Cloud formation condition
Consider a simplified model of cloub formation. Hot air in contact with the earth's surface contains water vapor. This air rises convectively till the water vapor content reaches its situation pressure. When this happens, the water vpor starts condensing and droplets are formed. We shall estimate the estimate the height a which this happens.We assume that the atmosphere consists of the diatomic gases oxygen and nitrogen in the mass proportion `21:79` respectively. We further assume that the atmosphere is an ideal gas, g the acceleration due to gravity is constant and air processes are adiabatic. Under these assumptions one can show that the pressure is given by
`p = p_(0) ((T_(0)-T_(z))/(T_(0)))^(alpha)`
Here `p_(0)` and `T_(0)` is the pressure and temperature respectively at seal level `(z = 0),T` is the lapse rate (magnitude of the change in temperature `T` with height `z` abive the earth's surface, i.e. `T gt 0)`.

Options

A. Obtain an expression for the lapse rate `T` in terms of `gamma, R,g` and `m_(a)`. Here `gamma` is the ratio of specific heat at constant pressure to specific heat at constant volume `R`, the gas constant , and `m_(a)`, the relevant molar mass.
B. Estimate the change in temperature when we ascend a height of one kilometer?
C. Show that pressure will depend on height as given by Eq.(1). Find an explicit expression for exponent `alpha` in terms of `gamma`
D. According to this model what is the height to which the atmosphere extends? Take `T_(0) = 300K` and `p_(0) = 1 atm`.

Answer

A

Explanation

(a) `dP = rhog dz P -dP`
`(dP)/(dz) = rho g`
`PT^((gamma)/(gamma-1))=P_(0)T_(0)^((gamma)/(1-gamma))="const"`
`P = (P_(0)T_(0)^((gamma)/(1-gamma)))/T^((gamma)/(1-gamma))`
`(dP)/(dz) =- P_(0)T_(0)^((gamma)/(1-gamma))(gamma)/(1-gamma) (1)/(T^((gamma)/(1-gamma)+1))(dT)/(dz)`
`=- (P_(0)T_(0)^((gamma)/(1-gamma)))/(T^((gamma)/(1-gamma))).(1)/(T).(gamma)/(1-gamma).(dT)/(dz)`
`(dP)/(dz) = (P)/(T) (gamma)/(1-gamma)(dT)/(dz)`
`rho g =- (P)/(T)(gamma)/(1-gamma) (dT)/(dz)`
`T = (dT)/(dz) =((rhoT)/(P)) g (gamma-1)/(gamma) =(m_(a))/(R)g((gamma-1))/(gamma)`
(b) Change in temperature `= (m_(a)g(gamma-1))/(Rgamma) xx 1000`
`m_(a) = (21 xx 23 +79 xx 28)/(21 +79) = 28.84 g//mol`
`gamma = 7//5`
Change in temperature `= 9.9` Kelvin
(c ) `P_(0)T_(0)^((gamma)/(1-gamma)) = PT^((gamma)/(gamma-1))`
Temperature at height `z, T = T_(0) - Tz`
`P_(0)T_(0)^((gamma)/(1-gamma)) = P (T_(0)-Tz)^((gamma)/(1-gamma))`
Given equation, `P = P_(0) ((T_(0)-Tz)/(T_(0)))^(alpha)`
Comparing `alpha = (gamma)/(gamma-1)`
(d) `T_(0) = 300 K & P_(0) = 1 atm`
`T_(min) = 0`
`T_(0) - Tz = 0`
`z = (T_(0))/(T) = (300)/(9.9) = 30.3 km`