Question

In analogy with the thermal conductivity, derive an approximate formula for the viscosity of an ideal gas in terms of its density, mean free path, and average thermal speed. Show explicitly that the viscosity is independent of pressure and proportional to the square root of the temperature. Evaluate your formula numerically for air at room temperature and compare to the experimental value quoted in the text.

Short answer

The experimental value quoted within the text is

$\eta =\frac{1}{2}\rho \overline{v}\ell ,$

$\eta =\frac{\sqrt{3mKT}}{8\pi r^2},$

${\eta }_{\text{air}}=3.88\times {10}^{-}5\mathrm{N}\cdot {\mathrm{m}}^{-2}\cdot \mathrm{s}$

Step-by-step solution

Path and Velocity

The atomic length is $r$, while the energy of $1$monomer is $m$. Suppose a skinny rectangular sheet of gas in between plates; electrons within a distance of such slab's midline can across it whether or not they are travelling within the same direction. The common half the molecules in each half are travelling towards the midpoint so if the average horizontal momentum of the molecules on side $i$is $p_i$for $i=1,2,$, then in a very time $\mathrm{\Delta }t$(that is, the time it takes a median molecule to travel the momentum transferred is:

$\begin{array}{c}\ell \approx \frac{1}{4\pi r^2}\frac{V}{N}\\ \end{array}$

$\overline{v}\approx \sqrt{\frac{3KT}{m}}$

$\mathrm{\Delta }p=\frac{1}{2}\left(p_1-p_2\right)$

but the momentum is that the mass multiplied by velocity,

$p_1=M{u}_{x,1},p_2=M{u}_{x,2}$

$\mathrm{\Delta }p=\frac{M}{2}\left(u_{x,1}-u_{x,2}\right)$

$p_1=M{u}_{x,1},p_2=M{u}_{x,2}$

Mean Path

The slabs' total force acting is:

$\frac{F_x}{A}=\frac{\mathrm{\Delta }p}{A\mathrm{\Delta }t}$

Once we transfer $F=\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}$from formula $\mathrm{\Delta }p$with , then need:

$\frac{F_x}{A}=\frac{1}{A\mathrm{\Delta }t}\left[\mathrm{\Delta }p\right]=\frac{1}{A\mathrm{\Delta }t}\left[\frac{M}{2}\ell \frac{d{u}_x}{dz}\right]$

$\frac{\ell }{\ell }$has been wont to scale, hence

$\frac{F_x}{A}=\frac{M{\ell }^2}{2\ell A\mathrm{\Delta }t}\frac{d{u}_x}{dz}$

Instead, the intensity $\rho$is capable the load $M$out over freed mean path scaled by the per sectional space, and $\mathrm{\Delta }t$ is that the time required the particle to maneuver one freedom meant line $l$, like in:

$\frac{M}{\ell A}=\rho \frac{\ell }{\mathrm{\Delta }t}=\overline{v}$

As a answer, equation are as chooses to follow:

$\begin{array}{c}\frac{F_x}{A}=\frac{1}{2}\left[\frac{M}{\ell A}\right]\left[\frac{\ell }{\mathrm{\Delta }t}\right]\ell \frac{d{u}_x}{dz}=\frac{1}{2}\rho \overline{v}\ell \frac{d{u}_x}{dz}\\ \end{array}$

$\frac{F_x}{A}=\frac{1}{2}\rho \overline{v}\ell \frac{d{u}_x}{dz}$

$\frac{F_x}{A}=\eta \frac{d{u}_x}{dz}$

$\eta =\frac{1}{2}\rho \overline{v}\ell$

Density

Average volume from its respective components. Its density of just an noble gas is obtained by multiplying the typical of an air molecule $m$but by full different molecules $N$out over given quantity $V$, or:

$\begin{array}{c}\rho =\frac{mN}{V}\\ \end{array}$

$\eta =\frac{1}{2}\frac{mN}{V}\overline{v}\ell$

$\begin{array}{c}\\ \eta =\frac{1}{2}\frac{mN}{V}\left[\ell \right]\left[\overline{v}\right]=\frac{1}{2}\frac{mN}{V}\left[\frac{1}{4\pi r^2}\frac{V}{N}\right]\left[\sqrt{\left.\frac{3KT}{m}\right]}\right.\\ \end{array}$

$\to \eta =\frac{\sqrt{3mKT}}{8\pi r^2}$

Average Mass

The dense of noble gas is up to average mass of surrounding air $m$ weighted by the tons. The dense of just an inert fluid is capable average mass of all its molecules. The concentration of wind is calculated based:

$\rho =\frac{mN}{V}$

$\begin{array}{c}\rho =\frac{4.81\times {10}^{-26}\times 6.022\times {10}^{23}}{0.028}=1.0345\mathrm{kg}\cdot {\mathrm{m}}^3\\ \end{array}$

$\rho =1.0345\mathrm{kg}\cdot {\mathrm{m}}^3$

Unless we've single mole of air under atmosphere pressure, the capacity of $1$ mole the least bit is $m=4.81\times {10}^{-26}\mathrm{kg}$, in addition because the molarity in one mole is indeed the Avogadro number , $1.9\times {10}^{-5}\mathrm{N}\cdot {\mathrm{m}}^{-2}$then thethe degreeis:

$\begin{array}{c}\eta =\frac{1}{2}\rho \overline{v}\ell =\frac{1}{2}\times 1.0345\times 500\times 1.5\times {10}^{-7}\\ \end{array}$

$=3.88\times {10}^{-}5{\mathrm{N}}^{-2}\cdot {\mathrm{m}}^{-2}\cdot \mathrm{s}$

$\begin{array}{c}\\ \to {\eta }_{\text{oir}}=3.88\times {10}^{-5}\mathrm{N}\cdot {\mathrm{m}}^{-2}\cdot \mathrm{s}\end{array}$

Are using data from Schroeder's book about ambient air.