- #1

- 12

- 0

## Homework Statement

At an altitude of 50 km, the average atmospheric temperature is 0C. what is the average number of air molecules per cm^3 of air at this atmosphere?

Given formula:

P(sub h) =

P(sub 0) x e^(-Mgh/RT) =

-(Mgh x 10^5)/2.303RT

Assuming P = 1 atm at sea level and h in km

M=28.97g/mol=28.97cm^3/mol

g = 9.81 m/s^2 = .0981 km/s^2

h = 50 km

R = 8.314 J/K•mol =

82.06cm^3•atm/K•mol

T = 0 C = 273.15 K

I'm having trouble because using the scale height form of the equation (second form) I get lost in the units, and I'm receiving an error on my TI using the first.

## The Attempt at a Solution

Using second equation:[/B]

**Log P(sub h) =**

(-Mgh x 10^5)/2.303RT =

-(28.97)(.0981)(50)(10^5)/

(2.303)(82.057)(273.15) =

-275.28

10^(-275.28) = 5.23... But the units after the conversion don't add up ... Suggestions?

(-Mgh x 10^5)/2.303RT =

-(28.97)(.0981)(50)(10^5)/

(2.303)(82.057)(273.15) =

-275.28

10^(-275.28) = 5.23... But the units after the conversion don't add up ... Suggestions?