Question

$A)$ Pressure decreases as one ascends the atmosphere. If the density of air is $p$, 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 $p_0$.
(The change in pressure $dp=-\rho gdh$ over a differential height $dh$)

$C)$ If $p_0=1.03\times {10}^{5}N/m^2,{\rho }_0=1.29kg/m^3$ and $g=9.8m/s^2$, at what height will the pressure drop to $\left(1/10\right)$ the value at the surface of the earth ?
(Pressure at height $h$ is given by $p=p_{o}exp(-\frac{{\rho }_{o}g}{p_o}h))$

$D)$ This model of the atmosphere (i.e., pressure is proportional to the density) works for relatively small distances. Identify the underlying assumption that limits the model.

Answer 1

$A)$ Let us consider a horizontal layer of air with cross-section $A$ and height $dh$.
Let, the Pressure at the top of the layer is $P$
As given pressure will decrease as we are going up the atmosphere.
Pressure at the bottom of the layer is $p+dp$.
Where, $dp$ is the change in pressure of topmost layer and bottom layer.
The layer is in equilibrium.
So, the net upward force must be balanced by the weight,
$(p+dp)A-pA=-mg$
$(m=\rho Adh)$
$\Rightarrow \left(dp\right)A=-\rho gAdh$
$dp=-\rho gdh$
Negative sign shows that pressure decreases with height.

Final answer: $dp=-\rho gdh$

$B)$ Let the density of air on the earth's surface be ${\rho }_o$,
As given pressure is proportional to the density.
$\frac{p}{p_o}=\frac{\rho }{{\rho }_o}$
$\Rightarrow \rho =\frac{{\rho }_o}{p_o}p$
So, change in pressure, $dp=-\rho gdh$
$dp=-\frac{{\rho }_{o}g}{p_o}pdh$
$\frac{dp}{p}=-\frac{{\rho }_{o}g}{p_o}dh$
Integrating on both sides,
${\int }_{p_o}^p\frac{dp}{p}=-\frac{{\rho }_{o}g}{p_o}{\int }_o^{h}dh$
$ln\frac{p}{p_o}=-\frac{{\rho }_{o}g}{p_o}h$
$p=p_{o}exp(-\frac{{\rho }_{o}g}{p_o}h)$

Final answer: $p=p_{o}exp(-\frac{{\rho }_{o}g}{p_o}h)$


$C)$ Given, pressure at surface of the earth, $p_0=1.03\times {10}^{5}N/m^2$
Density ${\rho }_0=1.29kg/m^3$
$g=9.8m/s^2$
Given, pressure at height $h$ is given by, $p=p_{o}exp(-\frac{{\rho }_{o}g}{p_o}h)$
Let the height be $h$ where pressure drop to $\frac{1}{10}$ the value at the surface of earth.
$\frac{p_0}{10}=p_0\left(exp(-\frac{{\rho }_{0}gh}{p_0})\right)$
$ln\left(\frac{1}{10}\right)=-\frac{{\rho }_{0}gh}{p_0}$
$h=-\frac{p_0}{{\rho }_{0}g}ln\left(\frac{1}{10}\right)$
$h=-\frac{1.03\times {10}^5}{1.29\times 9.8}(-2.303)$
$h=18.76km$

Final answer: $18.76km$

$D)$ Pressure is proportional to the density will be true only when the temperature of the air remains constant (i.e. isothermal). And temperature remains constant only near the surface of the earth not at greater heights.

Final answer: Isothermal case.