An ideal gas of molar mass `M` is located in the uniform gravitational field in which the free-fall acceleration is equal to `g`. Find the gas pressure as a function of height `h`. If `p = p_0` at `h = 0`, and the temperature varies with height as
(a) `T = T_0 (1 - ah)` ,
(b) `T = T_0 (1 + ah)`,
where `a` is a positive constant.

Explanation

(a) We know that the variation of pressure with height of a fluid is given by :
`dp = - rho g dh`
But from gas law `p = (rho)/(M) RT` or, `rho = (p M)/(RT)`
From these two Eqs.
`dp = - (p Mg)/(RT) dh` ….(1) ltrgt or, `(dp)/(p) = (-Mg dh)/(R T_0(1 - ah))`
Integrating, `int_(p_0)^p (dp)/(p) = (-Mg)/(R T_0) int_0^h (dh)/((1 - ah))`, we get
`1 n (p)/(p_0) = 1 n(1 - ah)^(Mg//aRT_0)`
Hence, `p = p_0(1 - ah)^(Mg//aRT_0)`, Obvionsly `h lt (1)/(a)`
(b) Proceed up to Eq. (1) of part (a). and then put `T = T_0 (1 + ah)` and proceed further in the same fashion to get
`p = (p_0)/((1 + ah)^(Mg//aRT_0))`.