[UWSteve]

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

[Borek]

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...

[UWSteve]

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).

[Borek]

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

[UWSteve]

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)

[Borek]

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

[UWSteve]

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

[Borek]

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

[UWSteve]

Thanks for the help.