Assuming the temperature and the molar mass of air, as well as the free-fall acceleration, to be independent of the height, find the difference in heights at which the air densities at the temperature `0 ^@C` differ.
(a) `e` times ,
(b) by `eta = 1.0 %`.

Explanation

We have `dp = - rho g dh` but from gas law `p = (rho)/(M) RT`,
Thus `dp = (d rho)/(M) RT` at const. temperature
So, `(d rho)/(rho) = (g M)/(RT) gh`
Integrating within limits `int_(rho_0)^rho (d rho)/(rho) = int_0^h (g M)/(RT) gh`
or, `1 n (rho)/(rho_0) = -(g M)/(RT) h`
So, `rho = rho_0 e^(-M gh//RT)` and `h = - (RT)/(M g) 1 n (rho)/(rho_0)`
(a) Given `T = 273^@K, (rho_0)/(rho) = e`
Thus `h = - (RT)/(M g) 1 n e^-1 = 8 km`.
(b) `T = 273^@ K` and
`(rho_0 - rho)/(rho_0) = 0.01` or `(rho)/(rho_0) = 0.99`
Thus `h = - (RT)/(M g) 1 n(rho)/(rho_0) = 0.09 km` on substitution.