# Molecule collisions in an Ideal Gas

• derravaragh
In summary: So, E(t) = ∞∫0 tP(t)dt. By the way, the integral from 0 to infinity of tP(t)dt is the same as the integral from 0 to infinity of tAe^(-bt)dt. In summary, the probability distribution governing the time between collisions is P(t) = Ae^(-bt). The value of A is calculated to be 1/b in order to correctly normalize P(t). The average time between collisions, t, can be found by taking the integral of t*P(t) from 0 to infinity. This can be rewritten in terms of the average time between collisions, tau, by replacing A and b with 1/tau.
derravaragh

## Homework Statement

Molecules in an ideal gas collide with each other at random times. The probability distribution governing the time between collisions is P(t) = Ae^(-bt).
(a) Find the value of A so that P(t) is correctly normalized.
(b) Find the average time between collisions, t. This time is traditionally called tau. Now re-write P(t) in terms of tau, without the original parameters A and b.
(c) Find the standard deviation of the collision times, σ_t.

## Homework Equations

∫(x^n)e^(-x/a)dx = n!a^(n+1) from 0 to ∞

## The Attempt at a Solution

I believe I have the right answer to (a), I normalized it to obtain P(t) = -be^(-bt), but I can't figure out what to do for part (b) on.
My attempt was to take the average time t to equal (ƩP(t))/n where n is the number of collisions and the sum goes from 0 to ∞, but that thought process got me no where. Any help would be appreciated.

I don't think the first part is right. The integral of P(t) from t=0 to t=infinity should be +1.

derravaragh said:
The probability distribution governing the time between collisions is P(t) = Ae^(-bt).

(b) Find the average time between collisions, t.

If you have the probability density function p(x) for some continuous variable X, what expression gives the average value of X?

For the first response, I rechecked my work on part (a) and realized my signs were off, using A = b the ∫Ae^(-bt) from 0 to ∞ gives me a value of +1.

For the second response, the only solution I can think of would be the Expected value E(x) which would give E(t) = ƩtP(t) = Ʃt(Ae^(-bt)) from t = 0 to ∞ which gives 0 + be^(-b) + 2be^(-2b) +...+ 0 which I don't know how to generalize or even how to apply to the next part. Quite frankly, I am lost on this problem after part (a).

The mean time is an expectation value. It's going to be the integral of t*P(t) from 0 to infinity.

derravaragh said:
Expected value E(x) which would give E(t) = ƩtP(t) = Ʃt(Ae^(-bt)) from t = 0 to ∞
That's for a discrete distribution. As Dick says, the equivalent for a continuous distribution is integration.

## 1. What is an ideal gas?

An ideal gas is a theoretical model used in chemistry and physics to describe the behavior and properties of gases. It assumes that the gas particles have no volume, do not interact with each other, and have perfectly elastic collisions.

## 2. What are molecule collisions in an ideal gas?

Molecule collisions in an ideal gas refer to the interactions between gas particles as they move around in a container. These collisions are responsible for the pressure, temperature, and volume of the gas.

## 3. How do molecule collisions affect the pressure of an ideal gas?

Molecule collisions in an ideal gas create a force on the walls of the container, resulting in pressure. The more frequent and energetic the collisions, the higher the pressure of the gas.

## 4. What is the relationship between temperature and molecule collisions in an ideal gas?

According to the kinetic theory of gases, temperature is directly proportional to the average kinetic energy of gas particles. Therefore, as temperature increases, the speed and frequency of molecule collisions also increase.

## 5. How do real gases differ from ideal gases in terms of molecule collisions?

In real gases, particles do interact with each other and have a finite volume, which affects the behavior of molecule collisions. These deviations from ideal behavior are more significant at high pressures and low temperatures.

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