- #1

FranzDiCoccio

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## Homework Statement

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1 mol of gas at temperature [itex]T[/itex] is contained in a cubic container of side [itex]L[/itex].

Estimate the number of collisions per second between the atoms in the gas and one of the walls of the cubic container.

My book gives this formula for that quantity

[tex]\frac{N_A}{6L}\sqrt{\frac{3 k T}{m}}[/tex]

Where [itex]N_A[/itex] is Avogadro's number, [itex]k[/itex] is the Boltzmann's constant and [itex]m[/itex] is the atomic mass.

I tried two different lines of reasoning, and came up with the same answer, which is different from that of the book

[tex]\frac{N_A}{2L}\sqrt{\frac{ k T}{m}}[/tex]

## Homework Equations

a) Root mean square velocity

[tex]v_{rms}=\sqrt{\frac{3 k T}{m}}[/tex]

b) average velocity along one direction:

[tex]\bar{v}_x=\frac{v_{rms}}{\sqrt{3}}=\sqrt{\frac{ k T}{m}}[/tex]

## The Attempt at a Solution

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A) Via the momentum transferred to a wall in a time [itex]\Delta t[/itex].

The average number of atoms hitting a wall in a time [itex]\Delta t[/itex] is

[tex]\bar{N} = \frac{\bar{F} \Delta t}{2 m \bar{v}_x}=\frac{p L^2 \Delta t}{2 m \bar{v}_x}=\frac{p V \Delta t}{2 L m \bar{v}_x}=\frac{N_A k T \Delta t}{2 L m \bar{v}_x}=\Delta t \frac{N_A}{2L}\sqrt{\frac{kT}{m}}[/tex]

B) Via the density of atoms.

The average velocity along direction x is [itex]\bar{v}_x[/itex]. On average, half of the particles goes in the direction of the wall I'm interested in.

The average number of atoms hitting a wall in a time [itex]\Delta t[/itex] is half of the number of particles contained in a volume [itex]\bar{V}=L^2 \Delta t \bar{v}_x[/itex].

Thus

[tex]\bar{N} =\frac{1}{2} \bar{V} \frac{N_A}{V}=\frac{1}{2} L^2 \Delta t \bar{v}_x \frac{N_A}{L^3}= \frac{N_A}{2L}\sqrt{\frac{kT}{m}} [/tex]

C) My book says "on average [itex]N_A/3[/itex] atoms hit the same face in the time [itex]\Delta t[/itex]", and takes

[tex]\Delta t = \frac{2 L}{v_{rms}}[/tex].

The statement about 1/3 of the particles sounds wrong to me.

Can someone help with this? Shouldn't it take into account my formula b) above? Isn't the square velocity along one direction 1/3 of the square modulus of the velocity?

Thanks a lot

Franz

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