Number of collisions/sec for a gas atom

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The discussion focuses on calculating the number of collisions per second for a gas atom in a cubic container. The proposed formula is (1/2)(l^2)(Vx)n, where n represents the number of gas atoms per unit volume. Participants question whether n should be included in the solution, noting that an increase in n would logically lead to more collisions per second with one of the container's faces. The relationship between atom density and collision frequency is emphasized, confirming that higher atom density results in increased collision rates. This highlights the importance of understanding gas behavior in confined spaces.
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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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