On a sunny day, the temperature and pressure of atmosphere varying with height z from sea level as $T = T_{0} (1 - \frac{z}{z_{0}}),$ and $P = P_{0} {(1 - \frac{z}{z_{0}})}^{α}$ . Where $T_{0}$ and $z_{0}$ , are $25^{\circ} C$ and $25 k m$ . The temperature, pressure and density of air at sea level is $T_{0} = 25^{\circ} C$ , $P_{0} = 1 a t m$ and $1.2 k g / m^{3}$ respectively. Assume that air follows ideal gas law in the given condition, i.e., $\frac{ρ T}{P} =$ constant, find the value of $α$ . $1 a t m = 10^{5} N / m^{2}$ and $g = 10 m / s^{2}$ (g is uniform)

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answer is 3.

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Detailed Solution

$d p = - ρ g d z$
$\Rightarrow \frac{d p}{d z} = - \frac{ρ_{0} g T_{0} P}{P_{0} T}$
$\Rightarrow \int_{P_{0}}^{P} \frac{d P}{P} = - \frac{ρ_{0} g}{P_{0}} \int_{0}^{z} \frac{d z}{(1 - \frac{z}{z_{0}})}$

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$\Rightarrow l n (\frac{P}{P_{0}}) = \frac{ρ_{0} g z_{0}}{P_{0}} l n (1 - \frac{z}{z_{0}})$
$\Rightarrow P = P_{0} {(1 - \frac{Z}{Z_{0}})}^{α},$ where
$α = \frac{ρ_{0} g z_{0}}{P_{0}} = \frac{1.2 \times 10 \times 25 \times 10^{3}}{10^{5}} = 3$