InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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 %`. |
|
Answer» 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. |
|