# Homework Help: Pressure on a wall due to n N2 molecules (close to ans)

1. Dec 8, 2008

### spaghed87

1. The problem statement, all variables and given/known data
4.60×1023 nitrogen molecules collide with a 20.0cm^2 wall each second. Assume that the molecules all travel with a speed of 450m/s and strike the wall head on.

What is the pressure on the wall in pascals (Pa)?

2. Relevant equations
p=(1/3)*(N/V)*(m*v_rms^2)

where,
N is the number of molecules: 4.60*10^23
V is the volume: 0.002m^2
m is the mass (diatomic): (14u*2*1.66*10^-27kg/u)
v is the rms speed 450m/s?

3. The attempt at a solution

p=(1/3)*(4.60*10^23/0.002)*(450^2)*(14*2*1.66*10^-27)=722*10^3Pa

Am I wrong on the v_rms? The v_rms is almost always within 10% of the v_avg.

2. Dec 8, 2008

### turin

You are not wrong on the v_rms, but you are wrong on the volume. Also, you should be able to determine exactly what is v_avg. Hint: the problem statement is sloppy in that it should probably specify that the molecules make elastic collisions with the wall.

For your own edification (i.e. not regarding the solution this particular problem), the collision of molecules with a wall can have the full range between perfectly elastic and perfectly inelastic. An example of a perfectly inelastic case would be an extremely cold wall, onto which the molecules literally stick to it. This is used, for example, in a cryo-pump for HVAC systems.

3. Dec 9, 2008

### spaghed87

I can't see how the volume is wrong though. 20cm^2/100^2= 0.002m^2. Another person who attempted the problem was using the same value for the volume as me. I've got the concept of elastic and inelastic collisions down. I learned it in my first calc. based physics class. More help would be appreciated very much. =)

4. Dec 9, 2008

### DrDan

Looks like Area not Volume

Looks like the problem is just giving you the area of one wall in square centimeters. Pressure will be force per unit area. To get the force you need to find the change in momentum per unit time for one particle and multiply times the number of particles. Then consider the total force over the area of the wall.

The equation that you used looks like it is for gas in a 3 dimensional volume of cubical symmetry (the 1/3 in that equation comes from considering one wall in a 3 dimensional cube). But here they are only talking about one wall it appears.

Just use the velocity given - you need to assume that every molecule has the same velocity by the way the question is stated.

DrDan

5. Dec 9, 2008

### turin

Because a) it's not a volume; it's an area, and b) the molecules are not contained inside of it, they just hit it. See DrDan's post for a more thorough explanation.

6. Dec 9, 2008

### spaghed87

Awesome, I figured it out. I used (2)*(number of particles)*(mass)*(velocity)/(Area). They only gave a formula for the 3-D volume in my book. Thanks for the help!