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Natural log of a pressure is ?

  1. Apr 28, 2005 #1
    Ok so i was doing a problem involving finding the pressure of mercury at its boiling point (630.05K) using Troutons rule and the final answer seems a bit strange to me.

    Integration of the Clausius-Clapeyron Equation:

    [tex]{ln(\frac{{P}_{2}}{{P}_{1}}) = -\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2-\frac{1}{T}_1)[/tex]

    So... you end up with a number, but it's unitless? What am I missing? Thanks.

    Troutons rule states that at standard T and P:

    [tex]{\Delta}_{vap}\overline{S} \approx {88} {J}\cdot{K}^{-1}\cdot{mol}^{-1}[/tex]

    and the change in entropy is related to the change in enthalpy by:

  2. jcsd
  3. Apr 28, 2005 #2


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    Nothing --- well, maybe you should take another peek at the expression (P2/P1).
  4. Apr 28, 2005 #3
    another hint is that you can never take logs or exponentials with anything that has units. it simply makes no sense.
  5. Apr 28, 2005 #4
    Yes, this was the heart of my question. We use Troutons rule to determine the pressure at the vaporization temperature, yet to do this we must take the exponential of the equation. So, for instance:

    [tex]{ln(\frac{{P}_{2}}{{P}_{1}}) = -\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2 -\frac{1}{T}_1)[/tex]

    [tex]{\frac{{P}_{2}}{{P}_{1}} = e^{-\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2 -\frac{1}{T}_1)}[/tex]

    [tex]{P}_{2} = e^{-\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2 -\frac{1}{T}_1)}{P}_{1}[/tex]

    Are units retained after taking the exponential of [itex]ln(\frac{{P}_2}{{P}_1})[/itex]? I mean I understand you're still taking the exponential of a unitless number because it's pascals/pascals, but my instructure mentioned that there was a special circumstance that kept the units around... Perhaps it is just simple math and he's just confusing us all :rofl:.
  6. Apr 28, 2005 #5
    I also realize that the title of my post is misleading, sorry about that. ;)
  7. Apr 28, 2005 #6

    if you take the exponential you will still have a unitless equation. like you said you will have p2/p1 so units will cancel out. Sure you can mutiply both sides by p1 to get p2=e^(stuff)p1 and you have an equation with pascals on the left and pascals on the right. But that just means you can still cancel out units. Don't get confused by playing around with the algebra. No matter how you play around with the equation, the units will always end up canceling out. Just ask yourself what is the ln(cm) or ln(bars) or e^cm or e^J. These unit quantities make no sense. I am 99.9% sure there are absolutely no cases where you can take the log or exponential of a quantity that has units.
  8. Apr 28, 2005 #7
    I was not confused by the algebra at all. Really what I wanted to find out was that my answer, which I obtained from the very equation for P2 I posted above, was correctly stated as a pressure. The basic algebra of the problem made sense to me, however my instructor circled the units on my answer and noted in class that the reason was due to the logarithm. I wanted to see if I could figure it out on my own and I came to the conclusion that my answer was right, just wanted to make sure I wasn't making a mistake.

  9. Apr 29, 2005 #8
    I really don't see the problem. You get an expression P2/P1 that is unitless for obvious reasons - the units cancel out on division. If P2/P1 = x then P2 = x times P1, P1 having a (arbitrary) unit of pressure like bar, kPa, atm, psi etc. and x being unitless. The only thing I wonder about is how to get P1 and T1... :uhh:
  10. May 1, 2005 #9
    i didn't see the problem either! i took my paper back and recieved full credit :wink:.

    as for finding P1 and T1, it's just more algebra.
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