To obtain the Maxwell speed distribution, we assumed a uniform temperature. kinetic-only energy of $\mathbf{E}=\left\{m\left(v_x^2+v_y^2+v_z^2\right)\right.$, and we assumed that we wished to find the average of an arbitrary function of X. Along the way, we obtained probability per unit height speed,$P\left(v\right)$.

a) Assuming a uniform temperature and an energy of$\left.E=\frac{1}{2}m\left(v_x^2+v_y^2+v_z^3\right)+mgy\right)$and assuming we wish to find the average of an arbitrary function of Y, obtain a probability per unit height,$P\left(y\right)$ .

b) Assuming a temperature of$300\text{K}$. how much less the density of the atmosphere'sat an altitude of(about$3000$ft) than at sea level'?

(c) What of the${\text{O}}_2$in the atmosphere?

Average of an unknown function $f\left(y\right):$

$\overline{f\left(y\right)}=\frac{\int \text{d}x\text{d}y\text{d}z\text{d}{v}_x\text{d}{v}_y\text{d}{v}_{z}f\left(y\right)e^{-E/k_{B}T}}{\int \text{d}x\text{d}yd{v}_x\text{d}{v}_y\text{d}{v}_z{e}^{-E/k_{B}T}}$

(a)

The average of an unknown function $f\left(y\right)$:

$\overline{f\left(y\right)}=\frac{\int \text{d}x\text{d}y\text{d}z\text{d}{v}_x\text{d}{v}_y\text{d}{v}_{z}f\left(y\right)e^{-E/k_{B}T}}{\int \text{d}x\text{d}yd{v}_x\text{d}{v}_y\text{d}{v}_z{e}^{-E/k_{B}T}}$

Next, since$f\left(y\right)$ depends only on Y all the integrals cancel except for the integration over Y. Since Y is restricted to $[0,\infty )$we get:

$\begin{array}{rcl}\overline{f\left(y\right)}& =& \frac{{\int }_0^{\infty }\text{d}yf\left(y\right)e^{-mgy/k_{B}T}}{{\int }_0^{\infty }\text{d}y{e}^{-mgy/k_{B}T}}\\ & =& \frac{{\int }_0^{\infty }\text{d}yf\left(y\right)e^{-mgu/k_{B}T}}{\frac{k_{B}T}{mg}{\int }_0^{\infty }\text{d}t{e}^{-t}}\\ & =& \frac{mg}{k_{B}T}\frac{{\int }_0^{\infty }\text{d}yf\left(y\right)e^{-\frac{mgy}{k_{B}T}}}{\Gamma \left(1\right)}\\ & =& \underset{0}{\overset{\infty }{\int }}dyf\left(y\right)\frac{mg}{k_{B}T}{e}^{-\frac{mgy}{k_{B}T}}\\ & & \end{array}$

The probability per unit height:

$P\left(y\right)=\frac{mg}{k_{B}T}{e}^{-\frac{\underset{¯}{mgy}}{k_{B}T}}$

(b)

The density $\rho \left(y\right)$is proportional to $P\left(y\right)$and so we have:

Put values for $N_2$in we get:

$\begin{array}{rcl}\frac{{\rho }_{N_2}(800\text{m})}{{\rho }_{N_2}\left(0\right)}& =& \exp \left(-\frac{\left(2·14.007·1.66·{10}^{-27}\text{kg}\right)·\left(9.81\text{m}/{\text{s}}^2\right)·(800\text{m})}{\left(1.38·{10}^{-23}\text{J}/\text{K}\right)·(300\text{K})}\right)\\ & \approx & \exp \left(-881.11·{10}^{-4}\right)\\ & \approx & 0.916\\ & & \end{array}$

(c)

Put values for $O_2$we get,

$\begin{array}{rcl}\frac{{\rho }_{O_2}(800\text{m})}{{\rho }_{O_2}\left(0\right)}& =& \exp \left(-\frac{\left(2·15.999·1.66·{10}^{-27}\text{kg}\right)·\left(9.81\text{m}/{\text{s}}^2\right)·(800\text{m})}{\left(1.38·{10}^{-23}\text{J}/\text{K}\right)·(300\text{K})}\right)\\ & \approx & \exp \left(-100.642·{10}^{-4}\right)\\ & \approx & 0.904\\ & & \end{array}$

This is expected because the $O_2$is heavier than the ${\text{N}}_2$and is therefore more tightly bound to the surface.