(a) Pressure decreases as one ascends the atmosphere.

Solution:

(a) As we move above, the pressure decreases because the thickness of the gas above us decreases. Let A be the cross-section of the horizontal layer of the air and height be dh.

Then P is the pressure on the top layer and p + dp is the pressure at the bottom.

Since the layer of the air is in equilibrium, the net upward force is balanced

That is net upward force = net downward force

Which is given as:

dp = -ρgdh

The negative sign indicates the decrease in pressure with height.

(b) Let ρo be the density of air on the surface of the earth such that pressure is proportional to the density.

$\rho \propto p$

$\frac{\text { pressure at some height }(p)}{\text { pressure at the surface of earth }\left(p_{0}\right)}=\frac{\rho}{\rho_{0}}$

Solving the above, we get pressure as:

$\left.p=p_{0} e^{(}-\frac{\rho_{0} g h}{p_{0}}\right)$

(c) We know that

$\left.p=p_{0} e^{(}-\frac{\rho_{0} g h}{p_{0}}\right)$

By substituting the values, we can calculate h = 18.43 km

(d) For the relation

$p \propto \rho$ to be valid, the temperature in the air should be uniform which is known as isothermal situation and this possible only near the surface of the earth and not at greater heights.