# Maxwell-Boltzmann law-urgent

• Mar 1st 2009, 06:02 PM
Yan
Maxwell-Boltzmann law-urgent
According to the Maxwell-Boltzmann law of theoretical physics, the probability density of V, the velocity of a gas molecule, is
f(v)=kv^2*e^(- βv^2), for v>0; =0, elsewhere.
Where β depends on its mass and the absolute temperature and k is an appropriate constant. Show that the kinetic energy E=(1/2)mV^2, where m the mass of the molecule is a random variable having a gamma distribution.
• Mar 1st 2009, 06:08 PM
mr fantastic
Quote:

Originally Posted by Yan
According to the Maxwell-Boltzmann law of theoretical physics, the probability density of V, the velocity of a gas molecule, is
f(v)=kv^2*e^(- βv^2), for v>0; =0, elsewhere.
Where β depends on its mass and the absolute temperature and k is an appropriate constant. Show that the kinetic energy E=(1/2)mV^2, where m the mass of the molecule is a random variable having a gamma distribution.

If you still have trouble after reading this, please show your working and where you get stuck.
• Mar 1st 2009, 06:29 PM
Yan
Quote:

Originally Posted by mr fantastic

If you still have trouble after reading this, please show your working and where you get stuck.

The problem is i don't understand what is the question ask for..
can you explain to me? and step by step???

Quote:

Originally Posted by mr fantastic

If you still have trouble after reading this, please show your working and where you get stuck.

In this problem, i'm confuse about the sentence<where m the mass of the molecule is a random variable having a gamma distribution.>what is the relationship between this sentence and the problem?
• Mar 1st 2009, 06:53 PM
mr fantastic
Quote:

Originally Posted by Yan
The problem is i don't understand what is the question ask for..
can you explain to me? and step by step???

You're given the pdf for V and asked to find the pdf of $\frac{m}{2} V^2$.

• Mar 1st 2009, 07:06 PM
Yan
Quote:

Originally Posted by mr fantastic
You're given the pdf for V and asked to find the pdf of $\frac{m}{2} V^2$.

G(y) = P (Y<y)
= P ( $\frac{m}{2} V^2$<y)
= 2 P(V< $\sqrt{\frac{2x}{M}}$)
= 2* ....

I just confuse about the m, because in the problem, it says that m is a random variable having a gamma distribution. what is this information use for???
• Mar 1st 2009, 10:24 PM
CaptainBlack
Quote:

Originally Posted by Yan
G(y) = P (Y<y)
= P ( $\frac{m}{2} V^2$<y)
= 2 P(V< $\sqrt{\frac{2x}{M}}$)
= 2* ....

I just confuse about the m, because in the problem, it says that m is a random variable having a gamma distribution. what is this information use for???

There is a comma missing after molecule, it should read:

"Show that the kinetic energy E=(1/2)mV^2, where m the mass of the molecule, is a random variable having a gamma distribution."

m is not a random variable, E (the kinetic energy) is the random variable in question.

CB