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What's the UNIT after you take the natural log?

  1. Mar 20, 2007 #1
    I was working on an experiment for the vapor pressure of water and I have the following formula

    ln (p) = -L/(RT) + ln (p_o)

    L=heat of vaporization of water
    R=Molar gas constant
    T=temperature in Kelvins

    I have some data points for the pressure p in units of mm Hg, when I take the natural log of it, ln (p), what will the units be? I am very confused...can someone please help me?

    When I analyse the units of -L/(RT), it seems like it's dimensionless, but the value of ln (p) definitely depend on the numerical value of p (i.e. having p in different units, e.g. Pa, atm, mm Hg, should give different values of ln (p) ), then how can ln (p) be dimensionless?

    Thanks!
     
    Last edited: Mar 20, 2007
  2. jcsd
  3. Mar 21, 2007 #2

    chroot

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    The units of a ln(p) would generally be referred to as "log Pa" or "log atm."

    Taking the logarithm doesn't actually change the dimension of the argument at all -- the logarithm of pressure is still pressure -- but it does change the numerical value, and thus "Pa" and "log Pa" should be considered different units.

    - Warren
     
  4. Mar 21, 2007 #3

    robphy

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    It seems that, strictly speaking,
    the units of a function like log or exp or cos should be dimensionless,
    and
    the argument of a function like log or exp or cos should be dimensionless.

    So, if p and p_o carry dimensions of pressure,
    one should NOT write
    ln (p) = -L/(RT) + ln (p_o),
    but instead write
    ln (p/p_ref) = -L/(RT) + ln (p_o/p_ref)
    where p_ref is any arbitrary but nonzero reference-pressure so that (p/p_ref) and (p_o/p_ref) are dimensionless. Of course, if you'd rather not introduce a p_ref, then the dimensionally correct way to write the intended relationship is
    ln (p/p_o) = -L/(RT)
    since
    ln (p/p_ref) = -L/(RT) + ln (p_o/p_ref)
    ln (p/p_ref) -ln (p_o/p_ref) = -L/(RT)
    ln( (p/p_ref) / (p_o/p_ref) )= -L/(RT)
    ln( p/p_o )= -L/(RT)


    This is a good example that distinguishes mathematical variables from physical variables. As long as you are consistent, one can work "formally".
     
  5. Mar 21, 2007 #4
    But I need to plot a graph of ln (p) v.s. 1/T, so I do need ln (p) to be isolated...
     
  6. Mar 21, 2007 #5

    Dick

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    I would guess you are making the graph in order to extract the slope. A change of units will only shift the line up and down, not change the slope. Pick whatever units are handy. As robphy points out, the relation is valid in any units.
     
  7. Mar 21, 2007 #6
    plot ln(p/1atm), ln(p/1Pa)...... whatever units you are using.
     
  8. Mar 21, 2007 #7
    I would concur with this explanation.
     
  9. Mar 22, 2007 #8

    Tom Mattson

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    I concur too. Kingwinner, just let [itex]p_{ref}[/itex] in robphy's post be 1 Pa. You can do that as follows.

    [tex]\ln(p)=-\frac{L}{RT}+\ln(p_0)[/tex]

    [tex]\ln(p) + 0=-\frac{L}{RT}+\ln(p_0) + 0[/tex]

    [tex]\ln(p) + \ln(1 Pa)=-\frac{L}{RT}+\ln(p_0) + \ln(1 Pa)[/tex]

    That last line is valid because [tex]\ln(1 Pa)=0[/tex].

    [tex]\ln(p/(1 Pa))=-\frac{L}{RT}+\ln(p_0/(1 Pa))[/tex]

    [tex]\ln(\overline{p})=-\frac{L}{RT}+\ln(\overline{p}_0)[/tex]

    Where the quantities with the bar are dimensionless. Once it is recognized that the mathematical manipulations are no different with the dimensionless quantities than they are with the dimensionful quantities, it is at once recognized that we can go on our merry way working with quantities such as [itex]\ln(p)[/itex] without worrying about the units.

    ETA:

    OK, I'm done editing now. :redface:
     
    Last edited: Mar 22, 2007
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