Given the Volume, Find the Number of Atoms

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Homework Help Overview

The discussion revolves around a problem involving the calculation of the number of atoms based on the volume of a cube, with a focus on the average distance between spheres (or atoms) within that volume.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between volume, number of atoms, and average distance, with one participant attempting to derive a formula. Others question the assumptions regarding the speed of the balls and its relevance to the problem.

Discussion Status

There is an ongoing exploration of concepts such as mean free path and its implications for the problem. Some participants have shared their reasoning and calculations, while others have raised questions about the assumptions made in the problem setup.

Contextual Notes

Participants note that the problem may lack clarity regarding the speed of the plastic balls and its effect on the interactions with the rubber balls, which could influence the interpretation of the problem.

asifion
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Homework Statement
A mad engineer builds a cube, 2.5 m on a side, in which 6.2-cm-diameter rubber balls are constantly sent flying in random directions by vibrating walls. He will award a prize to anyone who can figure out how many balls are in the cube without entering it or taking out any of the balls. You decide to shoot 6.2-cm-diameter plastic balls into the cube, through a small hole, to see how far they get before colliding with a rubber ball. After many shots, you find they travel an average distance of 2.8 m. How many rubber balls do you think are in the cube?
Relevant Equations
not sure but: N/V=p/(k_B*T)
pV=nRT
V/N = distance between atoms?
So, I was thinking that the total volume of the cube divided by the number of atoms (or rubber balls) should intuitively give the average distance between each ball.

What I did was:
N = number of balls
D = avg distance between balls

(2.5)^3 / N = D
(2.5)^3 / D = N
D = 2.8 - 2 * radius (I'm assuming the given average distance is from center to center of each sphere)

2.5^3/(2.8-.062) = 5.7 balls
 
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Have you covered the topic of "mean free path" of molecules in a gas?

I think the problem should have stated that the speed at which you shoot the plastic balls is similar to the average speed of the rubber balls. (For example, if the plastic balls were fired at an extremely slow speed, they would hardly go any distance inside the box before being hit by a rubber ball.)
 
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TSny said:
Have you covered the topic of "mean free path" of molecules in a gas?

I think the problem should have stated that the speed at which you shoot the plastic balls is similar to the average speed of the rubber balls. (For example, if the plastic balls were fired at an extremely slow speed, they would hardly go any distance inside the box before being hit by a rubber ball.)

Yes we have, but I don't think the professor went into detail. But when using the equation, I get the correct answer. Thanks.
 
Mean free path in a gas is related to the number of molecules per unit volume.
 
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