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

Discussion in 'Physics & Math' started by kingwinner, Mar 21, 2007.

  1. kingwinner Registered Senior Member

    Messages:
    796
    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 21, 2007
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  3. Pete It's not rocket surgery Moderator

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    10,166
    Try rearranging:
    L/RT = ln(p_o) - ln(p)

    Now simplify the right hand side.
     
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  5. kingwinner Registered Senior Member

    Messages:
    796
    It's ln(p_o/p), but how does this help?

    I want to make a table of ln(p) v.s. 1/T, but when specifying the units for ln(p), what units should I put?

    Say ln(p) = ln(760 mm Hg) = 6.63, but in what units?
     
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  7. Janus58 Valued Senior Member

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    1,790
    Just label it ln(p), as there is no specifc named unit for this quality.
     
  8. przyk squishy Valued Senior Member

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    3,151
    \(\frac{p_o}{p}\) is unitless.
     
  9. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    8,967
    kingwinner---

    You can always remember that arguments of things like log's, cosines, and exponentials HAVE to be unitless.
     
  10. kingwinner Registered Senior Member

    Messages:
    796
    But I need to plot a graph of ln (p) v.s. 1/T, so I do need ln (p) to be isolated...
     
  11. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    8,967
    So just make a graph of ln(p/p0) vs. 1/T. If your teacher marks it wrong send her to me

    Please Register or Log in to view the hidden image!

     
  12. Dinosaur Rational Skeptic Valued Senior Member

    Messages:
    4,588
    Are you sure that you copied the formula correctly? As mentioned in a previous post, the value of logs, exponentials, trig functions, roots, et cetera are dimensionless.

    The left side of your forumula is therefore dimensionless. However, the first term on the right side of the equals sign has dimensions.

    One of the ways to know that a formula is flawed is to analyze the dimension of each term. For example.
    • X = A*B + C - D
    In the above, X, A*B, C, & D must all have the same dimensions or there is something amiss.
     
  13. D H Some other guy Valued Senior Member

    Messages:
    2,257
    Because pressure often exhibits non-linear behavior, it is quite useful to look at the graph the logarithm of the pressure versus some the logarithm of some other quantity. The units are rather strange beasts like log atmospheres. You just have to be careful.

    Goofy units often appear in diffusion processes as well. Diffusion coefficients are often specified with units of length/square root(time).
     
  14. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

    Messages:
    8,967
    I mean, I remember making these graphs (my undergrad major was chemistry) but I always remember having the log of a pressure ratio to make the thing unitless. One could use, say p0 = 1 atm or something, so as to essentially divide out the units, but it has to be there because the log of a dimension doesn't make sense.

    Plus, you'd somehow have to get L/RT with dimensions of log(pressure)!

    The way that I think about it (and feel free to correct me) is that you can expand things like logs and sines and exponentials in a Taylor series, and each term in the Taylor series has to have the same dimensions. This means that they all have to be dimensionless, because otherwise you'd need an infinite number of coefficients with different dimensions.

    This seems odd...why not just define it L^2/t, proportional to, say, the surface area of a sphere at some time t?
     
  15. D H Some other guy Valued Senior Member

    Messages:
    2,257
    Mathematicians often complain about how loose physicists are with math. I suspect engineers are beneath contempt to them.

    I balked a bit when I first saw gyroscope specifications that specified gyro noise in units of degrees/sqrt hour (rate gyros measure attitude rate plus noise; integrating white noise results in a random walk).

    Here's a gyro with noise specs \(0.05^\circ/\text{s}/\sqrt{\text{Hz}}\quad\):

    http://www.ifirobotics.com/docs/analog-devices-yrg-yaw-rate-gyro-80degree-AD22304%20Rate%20Gyro%20Datasheet.pdf
     
  16. temur man of no words Registered Senior Member

    Messages:
    1,330
    You can just draw the graph of log(p), with p having whatever unit. The result will only be shifted if you change unit. As for the unit of log(p), it is dimensionless. To see this, just remember how you got the formula

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

    It does not make sense if you want to be really strict, actually what you have is

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

    with some constant p*. Take p* = 1 mm Hg, then you have "logarithm that can take mm Hg".
     
  17. James R Just this guy, you know? Staff Member

    Messages:
    30,360
    I agree with this.

    Physically meaningful equations involving logs, sines, cosines, exponentials and so on will always do so in such a way that the arguments of the trigonometric functions, exponentials or logarithms can be made dimensionless.

    I quite like temur's solution to the apparent problem here, but it might be just as easy to re-write the equation as:

    \(\ln \frac{p}{p_0} = -\frac{L}{RT}\)

    Either way, the dimensions are clear.
     
  18. Dinosaur Rational Skeptic Valued Senior Member

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    4,588
    James R: Looks like you have answered the question, assuming that L/RT is dimensionless, which it seems to be (I am not 100% sure of this).

    BTW: Is my post a ways back essentially correct?
     

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