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Homework Help: Help please - Change in temperature over altitude

  1. Dec 20, 2013 #1
    Help please -- Change in temperature over altitude

    1. The problem statement, all variables and given/known data
    Combine the equations dP/dT=-μgp/RT and the vapor pressure to find the rate dT/dz (use the chain rule).

    2. Relevant equations
    dP/dT=-μgp/RT, μ is the molecular weight

    3. The attempt at a solution
    My textbook says the definition of vapor pressure is p=p0e(-L/RT). However, to get there they used the Clausius-Clapeyron equation and one of the intermediate steps is 1/p(dp/dT)=L/RT2. This is the equation I used. Applying the chain rule, dT/dz=(dp/dz)(dT/dp) I found dT/dz=-μgT/L. However, when asked to find an actual value of dT/dz, I am given the L and Ï (density) to plug into the equation. Does μ have something to with the density? And what would I use for T. I think I may have done something wrong. I tried working out the equation in a different way using the original Clausius-Clapeyron equation for the vapor pressure: dP/dT=L/TΔV and found dT/dz=-μgpΔV/RL, but am not sure how to use density, Ï, and L, latent heat, with this equation either. Am I using the right equation for vapor pressure? Missing a step? How can I account for Ï and L? Thanks for any of your help!
  2. jcsd
  3. Dec 20, 2013 #2
    There are some things desperately wrong with this problem statement. For example, the equation dP/dT=-μgp/RT is incorrect. It should read: dP/dz=-μgp/RT. Also, it doesn't seem to make sense to say that you have a purely saturated vapor column whose temperature varies with z, and yet remains under saturation conditions without any liquid water present. This question is out of a math book, and not out of a physical chemistry or thermodynamics book, correct?
  4. Dec 20, 2013 #3
    RE: Help please -- Change in temperature over altitude

    I apologize! I did mean to write dP/dz=-μgp/RT. This is a thermodynamics problem. The change in the boiling temperature of water is supposed to be represented by dT. Let me write the full problem out:

    Combine dP/dz=-μgp/RT and the vapor pressure to find the rate dT/dz for the change in the boiling temperature of water at varying altitude above sea level. After solving it algebraically, assuming that the latent heat of vaporization of water L=2.4x106 J/kg and the density of water vapor Ï~0.6 kg/m3, find the rate of change of T in Kelvin per kilometer. (Hint: apply the chain rule)

    When I tried to plug in the numbers in to the first equation I found, I got a really weird answer and the unit was way off. Help :??
    Last edited: Dec 20, 2013
  5. Dec 20, 2013 #4
    Now this makes much more sense now.
    In your original post, you said that the equation you had written down was the Clausius Clapeyron equation, but that was not correct. The equation you wrote was the Clausius Equation. Clapeyron's contribution was to replace ΔV with the volume of the ideal gas. This leads to p=p0e(-L/RT), which is equivalent to the Clausius Clapeyron equation. I'm puzzled by the comment about the water vapor density. The density of water vapor is not constant; it decreases with altitude. Furthermore, this fact is already taken into account in your equation for dP/dz. You don't need to explicitly know the density of the water vapor. You need to substitute your Clausius Clapeyron equation p=p0e(-L/RT) into the equation for dP/dz.

  6. Dec 20, 2013 #5
    Thanks, but I am still somewhat confused. If I plug p=p0e(-L/RT) into dP/dz=-μgp/RT, where is the dT in the dT/dz equation I'm told to find?
    Last edited: Dec 20, 2013
  7. Dec 20, 2013 #6
    If I take dp/dz, I get:
    What do you get?

    What you are trying to determine is "what would the temperature profile of the atmosphere have to be if liquid water at the local temperature would boil at each altitude?" This may not bear much resemblance to the temperature profile that exists in the actual atmosphere.

  8. Dec 20, 2013 #7
    Ok, I did not differentiate the Clausius-Clapeyron equation last time... However, from there equating the two dp/dz's I got dT/dz=-μgT/L, which is the very first equation I received. I don't see how this can be correct. :// I apologize if I am asking bad questions, but I was given Ï and L. How can I use μ and T to find the value of dT/dz?
    Last edited: Dec 20, 2013
  9. Dec 20, 2013 #8
    What I was thinking is using the equation Ï=pμ/RT. I am not sure if it applies here, but plugging it in yields dT/dz=-gp/RLÏ. The only problem is, I am not sure what p is... wouldn't it vary over time?
  10. Dec 21, 2013 #9
    This equation is correct. The value of Ï they appear to be giving you is Ï at the surface of the earth. You can use that to calculate the temperature at the surface of the earth by applying the ideal gas law. Then you can integrate your differential equation from z = 0 to arbitrary z to obtain the vertical temperature profile. Once you know T vs z from this, you can substitute it into the differential equation again to get dT/dz as a function of z.

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