Pressure exerted by a gas (derivation using the kinetic theory of gases)

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In the derivation of finding pressure exerted by a gas using kinetic theory of gases I am not understanding why the time between two collisions is taken as the time for rate of change of momentum when a particle bounces back from the wall. please help me
 

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  • #2
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I think you misquote. Any reference ? The duration of the collision with the wall is the time in which the momentum changes.
 
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I think you misquote. Any reference ? The duration of the collision with the wall is the time in which the momentum changes.
I think you misquote. Any reference ? The duration of the collision with the wall is the time in which the momentum changes.
see this
 

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upload_2019-2-22_17-47-14.png


Something gave in in my neck when I tried to read this. Have to go to the hospital first ...

So where is this quote ? I see ##\Delta t = {2L\over |u|} ## for the time to go to the left wall and back again ...
 

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Here in the lower left corner its given that force is change in momentum upon time. where he writes the change in momentum and divides it by time between two collisions as given above it.
 

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They don't have a microscope here in the hospital to read your fine print. Can you be bothered to learn some ##\LaTeX## in order to post legibly ?
 
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  • #7
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But if I have to guess he now averages the change in momentum from the gas particles that bump into the wall, to calculate the average force on the wall.
 
  • #8
Its because yoy want to find average force exerted by the molecule. How u find average force? See u find the time in which this repears aka time period. The collision with wall PQRS repeats. Now find the average force in this time. If u were to find instantaneous force u have to divide by time of collision which is very small.

You are giving Maharashtra HSC? I am also giving it.
 
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Its because yoy want to find average force exerted by the molecule. How u find average force? See u find the time in which this repears aka time period. The collision with wall PQRS repeats. Now find the average force in this time. If u were to find instantaneous force u have to divide by time of collision which is very small.

You are giving Maharashtra HSC? I am also giving it.
yeah, me too thanks
 
  • #10
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But if I have to guess he now averages the change in momentum from the gas particles that bump into the wall, to calculate the average force on the wall.
thanks
 

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