How Does Doubling the Temperature Affect the Mean Free Path in a Gas?

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Doubling the temperature of a gas increases the average speed of the particles by a factor of √2, but the mean free path remains unchanged. This is because, although particles are moving faster, their relative positions and the distances between them during collisions do not change. The mean free path is determined by the frequency of collisions, which is also affected by the increased speed. Therefore, even with increased velocity, the average distance between collisions stays constant. The analogy of watching a video illustrates that speeding up or slowing down the motion does not alter the distances traveled between collisions.
Sarah0001
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Homework Statement
"The mean free path is independent of temperature at constant volume."
Relevant Equations
mean free path inversely proportional to density (=mass of gas/volume)
it is at constant volume density of gas remains same.
Assumption, all N particles in Volume V move with avg speed (rms speed) from avg KE for a given temperature T.
Say Temperature of a gas doubles, I do not understand how the average distance between particles (mean free path) is unaffected if they are traveling √2 times as fast in a fixed volume V. Root 2 as a factor of increase because T*2 --> KE*2 --> V*2 --> Vrms*√2
Is it because relative to one another the if both particles are moving at the same rate then, the distance between them is fixed, regardless of the rate at which distance is covered.
 
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Sarah0001 said:
how the average distance between particles (mean free path) is unaffected if they are traveling √2 times as fast
Imagine watching a video of them. If you speed ot up or slow it down the particles will move exactly the same distance between collisions.
 
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