Fyneman exercise I ch39-4

Q.

A)

Imagine a tall vertical column of gaseous or liquid fluid whose density varies with height. Show that the pressure as a function of height follows the differential equation ${dP \over dh}=-\rho(h)g$.

B)

Solve this differential equation for the case of a gaseous atmosphere of molecular weight $\mu$, in which the temperature is constant as a function of $h$.

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A.

A)

$\rho(h) = number\ of\ molecules\ in \ unit\ volume\ at\ height\ h$

Imagine the volume of unit area and infinitesimal height $dh$.

Number of molecules = $\rho(h)dh$

At equilibrium the force of gravity in the volume plus pressure difference is zero.

$dP + m g\rho(h)dh = 0\tag{1}$

, where $m$ is mass of one molecule.

For unit mass

${dP \over dh}=-\rho(h)g$

B)

For ideal gas

$PV=NkT$

, where $N$ is number of molecules in volume $V$.

Then, for unit volume

$P=nkT$

, where $n$ is number of molecules in unit volume, which means density.

$n={P \over kT}=\rho(h)$

From eq.(1)

${dP \over dh}=-mg{P \over kT}\tag{2}$

We need $m$.

$m= one\ molecule\ mass\ =\ {molecule\ weight \over Avogadro\ constant}={\mu \over N_A}$

Substituting $m$ to eq.(2) gives

${dP \over dh}=-{\mu g\over N_AkT}P$

$P(h)=P_0e^{-{\mu g\over N_AkT}h}=P_0e^{-{\mu g\over RT}h}$

, where $R=N_Ak$.

Reference

molecule weight
Avogadro constant