Kinetic Theory- Calculating Kinetic Energy

[FisiksIdiot]

Homework Statement

A quantity of helium gas is held in a square box 1 m length. The measured macroscopic pressure of the gas is 1000 Pa. An atom of helium makes 500 collisions per second with the wall of the container traveling at an average velocity perpendicular to a wall of a container. Assuming that collisions with the wall are elastic and there are no collisions or potential interaction between the atoms, calculate the kinetic energy of one atom in this system.

Homework Equations

P=(N m vmean^2)/3V

where
m=mass of one atom
vmean=mean speed
V=volume
N= I am unsure of exactly what this is, I assume it is the number of atoms striking the inside surfaces.

The Attempt at a Solution

1.There are 500 collisions per second on the wall of this cube. This means that there are 6*500 collisions overall in the cube: 3000 collisions per second overall and therefore 3000 atoms in total colliding with the inside surfaces of the cube.

2. The mass of an He atom is equal to 4.002602u= 6.644*10^-27 kg

3. The cubes volume is 1m*1m*1m=1*m^3

3. P=(N m vmean^2)/3V.
V=1
m=6.644*10^-27
N=3000
P=1000

1000=(3000*(6.64*10^-27)vmean^2)/3(1)

1000=1000*(6.64*10^-27)vmean^2)

1=(6.64*10^-27)vmean^2

1/(6.64*10^-27) = vmean^2= 1.51*10^26

therefore vmean= 1.23*10^13 m/s

1.23*10^13 m/s > c

oh dear.

 

[FisiksIdiot]

I think I worked out where I'm wrong. It's one atom colliding 300 times a second rather than 300 atoms colliding every second

 

[SNOTWY]

So how did you do it? I've got more or less the same problem but it also asks me to find the total mass of He...can you or someone give me a hand on this one?

 

[FisiksIdiot]

Basically, I had a mahoosive 'DUUUUUUH' moment so it's all sorted. It only comes down to half a page of working now.

First, have a look at this.
http://en.wikipedia.org/wiki/Kinetic_theory

We know pressure, volume and how many times an atom of He hits the wall of the container per second.

We have a formula given to us by the kinetic theory page above that relates speed to time.

t=2L/Vx
where L=the length of the container, Vx is the perpendicular speed and t is the time taken for one hit of the wall.
(Don't ask me why that is. I'm happy enough to know it's right atm. Ignorance is bliss sometimes)

The box is 1m in length so L=1
t is 1 second divided by 500 hits of the wall, so it becomes 1s/500=0.002s

So rearrange 2L/Vx=t
2L/t=Vx

Therefore Vx=2/0.002=1000m/s

You can assume this is the average velocity (vxbar) of the atoms seeing as if all the atoms strike the wall 500 times every second they must all have the same speed.
Right. Thats fantastic. Epic. Superb. However, we're not quite done with it because we've only assumed one plane of movement when it's in fact moving randomly. We use the following formula to find the actual velocity (v) from the velocity in one direction (Vx).

(Vx)^2=(V)^2)/3

so 1000^2=(V^2) /3

1x10^6=(V^2)/3

3x10^6=(V^2)

V=sqrt(3x10^6)

Noice. So we whack that into 1/2mv^2 to get our kinetic energy of a single atom

We got to find the mass of the atom though, but that's pretty easy to do. You know that the atomic mass of He is pretty much 4 (you can go more precisely but it doesn't REALLY make a difference) so the mass is 4u, where u=atomic mass unit.

so (1/2)(6.64215544 × 10^-27)(3x10^6)= 9.96x10^-21 J


Now for the bit where we work out the total number of atoms in the system.

Pressure is definitely related to kinetic energy right? Kinda makes sense seeing as if the particles hit the walls of the container with more energy, the more the walls are going to be pushed out. This formula links pressure and kinetic energy:

P=(Nmv^2)/3V

P= Pressure
v=velocity we worked out earlier
m=mass of one atom of He
V=volume.

Basically we're after Nm in this case- the number of atoms multiplied by the mass of each atom: the total mass of He in a container.

In this case, Volume is 1m^3, so you don't have to worry about it.

Rearrange:
3P=Nmv^2
3P/(v^2)=Nm.

so
3(1000)/(3x10^6)=Nm
Nm=0.001kg

Sorted!