What is the N2/O2 Ratio at 3255m Using Laplace's Law?

[Wolfmannm]

Homework Statement

The ratio of nitrogen to oxygen at sea level is 4.0:1. Use Laplace's law of
isothermal atmospheres to calculate the same ratio for 3255m above sea level
Assume that T = 273 K

Homework Equations

P_fN2 = P_iN2 * e^-\Big({\frac{M_N2 g}{R T}}*h\Big)
P_fO2 = P_iO2 * e^-\Big({\frac{M_O2g}{R T}}*h\Big)

P_i = 1 atm
h = 3255 meters
M_N2 = 28 g/mol
M_O2 = 32 g/mol
T = 273K
g = 9.8 m/s^2
R = 8.314 J/mol K

The Attempt at a Solution

I tried plugging in the numbers for both equations, but somehow that just makes my exponential go to 0 so I know that's not correct. I am wondering if I need to convert anything. I converted the initial pressure to Pa but that doesn't help me with an exponential of 0. For N2 I came up with approximately 393.52 for Mg/RT*h. e to -393.52 comes out zero on my calculator, I'm sure its not, because e to any power is never 0 but its too small for my calculator to read.[/B]

 

[SteamKing]

Your molecular weights must be expressed in kg / mole, rather than grams / mole.

Also, since you are given the ratio of nitrogen : oxygen at sea level, then the partial pressure of each element will not be 1.0 atm. That's the pressure of the whole atmosphere at sea level.