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Q87P

Expert-verifiedFound in: Page 447

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Show that the variation of atmospheric pressure with altitude is given by${P}{=}{{P}}_{{o}}{{e}}^{-ay}$, where a ${\alpha}{=}\frac{{\rho}_{o}g}{{P}_{o}}{,}$ ${{P}}_{{o}}$is atmospheric pressure at some reference level$y=0$, and role="math" localid="1663776653925" ${{\rho}}_{{o}}$ is the atmospheric density at this level. Assume the decrease in atmospheric pressure over an infinitesimal change in altitude (so that the density is approximately uniform over the infinitesimal change) can be expressed from Equation 14.4 asrole="math" localid="1663776670018" ${d}{p}{=}{-}{\rho}{g}{d}{y}$. Also assume the density of air is proportional to the pressure, which, as we will see in Chapter 20, is equivalent to assuming the temperature of the air is the same at all altitudes.**

It is proved that the variation of atmospheric pressure with altitude is given by $P={P}_{o}{e}^{-ay}$ .

**Pascal's law states that when pressure is applied to a confined fluid, the pressure is transmitted undiminished to every point in the fluid and to every point on the walls of the container.**

The pressure in a fluid at rest varies with depth h in the fluid according to the expression

$P={P}_{o}+\rho gh$

Where, $P$is the absolute pressure,${P}_{o}$ is the atmospheric pressure and $\rho $is the density of the fluid, assumed uniform, $g$is the gravitational acceleration constant, and $h$ is the height.

The incremental version of $P={P}_{o}+\rho gh$is $dP=-\rho gdy$.

Assume that the density of air is proportional to pressure, or $\frac{P}{\rho}=\frac{{P}_{0}}{{\rho}_{0}}$.

Combining these two equations you have,

$dP=-P\frac{{\rho}_{0}}{{P}_{0}}gdy$

Integrating both the above sides,

$\underset{{p}_{o}}{\overset{p}{\int}}\frac{dP}{P}=-g\frac{{\rho}_{0}}{{P}_{0}}\underset{0}{\overset{y}{\int}}dy\phantom{\rule{0ex}{0ex}}{\left[\mathrm{ln}\left(P\right)\right]}_{{p}_{o}}^{p}=-g\frac{{\rho}_{0}}{{P}_{0}}y\phantom{\rule{0ex}{0ex}}\mathrm{ln}\left(\frac{P}{{P}_{o}}\right)=-\frac{{\rho}_{0}gy}{{P}_{0}}\phantom{\rule{0ex}{0ex}}$

Defining $\alpha =\frac{{\rho}_{0}g}{{P}_{0}}$ then gives $P={P}_{o}{e}^{-ay}$.

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