# Homework Help: Help please - Change in temperature over altitude

1. Dec 20, 2013

### yo56

Help please -- Change in temperature over altitude

1. The problem statement, all variables and given/known data
Combine the equations dP/dT=-Î¼gp/RT and the vapor pressure to find the rate dT/dz (use the chain rule).

2. Relevant equations
dP/dT=-Î¼gp/RT, Î¼ is the molecular weight
dT/dz

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=-Î¼gT/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 Î¼ 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/TÎ”V and found dT/dz=-Î¼gpÎ”V/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. Dec 20, 2013

### Staff: Mentor

There are some things desperately wrong with this problem statement. For example, the equation dP/dT=-Î¼gp/RT is incorrect. It should read: dP/dz=-Î¼gp/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?

3. Dec 20, 2013

### yo56

RE: Help please -- Change in temperature over altitude

I apologize! I did mean to write dP/dz=-Î¼gp/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=-Î¼gp/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
4. Dec 20, 2013

### Staff: Mentor

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 Î”V 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.

Chet

5. Dec 20, 2013

### yo56

Thanks, but I am still somewhat confused. If I plug p=p0e(-L/RT) into dP/dz=-Î¼gp/RT, where is the dT in the dT/dz equation I'm told to find?

Last edited: Dec 20, 2013
6. Dec 20, 2013

### Staff: Mentor

If I take dp/dz, I get:
$$\frac{dp}{dz}=p_0\frac{L}{RT^2}e^{-L/RT}\frac{dT}{dz}=p\frac{L}{RT^2}\frac{dT}{dz}$$
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.

Chet

7. Dec 20, 2013

### yo56

Ok, I did not differentiate the Clausius-Clapeyron equation last time... However, from there equating the two dp/dz's I got dT/dz=-Î¼gT/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 Î¼ and T to find the value of dT/dz?

Last edited: Dec 20, 2013
8. Dec 20, 2013

### yo56

What I was thinking is using the equation Ï=pÎ¼/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?

9. Dec 21, 2013

### Staff: Mentor

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.

Chet