If KEavg is same for all mole of any gas at same temp, and gases have different mol weights, does this just mean that the more massive gases have slower velocities, and vice versa? I understand this as an application of conservation of momentum. Is that an accurate way to envision avg KE of gases?
Avg KE question
- Thread starter dougkaye
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That's correct. The equation is v2/v1 = sqrt(m1/2). Don't forget that it's the AVERAGE velocity of the gas. Individual molecules can have velocities above or below the average.
To envision it, think of a container with a small hole in it. If 2 gases with different masses are on the same side, the object with the smaller mass with diffuse to the other side at a faster rate.
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As a point to remember, they do not have the same momentum.
I ran into that train of thought earlier, thinking if if mass is increased, v would decrease by the same amount = same momentum.
Using the equation dlouis provided, for a gas: sqrt(m) ~ 1/v or m ~ 1/(v^2). So for example, increasing m by a factor of 4 will decrease v by a factor of only 2. Increase m by 16 and only decrease v by 4. So the bigger mass will have the bigger momentum.
Not to waste your time or insult your intelligence if you already knew that, but maybe someone confused like I was can benefit 👍
I ran into that train of thought earlier, thinking if if mass is increased, v would decrease by the same amount = same momentum.
Using the equation dlouis provided, for a gas: sqrt(m) ~ 1/v or m ~ 1/(v^2). So for example, increasing m by a factor of 4 will decrease v by a factor of only 2. Increase m by 16 and only decrease v by 4. So the bigger mass will have the bigger momentum.
Not to waste your time or insult your intelligence if you already knew that, but maybe someone confused like I was can benefit 👍
Commissioner Rabo approves. It's important to know that momentum and kinetic energy are different. Same temperature means same kinetic energy, and conservation of momentum has nothing to do with it.
