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Units Question: Avagadro's Constant

  1. Aug 20, 2011 #1
    Generally, I see Avogadro's constant being given with units mol-1. Now to me that doesn't seem very... descriptive. But as they say there's more than one way to skin the cat, so I was wondering if these other ways of thinking of it are correct:
    • The conversion factor from grams to amu. The mass of an atom/molecule in amu divide by Avogadro's constant is it's mass in grams?
    • Molar mass is the ratio of of mass per particle? IE: amu/particle, 12amu/particle for carbon
    • Equivalently, Avogadro's Constant is the number of particles in a mol, so rather than mol-1 it's particles*mol-1?
    • There is an invisible unit in A's constant we just don't bother writing?

    Whenever I do math with Avogadro's constant my numbers seem to come out, but I guess I just never wrapped my head around it all the way.
     
  2. jcsd
  3. Aug 20, 2011 #2
    And also in general, whenever we have something that represents the number of things, we don't bother writing a unit for the number of things? Is there a reason for that?
     
  4. Aug 20, 2011 #3

    cjl

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    In general counts of things are unitless. So, number density (for example) has units of m-3, rather than particles (or things) per cubic meter. I'm not completely sure why this is, but it's definitely consistent across a pretty wide range of applications.
     
  5. Aug 20, 2011 #4

    K^2

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    Mols are not units. They are a scaling factor. So technically, Avogadro's number is unitless. It's just a number. I mean, you don't ask what units Pi has. It's a number.

    But sometimes, chemists like to treat mol as a unit. In that case, you write Avogadro's number as mol-1. But that's just silly chemists being silly.
     
  6. Aug 20, 2011 #5

    Borek

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    To add to what cjl wrote: Hz is a number of events per sec - but it is given as s-1, that is, "event" is unitless. Same convention.
     
  7. Aug 20, 2011 #6
    Perhaps someone would explain to me how you can do themodynamics without using the mol as a unit?

    Would the entropy or enthalpy of fusion be dependent upon the number of mols present?

    Of course in the old days we used to call them gram-moles.
     
  8. Aug 20, 2011 #7

    K^2

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    By using N instead of n and kB instead of R. E.g. PV = nRT = NkBT, where N is total number of particles. Like I said, it's just a scaling factor.
     
  9. Aug 20, 2011 #8
    What does PV=NRT have to do with the enthalpy or entropy of fusion ?
     
  10. Aug 20, 2011 #9

    Redbelly98

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    In practice, I prefer to either have no units on Avogadro's number, or to make the units be particles/mole. Here, "particles" means the number of particles (atoms, molecules, or whatever particle or thing is implied by context).

    Because it's just a number. Still, it is sometimes helpful to include "things" as the unit in order to double check that the calculation was done properly, especially when there is more than 1 type of thing involved in the problem.

    Example: how many electrons are there in 2.5 grams of helium?

    Solution:
    [tex]\rm 2.5 \ g \ \cdot \ \frac{1 \ mol \ He}{4.0 \ g} \ \cdot \ 6.02 \cdot 10^{23} \frac{He \ atoms}{1 \ mol \ He} \ \cdot \ 2 \frac{electrons}{He \ atom} \ = \ xxx \ electrons[/tex]
     
  11. Aug 20, 2011 #10

    Redbelly98

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    I prefer to specify what the event is ... is it 1 cycle, or 1 radian?
     
    Last edited: Aug 20, 2011
  12. Aug 20, 2011 #11
    Instead of using kilo Joules per mol for enthalpy of fusion, couldn't you use kilo Joules for every so many particles of that substance (I know I'm really just saying the same thing)? When you use mol for enthalpy of fusion, it is understood that you are saying these many kilo Joules required for these many particles. It's just a matter of convention.
     
  13. Aug 20, 2011 #12

    Borek

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    I never said I like this convention.
     
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