Solution Found!
The speed of a molecule in a uniform gas at equilibrium is
Chapter 6, Problem 29E(choose chapter or problem)
The speed of a molecule in a uniform gas at equilibrium is a random variable \(V\) whose density function is given by
\(f(v)=a v^{2} e^{-b r^{2}}, v>0\)
where \(b=m / 2 k T\) and \(k, T\), and \(m\) denote Boltzmann's constant, the absolute temperature, and the mass of the molecule, respectively.
a. Derive the distribution of \(W=m V^{2} / 2\), the kinetic energy of the molecule.
b Find \(E(W)\).
Equation Transcription:
Text Transcription:
V
f(v)=av^2e^-br^2, v>0
b=m/2kt
k,T
m
W=mV^2/2
E(W)
Questions & Answers
QUESTION:
The speed of a molecule in a uniform gas at equilibrium is a random variable \(V\) whose density function is given by
\(f(v)=a v^{2} e^{-b r^{2}}, v>0\)
where \(b=m / 2 k T\) and \(k, T\), and \(m\) denote Boltzmann's constant, the absolute temperature, and the mass of the molecule, respectively.
a. Derive the distribution of \(W=m V^{2} / 2\), the kinetic energy of the molecule.
b Find \(E(W)\).
Equation Transcription:
Text Transcription:
V
f(v)=av^2e^-br^2, v>0
b=m/2kt
k,T
m
W=mV^2/2
E(W)
ANSWER:
Solution :
Step 1 of 2:
Let the speed of a molecule in a uniform gas at equilibrium is a random variable V whose density function is given by
Where b=and k, T and m denote Boltzmann's constant.
Our goal is:
a). We need to derive the distribution of , kinetic energy of the molecule.
b). We need to find E(W).
Now we have derive the distribution of , kinetic energy of the molecule.
We know that ,
and
.
Then,
,W
0.
The above expression in the form of a gamma density, so the constant must be chosen so that the density integrate to 1, or simply
Therefore is
.