According to the Maxwell–Boltzmann law of theoret-ical physics, the probability density of V, the velocity of a gas molecule, isf(v) =⎧⎨⎩kv2e−βv2for v > 00 elsewhere where β depends on its mass and the absolute tem-perature and k is an appropriate constant. Show that the kinetic energy E = 1 2mV2, where m the massof the molecule is a random variable having a gammadistribution.

According to the Maxwell–Boltzmann law of theoret-ical physics, the probability density of V, the velocity of a gas molecule, isf(v) =⎧⎨⎩kv2e−βv2for v > 00 elsewhere where β depends on its mass and the absolute tem-perature and k is an appropriate constant. Show that the kinetic energy E = 1 2mV2, where m the massof the molecule is a random variable having a gammadistribution.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.8: Probability

Problem 32E

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Question

According to the Maxwell–Boltzmann law of theoret-
ical physics, the probability density of V, the velocity of a

gas molecule, is
f(v) =
⎧
⎨
⎩
kv2e−βv2
for v > 0
0 elsewhere

where β depends on its mass and the absolute tem-
perature and k is an appropriate constant. Show that

the kinetic energy E = 1

2mV2, where m the mass
of the molecule is a random variable having a gamma
distribution.

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