# How Do You Calculate the Probability Density p(E) for an Ideal Gas?

• ehrenfest
Now you can integrate over v to get \rho(E).So, in summary, the correct answer is p(E) = 4 \pi (2 \pi \sigma^2)^{-3/2} \exp\left({\frac{E}{m \sigma^2}}\right) E^2 dE.
ehrenfest

## Homework Statement

The following is a probability density distribution for an ideal gas:
$$p(v_x,v_y,v_z) = (2 \pi \sigma ^2)^{-3/2} \exp\left({\frac{v_x^2 +v_y^2+v_z^2}{2 \sigma^2}}\right)$$

$$E = 0.5 m |\vec{v}|^2$$

Find p(E).

## The Attempt at a Solution

I want to just plug in what E but that gives the wrong answer and I cannot figure out why.
$$p(E) = (2 \pi \sigma ^2)^{-3/2} \exp\left({\frac{E}{m \sigma^2}}\right)$$
Why is that wrong? It makes no sense, but I just plugged in what E was! This is really bothering me.

ehrenfest said:

## Homework Statement

The following is a probability density distribution for an ideal gas:
$$p(v_x,v_y,v_z) = (2 \pi \sigma ^2)^{-3/2} \exp\left({\frac{v_x^2 +v_y^2+v_z^2}{2 \sigma^2}}\right)$$

$$E = 0.5 m |\vec{v}|^2$$

Find p(E).

## The Attempt at a Solution

I want to just plug in what E but that gives the wrong answer and I cannot figure out why.
$$p(E) = (2 \pi \sigma ^2)^{-3/2} \exp\left({\frac{E}{m \sigma^2}}\right)$$
Why is that wrong? It makes no sense, but I just plugged in what E was! This is really bothering me.

What is the correct answer? It's a probability *density* so there is also a factor $$dv_x dv_y dv_z$$ that must be multiplied to get a probability. If you want the probability density in terms of the energy you wil have to relate the above factors to the infinitesimal $$dE$$

EDIT: it's a density in speed so the factor to convert is probably dv

Interesting. BTW, is it generally true that if y=y(x) and x=x(z), then y=y(x(z))? That is, you can write y as a function of z just by plugging in x as a function of z. That rule seems to fail here. But in general, when is it true?

Last edited:
ehrenfest said:
Interesting. BTW, is it generally true that if y=y(x) and x=x(z), then y=y(x(z))? That is, you can right y as a function of z just by plugging in x as a function of z. That rule seems to fail here. But in general, when is it true?

It is true in general, yes.

The problem here is that what you are looking for is *not* simply y(x(z))!

what you are looking for is the probability density in energy so the probability of measuring an energy between E and E+dE.. If you just plug in v in terms of energy, you are still giving the probability density in *speed*, so you are giving th eprobability of measuring the speed between v and v+dv. This is not what you are asked to find!

So, back to the original problem. $p(v_x, v_y, v_z)dv_x dv_y dv_z[/tex] gives the probability of finding the particle's velocity in the box with side lengths [itex]dv_x, dv_y, dv_z$ with the one corner at $(v_x, v_y, v_z)$.

And I want to find p(E) such that $p(E)dE[/tex] gives the probability of finding the particle's energy between E and E+dE. We know that [itex]dE = 1/2m(2v_x dv_x + 2v_y dv_y+2v_z dv_z)$, but I still don't really know how to proceed. Do express each of dv_x dv_y dv_z as a function of dE somehow?

You said something about just using dv, but I am not sure what you meant. Can you elaborate?

ehrenfest said:
So, back to the original problem. $p(v_x, v_y, v_z)dv_x dv_y dv_z[/tex] gives the probability of finding the particle's velocity in the box with side lengths [itex]dv_x, dv_y, dv_z$ with the one corner at $(v_x, v_y, v_z)$.

And I want to find p(E) such that $p(E)dE[/tex] gives the probability of finding the particle's energy between E and E+dE. We know that [itex]dE = 1/2m(2v_x dv_x + 2v_y dv_y+2v_z dv_z)$, but I still don't really know how to proceed. Do express each of dv_x dv_y dv_z as a function of dE somehow?

You said something about just using dv, but I am not sure what you meant. Can you elaborate?

You can't directly get from the distribution in v_x, v_y an dv_z to the density in Energy since there is more information in the first than in the second (since E depends only on v^2). So you need to find the probability density in the speed first. So you integrate over all directions using $$dv_x dv_y dv_z = v^2 dv d\Omega$$. But the result of the angular integral is simply $$4 \pi v^2 dv$$ times the probability density in v_x, v_y and v_z so

$$\rho(v) dv = 4 \pi \rho(v_x,v_y,v_z) v^2 dv$$
where it is understood that in the sceon term you rewrite everything in terms of the speed v.

## 1. What is an ideal gas?

An ideal gas is a theoretical gas that follows the kinetic theory of gases, which assumes that the gas particles have negligible volume and do not interact with each other. It also assumes that the gas particles are in constant, random motion and that the average kinetic energy of the particles is directly proportional to the temperature of the gas.

## 2. How is p(E) defined for an ideal gas?

p(E) is the probability of a gas particle having a certain amount of kinetic energy, E. It is defined as the number of particles that have kinetic energy between E and E+dE, divided by the total number of particles in the gas.

## 3. How is p(E) calculated for an ideal gas?

p(E) can be calculated using the Maxwell-Boltzmann distribution, which is a probability distribution function that describes the distribution of kinetic energies of particles in an ideal gas. It takes into account the temperature, mass, and number of particles in the gas.

## 4. What factors affect p(E) for an ideal gas?

The main factors that affect p(E) for an ideal gas are temperature, mass, and number of particles. An increase in temperature will result in a higher average kinetic energy and a wider distribution of kinetic energies, causing p(E) to shift towards higher values. A higher mass of gas particles will result in a lower average kinetic energy and a narrower distribution, shifting p(E) towards lower values. An increase in the number of particles will result in a higher overall p(E) value.

## 5. How does p(E) relate to the behavior of an ideal gas?

p(E) is directly related to the average kinetic energy and temperature of an ideal gas. As p(E) increases, the gas particles have a higher probability of having a certain amount of kinetic energy, indicating that the gas is more energetic and has a higher temperature. In addition, the shape of p(E) can also reveal information about the behavior of the gas, such as the average speed and average kinetic energy of the particles.

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