Timeline for According to KMT, is the velocity of an ideal gas always sqrt(3RT/M)?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| S Jan 27, 2020 at 17:02 | history | suggested | CommunityBot | CC BY-SA 4.0 |
Small math symbol edit
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| Jan 27, 2020 at 16:42 | review | Suggested edits | |||
| S Jan 27, 2020 at 17:02 | |||||
| Jan 27, 2020 at 14:14 | comment | added | Cyclopropane | Thank you so much! I understand it now :) Your comment was very helpful. | |
| Jan 27, 2020 at 5:46 | comment | added | theorist | @DrPepper No, the kinetic energy is the sum of the translational, vibrational, and rotational kinetic energy. The speed, however, is due only to the translational (as opposed to rotational or vibrational) kinetic energy. This is because translational kinetic energy is the energy the molecules have as a result of how fast they move (i.e., translate) through space (and their mass). | |
| Jan 27, 2020 at 5:34 | comment | added | Cyclopropane | so the kinetic energy only takes into account the translational modes? | |
| Jan 27, 2020 at 4:39 | history | answered | theorist | CC BY-SA 4.0 |