Number of collisions/sec for a gas atom

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SUMMARY

The discussion focuses on calculating the number of collisions per second for a single ideal gas atom in a cubic container with side length l. The established formula is (1/2)(l^2)(Vx)n, where Vx represents the velocity component of the gas atom and n denotes the number of gas atoms per unit volume. It is confirmed that the inclusion of n in the equation is essential, as an increase in n directly correlates with an increase in the number of collisions per second with one of the container's faces.

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Homework Statement


Consider a single ideal gas atom bouncing between two opposite faces of a cubic container, with sides of length l.
Using m for the mass of the gas atom and Vx (V sub x) for the velocity component, give an expression for the number of collisions per second the atom makes with one of the faces.


Homework Equations





The Attempt at a Solution



Number of collisions per second with one of the faces = (1/2)(l^2)(Vx)n

Where n is the number of gas atoms per unit volume.

Should n be in the solution?
If the number of atoms per unit volume, n, increases then according to the solution so do the number of collisions with one of the faces. Would that be correct? The higher the number of gas atoms per unit volume, the more collisions there are per second with one of the faces?

Thank you
 
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ZedCar said:
Consider a single ideal gas atom bouncing between two opposite faces of a cubic container, with sides of length l.
Using m for the mass of the gas atom and Vx (V sub x) for the velocity component, give an expression for the number of collisions per second the atom makes with one of the faces.


The Attempt at a Solution



Number of collisions per second with one of the faces = (1/2)(l^2)(Vx)n

Where n is the number of gas atoms per unit volume.

Should n be in the solution?


Read the problem text carefully.

ehild
 

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