If air is assumed to be at temperature 290 K with molar mass 29 gm/mole and $g = 10 m / s^{2}$ then change in pressure at height 8.314 km and ground level as a fraction of pressure at ground level is given by :

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$\begin{aligned} \frac{1}{e} \end{aligned}$

$\begin{aligned} (e - 1) \end{aligned}$

$\frac{e}{e - 1}$

$\begin{aligned} \frac{e - 1}{e} \end{aligned}$

answer is C.

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Detailed Solution

It is known that,

$P = P_{0} e^{- \frac{M g h}{R T}}$

Change in pressure, $Δ P = P - P_{0}$

$\begin{aligned} \Rightarrow \frac{Δ P}{P_{0}} = e^{- \frac{M g h}{R T}} - 1 \end{aligned}$

$\frac{Δ P}{P_{0}} = e^{- \frac{29 \times 10^{- 3} \times 10 \times 8.314 \times 10^{3}}{8.314 \times 290}} - 1$

$\Rightarrow e^{- 1} - 1 = \frac{1 - e}{e} < 0$

So, $\frac{e - 1}{e}$ .

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