Timeline for According to KMT, is the velocity of an ideal gas always sqrt(3RT/M)?
Current License: CC BY-SA 4.0
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| S Sep 5, 2020 at 10:29 | history | suggested | sai-kartik | CC BY-SA 4.0 |
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| Sep 5, 2020 at 7:40 | review | Suggested edits | |||
| S Sep 5, 2020 at 10:29 | |||||
| Jan 29, 2020 at 18:42 | history | edited | Poutnik | CC BY-SA 4.0 |
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| Jan 28, 2020 at 18:55 | history | edited | Poutnik | CC BY-SA 4.0 |
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| Jan 28, 2020 at 10:28 | history | edited | Poutnik | CC BY-SA 4.0 |
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| Jan 27, 2020 at 14:25 | vote | accept | Cyclopropane | ||
| Jan 27, 2020 at 14:24 | comment | added | Poutnik | @DrPepper No, I do not say that. KE >= translational KE =3/2kT. ( Al as mean energy) | |
| Jan 27, 2020 at 14:10 | comment | added | Cyclopropane | I am not too sure about your statement. Consider an ideal gas; the total internal energy is defined as $U = \text{PE} + \text{KE}$, where $\text{PE}$ and $\text{KE}$ are the potential and kinetic energies, respectively. Since in an ideal gas, there are no attractive or repulsive forces, $\text{PE} = 0 $. Are you saying that $U = \text{PE} + \text{KE} = 0 + 3/2kt$ for all gases? Surely this can't be true.... | |
| Jan 26, 2020 at 19:24 | history | edited | Poutnik | CC BY-SA 4.0 |
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| Jan 26, 2020 at 19:05 | history | edited | Poutnik | CC BY-SA 4.0 |
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| Jan 26, 2020 at 16:20 | history | answered | Poutnik | CC BY-SA 4.0 |