10_1425_web_Lec_31_KineticTheoryGases

10_1425_web_Lec_31_KineticTheoryGases - Kinetic Theory of...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Kinetic Theory of Gases Physics 1425 Lecture 31 Michael Fowler, UVa
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Bernoulli’s Picture Daniel Bernoulli, in 1738, was the first to understand air pressure in terms of molecules— he visualized them shooting around very rapidly in a closed container, supporting a weight as shown by constantly bouncing off the underside of the piston. Given more room, they would rush in to fill the new space, just as a gas is observed to do. No-one believed him. Applet here.
Background image of page 2
One Dimensional, One Molecule Gas The molecule roundtrips in time 2 L / v , so it bounces off the piston v /2 L times per sec, each time delivering momentum 2 mv , so the piston will pick up momentum from this “gas” at rate 2 x v /2 L per second. Force from gas on piston: F = rate of change of momentum = 2 / L . An equal opposite force must be supplied from outside to keep the piston at rest. Animation! V v L 1-D gas: molecule bounces between ends of cylinder.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Molecule in a Two-Dimensional Box Assume perfectly elastic collisions with all walls. The molecule will follow a zigzag path, the time between collisions with the same end, say the end at x = L , is now 2 L / v x , and the momentum transferred per collision is 2 mv x , so the average force on the end is x 2 / L . This will still hold good in three dimensions. a L v x v y 0 x y
Background image of page 4
N Molecules in an L x L x L Cube Assume first that we have a very large number N of molecules bouncing around, so small that they don’t hit each other, each follows its own zigzag path. The force on the right-hand wall at x = L is just the sum of the forces from each one, so F = mv x 1 2 / L + x 2 2 / L + x 3 2 / L + … + xN 2 / L .
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Bar Means Average The force on the wall depends on the sum We don’t need the individual values v x 1 2 , etc., just the average , written with a bar : So the force on the wall is: 22 2 2 123 x x x xN vv v v + + ++ 2 2 2 x x x xN x v v v N + + = 2 x Nmv F L =
Background image of page 6
Gas Molecules Have Random Velocities…
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 22

10_1425_web_Lec_31_KineticTheoryGases - Kinetic Theory of...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online