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Natural log of units

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data
    We're doing a lab and we basically had to find ln(Pv), where Pv is vapor pressure of isopropanol. Well, the pressure is initially measured in kPa, but what happens to the units if you take a natural log of that whole thing?
    For example, if I take ln(5 kPa), do the units remain or disappear, or what happens to them? It'd be really nice if they disappeared :) But I'm doubting that possibility

  2. jcsd
  3. Oct 24, 2008 #2


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    This is a really common abuse. You should technically only take the log of a dimensionless quantity. But in your problem you are only considering differences of logs. So e.g. ln(5 kPa)-ln(3 kPa)=(ln(5)+ln(kPa))-(ln(3)+ln(kPa))=ln(5)-ln(3). The expression ln(kPa) is complete nonsense. Luckily, the nonsense cancels.
  4. Oct 24, 2008 #3
    Thanks! We used it in slope calculations and they were subtracted and I carried the ln(kPa) through the whole calculation only to realize that in the next step I'd have to do something with them. Thanks again :) Now I can finish my write-up.
  5. Mar 28, 2011 #4
    I'm confused. Let's look at the thermodynamic relationship [tex]\Delta[/tex]G=-RTlnKeq. In UV thermal melting of non-self complementary DNA duplexes this is defined as Keq=[tex]\frac{2(1-\Theta)}{\Theta^{2}Ct}[/tex] where [tex]\Theta[/tex] is the fraction of broken basepairs observed experimentally and Ct is the concentration of the duplex (Molarity). The gas constant has units of Joules*Mol[tex]^{-1}[/tex] *K[tex]^{-1}[/tex], and the temperature is obviously is Kelvin to cancel out. In these cases the calculated [tex]\Delta[/tex]G is reported in Joules*Mol[tex]^{-1}[/tex]. What happened to the unit of molarity from the concentration term from the equilibrium constant?
  6. Mar 28, 2011 #5


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    ln is always unitless, so there shouldn't be any unit of molarity.

    I think the confusion arises from the fact that, intuitively, it doesn't make sense to take the log of a non-dimensionless quantity, because the ln of a quantity depends on what units you choose. And indeed, this is the case: -RTlnKeq is physically meaningless. Fortunately, whenever we need to calculate an actual physical quantity--for example, the change in Gibbs free energy from a given reaction--we subtract one standard delta_G from another and get -RT*ln(Keq1/Keq2). Keq1/Keq2 is now a dimensionless quantity, and the log of it makes physical sense.
  7. Mar 28, 2011 #6


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    Same answer as before. If you only care about differences of the quantity, then log(a)-log(b)=log(a/b). Whatever units are in the argument of the log will cancel out.
  8. Mar 29, 2011 #7


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    This is not a correct definition of Keq. Keq equals equilibrium value of reaction quotient, reaction quotient is built using not concentrations but activities - and activities are dimensionless. Definition you are using is a simplified one, derived from the full one - and it nonchalantly ignores exact approach.
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