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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: UWSteve on April 16, 2008, 03:41:52 PM

Title: Boiling point relative to altitude problem
Post by: UWSteve on April 16, 2008, 03:41:52 PM

So, I have a homework question:

"What is the boiling point of water 2 miles above sea level?  Assume that the atmosphere follows the barometric formula with M = .0289 kg/mol and T = 300 K. Assume the enthalpy of vaporization of water is 44.0 kJ/mol independent of temperature."

We're also supposed to assume that the pressure at sea level is 1 ATM.

So, barometric formula:

P = Pstandard e^-(g * M * h / (R * T))

I calculated the pressure at 2 miles (~3219 m) above sea level using the barometric formula (I got .365 ATM).

Then, I used dP/dT = Heat of vaporization / (T(molar volume steam - molar volume water)) to solve for dP/dT.

I then inversed dT/dP in units of K/Pa

I converted this value to units in K/ATM and then multiplied this value by the change in pressure (calculated from barometric formula) to get how much the temperature changed.

Then I can just calculate the new BP by taking into account dT/dP.

Now, unless my book is wrong (the book answer is 90.6 degrees C), I'm slipping up somewhere along the way.

I was wondering if there was a blatant error in the logic I used for this problem.

Thanks

Title: Re: Boiling point relative to altitude problem
Post by: Borek on April 16, 2008, 03:56:28 PM

Quote from: UWSteve on April 16, 2008, 03:41:52 PM

I calculated the pressure at 2 miles (~3219 m) above sea level using the barometric formula (I got .365 ATM).

Sounds more like Mount Everest level pressure to me...

Title: Re: Boiling point relative to altitude problem
Post by: UWSteve on April 16, 2008, 04:10:35 PM

Quote from: Borek on April 16, 2008, 03:56:28 PM

Quote from: UWSteve on April 16, 2008, 03:41:52 PM

I calculated the pressure at 2 miles (~3219 m) above sea level using the barometric formula (I got .365 ATM).

Sounds more like Mount Everest level pressure to me...

Hrmm, must have plugged something into the calculator incorrectly.

New pressure is .694 ATM

However, my final answer is still off by a pretty significant amount (~2 degrees).

Title: Re: Boiling point relative to altitude problem
Post by: Borek on April 16, 2008, 04:29:41 PM

Do I understand correctly that you have approximated dP/dT by delta P/delta T instead of using integrated formula?

Title: Re: Boiling point relative to altitude problem
Post by: UWSteve on April 16, 2008, 04:43:19 PM

I just used the Clapeyron equation, which I arrived at by...

Using a coexistence line.

Therefore,

dchemicalpotential(1) = dchemicalpotential(2)

And since dchemical potential is equal to dG/n (i'm just going to write molar quantities as underlined for ease).

V(1)dP - S(1)dT = V(2)dP - S(2)dT

=> dP/dT = (S(2) - S(1))/(V(2)-V(1)) = deltaS/deltaV = deltaH/(TdeltaV)

Title: Re: Boiling point relative to altitude problem
Post by: Borek on April 16, 2008, 04:57:12 PM

Show how you have plugged numbers into formula to get your answer.

Title: Re: Boiling point relative to altitude problem
Post by: UWSteve on April 16, 2008, 05:03:32 PM

dP/dT = (44000 J/mol) / (300 K)(30.180e-3 m^3 mol^-1)

dP/dT = 4860 Pa/K

dT/dP = 1/4860 k/Pa = 2.058e-4 K/Pa

2.058e-4 K/Pa * 1 Pa / 9.87e-6 ATM = 20.85 K/ATM

20.85 K/ATM * (-(1-.694)) ATM = -6.4 K

373 K - 6.4 K = 366.6 K or 93.6 C

Title: Re: Boiling point relative to altitude problem
Post by: Borek on April 16, 2008, 05:33:21 PM

Basically what you did is

T1 = T0 + dT/dP * delta P

This is wrong, as dT/dP is not constant.

http://en.wikipedia.org/wiki/Clausius-Clapeyron_relation

Title: Re: Boiling point relative to altitude problem
Post by: UWSteve on April 16, 2008, 07:06:49 PM

Thanks for the help.