Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: Fundamental physical properties



Roger P. asks:

David, what units are you using to measure the mass of a C-12 atom?

kilograms.

If you
change the definition of a kilogram, will this have any effect on the number
of atoms in a mole?

Yes.

The mass of the C-12 atom will also be defined in this
new unit too.

The mass of a C-12 atom is independent of anyone's choice of units. The
numerical value which represents the ratio of that mass to the standard unit
of mass (i.e. the ratio of the mass of a C-12 atom to the mass of the
standard kilogram) will change when the standard unit changes.

number of atoms in a mole=mass of a gram of atoms/mass of a single atom in
grams.

Not quite. The LHS of your equation is Avagadro's number, and the RHS is
the number of atoms in a gram of a given type of substance. There is no
element for which the above equation is true. It is close to being true for
the protium isotope of hydrogen (H-1) however. (In this H-1 case the RHS =
= 0.9922357 X Avagadro's number = 0.9922357 moles = 0.9922357 * LHS. In the
case of C-12 the RHS = (1/12) * LHS)

Which is more fundamental? Defining the number in a mole, or defining a
kilogram. Aren't both somewhat arbitrary?

Currently the kilogram is more fundamental than the mole in that the concept
of the kilogram is needed first before Avagadro's number (and hence the mole)
can be defined. Let m_12 be the mass of a C-12 atom in kg (i.e. m_12 is the
ratio of the mass of a C-12 atom to the mass of a standard kilogram). By
definition, Avagadro's number (and hence the quantity in a mole) is *exactly*
N_A = 0.012 kg / m_12. Using current definitions the measured value of m_12
= 1.9926482 X 10^(-26) kg to an accuracy of 0.59 ppm. The size of the mole
depends on the size of the kilogram since if the definition of the kilogram
is changed (to some other macroscopic *non*atomic standard) the numerical
value of m_12 would change while the numerator in the above formula for N_A
would not change. This would change the quotient and hence N_A. OTOH, if
(as some people keep suggesting) the kilogram's definition becomes based on
the mass of some fundamental particle such as the proton or the electron
(for instance the kilogram could be defined as exactly 5.978633 x 10^26 times
the mass of a proton), then both the kilogram and Avagadro's number would be
on the same level as equally fundamental (or equally nonfundamental) since
nature has already given us a prefixed ratio between the mass of a C-12 atom
and the mass of a proton.

It is true, though, that both the kilogram and the mole are somewhat
arbitrary. That's why neither of them are very fundamental as concepts.

David Bowman
dbowman@gtc.georgetown.ky.us