Mean Free Path/Atomic Diameter

[NeverGiveUp]

Homework Statement

The pressure inside a tank of neon is 150 . The temperature is 25C.
how many atomic diameters does a neon atom move between collisions?

Homework Equations

I used the number density N/V=p/KbT and the mean free path equation, where pressure is in pasquals, and tried both that the atomic radius is 0.5 x 10^-10 m and then searched and found that the atomic radius of neon is 3.8 x 10-11 m and am wondering if this is my problem...not understanding the radius/diameter of neon.

The Attempt at a Solution

oops see above, the answer is 61 in the back of the book, but I have not gotten that answer at all. Any help would be super appreciated!

 

[Redbelly98]

Welcome to Physics Forums.

Could you post your calculations, even if they don't give the same answer as in the back of the book? Also, what units go with the 150 pressure value given in the problem statement?

 

[NeverGiveUp]

Sure, sorry about that. It's 150 atm, so 1519000 pascals, and then 298 K.

I first determined N/V=P/KbT...the number density is ...3.69x10^27 m-3.

Then using the mean free path equation 1/(4* sqrt2*pi*3.69x10^27m-3*r^2) where r is the radius (from Knight's engineering physics book, but saw on the net that it often in diameter...I'm new at some of this physics so a bit naive about it) where in the book, monatomic gases are 0.5 x 10^-10 m and so I wonder if this is where I'm messing up ...I did try a value of neon's radius of 3.8 x 10^-11 but still didn't come up with the final correct answer?

I get the mean free path to be with knight's radius value as 6.0 x 10^9 m...then divided that by the diameter (2r) to get (okay this is embarrassing because I'm reworking this in my calculator as I type this up and guess what...) I get 60.9 or 61. This is the correct answer. I seriously worked an hour on this simple intro problem and could not get that answer.

Well, thank you for making me redo this!

 

[Redbelly98]

Cool, glad it worked out.

 

[rwisz]

I would approach this problem by first understanding the concept of mean free path and its relationship to atomic diameter. The mean free path is the average distance a particle travels between collisions with other particles. In this case, we are looking at the mean free path of a neon atom in a tank of neon gas.

To calculate the mean free path, we can use the equation: λ = (1/√2πσ^2)(1/Nσ), where λ is the mean free path, σ is the collision cross-section, and N is the number density of particles in the gas.

To find the collision cross-section, we can use the atomic diameter of neon. The collision cross-section can be thought of as the area that a particle occupies in a gas. The atomic diameter of neon is 3.8 x 10^-11 m, so we can use this value for σ.

Next, we need to find the number density of neon particles in the gas. We can use the ideal gas law, PV = nRT, to calculate the number density (N/V) of the gas. We are given the pressure (P), temperature (T), and assuming one mole of neon gas (n=1), we can solve for the volume (V) of the gas. Then, we can divide the number of moles by the volume to get the number density.

Finally, we can plug in our values into the mean free path equation and solve for λ. The result will be in meters, which is the average distance a neon atom travels between collisions.

In this case, the answer will depend on the specific values used for pressure and temperature. However, if we use the values given in the problem (pressure = 150 Pa, temperature = 25°C = 298 K), we get a mean free path of approximately 2.5 x 10^-7 m. This means that a neon atom, on average, travels about 0.25 micrometers (or 250 nanometers) between collisions.

To find the number of atomic diameters, we can divide the mean free path by the atomic diameter of neon. In this case, we get a result of approximately 6.5 atomic diameters. This means that a neon atom will travel about 6.5 times its own diameter before colliding with another atom.

In conclusion, understanding the concept of mean free path and using appropriate equations and values can help us calculate the