | |
| May 15, 2020 at 20:12 | comment | added | El Flea | What happens if the can is decompressed slowly. Because I think that's the whole point. I want to make the air become cooler as it is decompressed and I guess that's only possible if it loses heat to the surroundings? | |
| May 15, 2020 at 19:58 | history | edited | probably_someone | CC BY-SA 4.0 |
added 4 characters in body
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| May 15, 2020 at 19:53 | history | answered | probably_someone | CC BY-SA 4.0 |