FranzDiCoccio
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Homework Statement
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1 mol of gas at temperature T is contained in a cubic container of side L.
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
\frac{N_A}{6L}\sqrt{\frac{3 k T}{m}}
Where N_A is Avogadro's number, k is the Boltzmann's constant and m 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
\frac{N_A}{2L}\sqrt{\frac{ k T}{m}}
Homework Equations
a) Root mean square velocity
v_{rms}=\sqrt{\frac{3 k T}{m}}
b) average velocity along one direction:
\bar{v}_x=\frac{v_{rms}}{\sqrt{3}}=\sqrt{\frac{ k T}{m}}
The Attempt at a Solution
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A) Via the momentum transferred to a wall in a time \Delta t.
The average number of atoms hitting a wall in a time \Delta t is
\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}}
B) Via the density of atoms.
The average velocity along direction x is \bar{v}_x. 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 \Delta t is half of the number of particles contained in a volume \bar{V}=L^2 \Delta t \bar{v}_x.
Thus
\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}}
C) My book says "on average N_A/3 atoms hit the same face in the time \Delta t", and takes
\Delta t = \frac{2 L}{v_{rms}}.
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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