**Question:**

**(a) Pressure decreases as one ascends the atmosphere. If the density of air is ρ, what is the change in pressure dp over a differential height dh?**

**(b) Considering the pressure p to be proportional to the density, find the pressure p at a height h if the pressure on the surface of the earth is ρo.**

**(c) If po = 1.03 × 105 N/m2, ρo = 1.29 kg/m3 and g is 9.8 m/s2 at what height will the pressure drop to (1/10) the value at the surface of the earth?**

**(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.**

**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.

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