Kinetic Gas Theory Calculations

Click For Summary
SUMMARY

The discussion focuses on Kinetic Gas Theory calculations, specifically addressing momentum and pressure equations. Key equations include momentum (P = mv), pressure (P = F/A), and the relationship between force and momentum change (F = ΔP/Δt). The participant successfully solved initial questions but expressed confusion regarding the time variable in calculating average force. The correct answers were verified against textbook solutions, with an extension question yielding a pressure of 150 kPa, while the participant calculated 104 kPa, leading to further exploration of the factors affecting pressure changes with velocity.

PREREQUISITES
  • Understanding of Kinetic Gas Theory principles
  • Familiarity with momentum and pressure equations
  • Basic knowledge of calculus for rate of change concepts
  • Experience with physics problem-solving techniques
NEXT STEPS
  • Study the derivation of the ideal gas law and its applications
  • Learn about the relationship between velocity and pressure in gas dynamics
  • Explore advanced topics in fluid mechanics, particularly Bernoulli's principle
  • Investigate the implications of molecular motion on pressure and temperature
USEFUL FOR

Students studying physics, particularly those focusing on thermodynamics and fluid mechanics, as well as educators seeking to clarify concepts in Kinetic Gas Theory.

Jimmy87
Messages
692
Reaction score
19

Homework Statement


Questions are attached to this thread.

Homework Equations


P =mv (P is momentum)
P = F/A (P is pressure)
F = deltaP/delta t (P is momentum)

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!
 

Attachments

Physics news on Phys.org
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:
  • Like
Likes   Reactions: Jimmy87

Similar threads

Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
994
  • · Replies 54 ·
2
Replies
54
Views
11K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 39 ·
2
Replies
39
Views
5K