# Kinetic Gas Theory Calculations

1. Nov 30, 2015

### Jimmy87

1. The problem statement, all variables and given/known data
Questions are attached to this thread.

2. Relevant equations
P =mv (P is momentum)
P = F/A (P is pressure)
F = deltaP/delta t (P is momentum)

3. The attempt at a solution
I have managed to do questions a-e which I was very happy with but then moving onto the extension has made me doubt everything despite having the correct answers. The answer to (a) is simply distance/speed = 1m/500m/s = 0.002s. (b) involves calculating the change in momentum as 2mv = 5 x 10^-23. (c) involves finding the average force which is answer to part b divided by answer to part a which gives 2.5 x 10^-20 N. All of these answers are correct as the answers are in the back of the book. However, looking back at it I don't understand why part (c) can be correct. The rate of change of momentum (force) is the change in momentum divided by the time taken for the momentum to change. The time in part (a) is not this time. Part (a) is the time between the two opposite walls. Surely the time you need for part (c) needs to be the time it takes for the momentum to change by 2mv which is the time it is in contact with the wall for? The answer to the extension is 150kPa. I got an answer to part (e) of 104kPa so factor is roughly 1.4 but I can't see where that comes from. An increase from 500 m/s to 600m/s is a factor of 1.2. The only thing I can think of is if the speed increases by a factor of 1.2 then the time will decrease by a factor of 1.2 giving an increase in pressure of 1.4? I doubt myself with this because when you double the velocity of an object the momentum doubles not quadruples?

Thanks for any help offered!

#### Attached Files:

• ###### Kinetic Theory of Gases Questions.docx
File size:
708.1 KB
Views:
58
2. Nov 30, 2015

### Andrew Mason

Why not show us what you did for part e) and use your answers to a-d. Then we might be able to help you.

Keep in mind that you cannot have all the molecules traveling in a direction perpendicular to x moving in the same direction at any given time. The reason $\bar{F} = \Delta p/\Delta t = 2mv/\Delta t$ where $\Delta t = .002 sec$ is that you are determining the average rate of change of momentum per molecule.

AM

Last edited: Nov 30, 2015