Mean Free Path — Question 10

Section 19-5 · Physics pages 3130–3131

Question 10 A channel size sets the pressure needed for a chosen Knudsen number At 300 K, a dilute gas has λp = 6.00 × 10−3 Pa m. A channel has characteristic width L = 10.0 µm. Define Kn = λ/L. For this exercise adopt the design criterion Kn ≤.0100 for a continuum approximation. L Schematic; labels, not drawing lengths, specify the data. (a) Find the minimum pressure satisfying the criterion and the Knudsen number at 100 Pa. (b) For a channel one-tenth as wide, find the new minimum pressure. Explain whether reducing pressure improves the adopted continuum approximation.
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Question 10 - Solution A channel size sets the pressure needed for a chosen Knudsen number L (a) Method and calculation Kn = 6.00 × 10−3 p(10−5) = 600 Pa p . pmin = 600/.0100 = 60000 Pa, Kn(100 Pa) = 6.00. (b) Interpretation and further calculation L →L/10 ⇒pmin →10pmin = 600000 Pa. p ↓⇒λ ↑⇒Kn ↑. Lower pressure makes molecule-wall travel more important relative to intermolecular collisions. The numerical cutoffis the stipulated design convention, not a sharp universal phase boundary. Physical check Any high-pressure extrapolation also requires the given dilute-gas relation to remain applicable.

Question 10

A channel size sets the pressure needed for a chosen Knudsen number

At 300 K, a dilute gas has λp=6.00×10−3Pam\lambda p=6.00\times10^{-3}\,\mathrm{Pa\,m}. A channel has characteristic width L=10.0μL=10.0\,\mum. Define Kn=λ/L\mathrm{Kn}=\lambda/L. For this exercise adopt the design criterion Kn≤.0100\mathrm{Kn}\le.0100 for a continuum approximation.

Diagram shown in the original SVG worksheet.

  1. Find the minimum pressure satisfying the criterion and the Knudsen number at 100 Pa.

  2. For a channel one-tenth as wide, find the new minimum pressure. Explain whether reducing pressure improves the adopted continuum approximation.

Question 10 - Solution

A channel size sets the pressure needed for a chosen Knudsen number

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Kn=6.00×10−3p(10−5)=600Pap.\mathrm{Kn}=\frac{6.00\times10^{-3}}{p(10^{-5})}=\frac{600\,\mathrm{Pa}}p.pmin=600/.0100=60000Pa,Kn(100Pa)=6.00.p_{\min}=600/.0100=60000\,\mathrm{Pa},\quad\mathrm{Kn}(100\,\mathrm{Pa})=6.00.

(b) Interpretation and further calculation

L→L/10⇒pmin→10pmin=600000Pa.L\to L/10\ \Rightarrow\ p_{\min}\to10p_{\min}=600000\,\mathrm{Pa}.p↓⇒λ↑⇒Kn↑.p\downarrow\ \Rightarrow\lambda\uparrow\ \Rightarrow\mathrm{Kn}\uparrow.Lower pressure makes molecule-wall travel more important relative to intermolecular collisions. The numerical cutoff is the stipulated design convention, not a sharp universal phase boundary.

Physical check

Any high-pressure extrapolation also requires the given dilute-gas relation to remain applicable.