Speed distribution for a sample of N gas particles

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SUMMARY

The discussion focuses on calculating the average speed ratio vavg/v0 for a sample of N gas particles based on a triangular speed distribution. The probability density function P(v) is defined as P(v) = av/v0 for speeds within the range of 0 to v0. Participants clarify that P(0) equals 0 and P(v0) equals a, confirming the linearity of the function in the specified range. The key takeaway is understanding how to derive the average speed from the given probability distribution.

PREREQUISITES
  • Understanding of probability density functions
  • Familiarity with integrals and average calculations
  • Knowledge of gas particle behavior in statistical mechanics
  • Basic graph interpretation skills
NEXT STEPS
  • Study the derivation of average values from probability distributions
  • Learn about triangular distributions in statistical mechanics
  • Explore the concept of normalization in probability density functions
  • Investigate the implications of speed distributions on gas behavior
USEFUL FOR

Students in physics or engineering, particularly those studying thermodynamics and statistical mechanics, will benefit from this discussion.

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Homework Statement


http://www.fileden.com/files/2006/11/15/381656/19.39.png

The figure shows a hypothetical speed distribution for a sample of N gas particles (note that P(v) = 0 for speed v > 2v0).

(b) What is the value of vavg/v0?

Homework Equations



vavg = int[vP(v)]dv

The Attempt at a Solution



The solutions manual says that, for the triangular portion of the distribution, P(v) = av/v0.
This may be a stupid question, but how does one get that?

(only need help with that portion of the problem for now)
 
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Well, for the given function P(v), P(0)=0, P(v0)=a, and the function is linear in v. So it looks like your graph.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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