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For the equation $E_k=\frac{1}{2}mu^2$ what does the $m$ variable represent?

I would say mass but the textbook does not say explicitly that it is anything.

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$$\ce{Kinetic ~Energy ~= ~1/2 (mass~of ~the~object~ [kg]) * (velocity~of~the~object~[m/sec])^2}$$

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    $\begingroup$ I'm not sure if the $u$ should be $\bar{u}$ and so the $m$ should be the mass of one molecule. $\endgroup$ – jonsca Jul 9 '14 at 17:38
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    $\begingroup$ (if you were keeping that under your hat for the time being, I apologize :) ) $\endgroup$ – jonsca Jul 9 '14 at 17:41
  • $\begingroup$ I cannot overcome a temptation to post this old anecdote: A physicist was hospitalized after a car accident. He lies and groans: - Ahh it's so good it's half. So nice it's half... - What's about this "half" thing? - the doctor asks. - It's amazing that the kinetic energy is only half em vee squared! - says the poor guy with a smile:) $\endgroup$ – andselisk Oct 15 '17 at 10:22
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$$E_k = \frac{1}{2}mu^2$$

where, $m$ is molar mass (unit - kg/mol), and $u$ is root mean square speed.

According to wikipedia,

The molar mass of atoms of an element is given by the atomic mass of the element multiplied by the molar mass constant.

See this page (from wikipedia) for complete explanation.

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Because this is Chemistry, I'll assume you're dealing with some form or shape of Kinetic-Molecular Theory. So, with

$$E_k = \frac{1}{2}mv^2$$ $E_k$ is the average kinetic energy, m is the molecular mass and $v^2$ is the average of the squares of the molecular speeds. (if you're dealing with one particle, its just the velocity of said particle)

Gases at the same temperature have the same average kinetic energy, so, on average, molecules with a lower speed have a higher mass. If the gas is pure, all molecules will have the same mass, but at any instant the direction of movement will be different for each.

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