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

AI Thread Summary
The discussion centers on understanding the derivation of gas pressure using the kinetic theory of gases, specifically the time between collisions and its relation to momentum change. Participants clarify that the time taken for a particle to collide with the wall and bounce back is crucial for calculating the average force exerted on the wall. The average force is derived from the momentum change of gas particles during repeated collisions, emphasizing the importance of the time period for these events. There is also a mention of the need for clear communication and legibility in mathematical expressions. Overall, the conversation highlights the relationship between collision timing and force calculations in gas behavior.
Bheshaj
Messages
20
Reaction score
0
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
 
Physics news on Phys.org
I think you misquote. Any reference ? The duration of the collision with the wall is the time in which the momentum changes.
 
  • Like
Likes Bheshaj
BvU said:
I think you misquote. Any reference ? The duration of the collision with the wall is the time in which the momentum changes.
BvU said:
I think you misquote. Any reference ? The duration of the collision with the wall is the time in which the momentum changes.
see this
 

Attachments

  • IMG_20190222_220649.jpg
    IMG_20190222_220649.jpg
    28.7 KB · Views: 413
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 ...
 

Attachments

  • upload_2019-2-22_17-47-14.png
    upload_2019-2-22_17-47-14.png
    38.5 KB · Views: 503
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.
 

Attachments

  • IMG_20190223_002948.jpg
    IMG_20190223_002948.jpg
    29.9 KB · Views: 390
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 ?
 
Last edited:
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.
 
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.
 
  • Like
Likes Bheshaj
Mohammed Ayaz Quadri said:
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
BvU said:
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
 
Back
Top