Question

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 p₀.

Answer 1

Let p₀ be the density of air on the surface of the earth.

As per the question, pressure ∝ density

⇒ ${\displaystyle \frac{p}{p_0}=\frac{\rho }{{\rho }_0}}$

⇒ ${\displaystyle \rho =\frac{{\rho }_0}{p_0}p}$ 

∴ ${\displaystyle dp=-\frac{{\rho }_{0}g}{p_0}pdh}$  ......[∵ dp = – ρgdh]

⇒ ${\displaystyle \frac{dp}{p}=-\frac{{\rho }_{0}g}{p_0}dh}$

⇒ ${\displaystyle {\int }_{p_0}^p\frac{dp}{p}=-\frac{{\rho }_{0}g}{p_0}{\int }_0^{h}dh}$  .....${\displaystyle \left[\begin{array}{cc}\because at h=p& r=p_0\\ \text{and}at h=h& p=p\end{array}\right]}$

⇒ In ${\displaystyle \frac{p}{{\rho }_0}=-\frac{{\rho }_{0}g}{p_0}h}$

By removing log, ${\displaystyle p=p_{0}e(-\frac{{\rho }_{0}gh}{p_0})}$