Gold Atom Count in O-Ring: 10.4x10^21

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Discussion Overview

The discussion revolves around calculating the number of gold atoms in a golf O-ring used in a high-vacuum chamber. Participants explore the necessary calculations involving volume, density, and unit conversions, focusing on the correct application of formulas and the implications of unit discrepancies.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates the volume of the O-ring using the formula for the volume of a cylinder and arrives at a value of 176.7, but later realizes that the volume is expressed in mm2 instead of mm3.
  • Another participant points out the need for unit consistency, questioning the units used for volume and density, and suggests that the volume should be in cubic units.
  • There is a discussion about converting volume from mm3 to cm3 using the appropriate conversion factor, specifically mentioning the relationship between cm and mm.
  • A participant acknowledges the need to convert the volume correctly to avoid errors in the final calculation of the number of atoms.
  • One participant emphasizes the importance of keeping track of units throughout the calculations to prevent mistakes.

Areas of Agreement / Disagreement

Participants generally agree that there is a unit conversion issue that needs to be addressed, but there is no consensus on the final resolution of the problem as the discussion is still ongoing.

Contextual Notes

Participants have identified limitations in their calculations related to unit conversions and the representation of volume, but the specific steps to resolve these issues are still being discussed.

shreddinglicks
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Homework Statement


A golf O-ring is used to form a gastight seal in a high-vacuum chamber. The ring is formed from a 100-mm length of 1.5-mm-diameter wire. Calculate the number of gold atoms in the O-ring.

Homework Equations


Volume of cylinder = PI*h*r^2

The Attempt at a Solution


I got the volume of the cylinder:
PI*100*.75^2 = 176.7

I used the formula m = D*V:

density gold = 19.28 g/cm^3
amu gold = 196.97 grams

19.28*176.7 = 3406.776

I then multiplied by avogadro's number over amu.

3406.776 *(6.02x10^23 / 196.97) = 1.04x10^25 atoms

The answer should be 10.4x10^21 atoms. I'm thinking I messed up some unit conversion. I am unsure of where.
 
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shreddinglicks said:

Homework Statement


A golf O-ring is used to form a gastight seal in a high-vacuum chamber. The ring is formed from a 100-mm length of 1.5-mm-diameter wire. Calculate the number of gold atoms in the O-ring.

Homework Equations


Volume of cylinder = PI*h*r^2

The Attempt at a Solution


I got the volume of the cylinder:
PI*100*.75^2 = 176.7

I used the formula m = D*V:

density gold = 19.28 g/cm^3
amu gold = 196.97 grams

19.28*176.7 = 3406.776

I then multiplied by avogadro's number over amu.

3406.776 *(6.02x10^23 / 196.97) = 1.04x10^25 atoms

The answer should be 10.4x10^21 atoms. I'm thinking I messed up some unit conversion. I am unsure of where.

What are the units on your volume? What are the units on your density?
 
The density is g/cm^3

the volume is mm^2

I see the issue, how do I convert this? Is there a way to convert so I can cancel out a unit?
 
shreddinglicks said:
The density is g/cm^3

the volume is mm^2

I see the issue, how do I convert this? Is there a way to convert so I can cancel out a unit?

Volume should be cubic something. mm^2 is an area.

To convert a volume in mm^3 to cm^3 think about the following:

Conversion factor * X mm^3 = Y cm^3

The conversion factor needs to have units of cm^3 on the top and mm^3 on the bottom. What about [( 1 cm/ 10 mm)]^3 ? I.e. can you use the conversion between cm and mm to come up with a conversion factor between mm^3 and cm^3 ?
 
I see, so I have [176.7 mm^3] * [1cm/10mm]^3

basically

176.7 / 10^3 = .1767
 
Numbers are OK, but in the final formula you again ignored units. That's the simplest way of making mistakes, as you have already seen. Don't do that.
 
.1767cm^3
 
Now use this number to answer the problem.
 

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