Mean Free Path — Question 4

Question 4

Transmission through a gas column can infer its pressure

For identical dilute hard-sphere molecules of diameter $d$ and number density $\nu$, use $\lambda=1/(\sqrt2\pi d^2\nu)$. An ideal gas at 300 K has molecular diameter .300 nm. A supplied attenuation model gives collision-free transmission $S=e^{-L/\lambda}$ over $L=1.00\,\mu$m. Measured $S=e^{-2}$. Use $k_B=1.381\times10^{-23}$ J/K and $\nu=p/(k_BT)$.

1. Infer mean free path and pressure.

2. At the same temperature, what pressure would double transmission? Explain why this does not halve pressure.

Question 4 - Solution

Transmission through a gas column can infer its pressure

(a) Method and calculation

$$-\ln S=L/\lambda=2\ \Rightarrow\lambda=.500\,\mu\mathrm m.$$$$p=\frac{k_BT}{\sqrt2\pi d^2\lambda}=\frac{1.381\times10^{-23}(300)}{\sqrt2\pi(.300\times10^{-9})^2(.500\times10^{-6})}\approx2.072\times10^4\,\mathrm{Pa}.$$

(b) Interpretation and further calculation

$$S_{\mathrm{new}}=2e^{-2},\quad\frac{p_{\mathrm{new}}}p=\frac{-\ln S_{\mathrm{new}}}{-\ln S}=\frac{2-\ln2}{2}\approx.653426.$$$$p_{\mathrm{new}}\approx.653426p.$$Transmission is exponential in pressure at fixed geometry, so a factor of two in transmission corresponds to an additive change in optical depth.

Physical check

The inferred pressure uses the supplied effective exponential model, not an assumption that every molecular trajectory has the same speed.