It’s possible to use the ideal-gas law to show that the

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a.)

We have to determine the value of \(z_{a}\), if the density of the earth's atmosphere decreases exponentially with height, that is \(\rho=\rho_{o} \exp \left(-z / z_{0}\right)\) where, \(z\) is the height above sea level and \(\rho_{0}\) is the density at sea level

The atmospheric pressure at sea level is \(1.013 \times 10^{5} \mathrm{~Pa}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) that is the weight of the air column with a \(1 \mathrm{~m}^{2}\) cross section is \(1.013 \times 10^{5} \mathrm{~N}\).

\(\rho_{o}=1.28 \mathrm{~kg} / \mathrm{m}^{3}\)

Consider the weight of \(1 \mathrm{~m}^{2}\) slice of air column with thickness \(d z\) at a height \(z\). This slice has a volume \(d V\) given by

\(d V =A z\)

\(=\left(1 m^{2}\right) d z\)

and its weight is given by

\(d w =(\rho d V) g\)

\(=\rho_{o} \exp \left(-z / z_{o}\right)\left(1 \mathrm{~m}^{2}\right) d z g\)

Thus, the total weight of the air column can be found by integrating \(d w\) from \(z=0\) to \(z=\infty\)

\(w=\int_{0}^{\infty} \rho_{o} \exp \left(-z / z_{o}\right)\left(1 \mathrm{~m}^{2}\right) g d z\)

\(=\left[-\rho_{o} g\left(1 \mathrm{~m}^{2}\right) z_{o}\right]\left[e^{-z_{o} / z}\right]_{0}^{\infty}\)

\(=\rho_{o} g\left(1 \mathrm{~m}^{2}\right) z_{o}\)

Hence,

\(z_{o}=\frac{w}{\rho_{o} g}\)

\(=\frac{101,300}{1.28 \times 9.80}\)

\(=8.1 \times 10^{3} \mathrm{~m}\)

Therefore, the value of \(z_{0}=8.1 \times 10^{3} \mathrm{~m}\)