The Distribution of Molecular Speeds — Question 9

Section 19-6 · Physics pages 3148–3149

Question 9 A ratio of speed densities can determine temperature A Maxwell gas of molecular mass m = 4.00 × 10−26 kg has measured speed-density ratio f(600 m/s)/f(300 m/s) = 2.00. Use kB = 1.381 × 10−23 J/K. These are density values or equal-width narrow-bin limits, not unequal-width raw counts. v f(v) 300 600 Axes and labels specify the model; the sketch is schematic. (a) Infer temperature while retaining the v2 prefactor in the density. (b) For these two speeds, determine the allowable ratio range for finite positive temperature and interpret a reported ratio of 5.00.
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Question 9 - Solution A ratio of speed densities can determine temperature v f(v) 300 600 (a) Method and calculation r = (600 300 )2 exp [ −m(6002 −3002) 2kBT ] = 4e−A/T . A = 4.00 × 10−26(270000) 2(1.381 × 10−23) ≈391.021 K, T = A ln(4/2) ≈564.124 K. (b) Interpretation and further calculation T > 0 ⇒0 < e−A/T < 1 ⇒0 < r < 4. r = 5 ⇒ln(4/r) < 0, which cannot give a positive Maxwell temperature. Such data require checking the model, detector bias or bin-width conversion. Physical check Omitting the v2 factor would reverse the interpretation of a density that rises over the selected speed interval.

Question 9

A ratio of speed densities can determine temperature

A Maxwell gas of molecular mass m=4.00×10−26m=4.00\times10^{-26} kg has measured speed-density ratio f(600m/s)/f(300m/s)=2.00f(600\,\mathrm{m/s})/f(300\,\mathrm{m/s})=2.00. Use kB=1.381×10−23k_B=1.381\times10^{-23} J/K. These are density values or equal-width narrow-bin limits, not unequal-width raw counts.

Diagram shown in the original SVG worksheet.

  1. Infer temperature while retaining the v2v^2 prefactor in the density.

  2. For these two speeds, determine the allowable ratio range for finite positive temperature and interpret a reported ratio of 5.00.

Question 9 - Solution

A ratio of speed densities can determine temperature

Diagram shown in the original SVG worksheet.

(a) Method and calculation

r=(600300)2exp⁡[−m(6002−3002)2kBT]=4e−A/T.r=\left(\frac{600}{300}\right)^2\exp\left[-\frac{m(600^2-300^2)}{2k_BT}\right]=4e^{-A/T}.A=4.00×10−26(270000)2(1.381×10−23)≈391.021K,T=Aln⁡(4/2)≈564.124K.A=\frac{4.00\times10^{-26}(270000)}{2(1.381\times10^{-23})}\approx391.021\,\mathrm K,\quad T=\frac A{\ln(4/2)}\approx564.124\,\mathrm K.

(b) Interpretation and further calculation

T>0⇒0<e−A/T<1⇒0<r<4.T>0\ \Rightarrow\ 0<e^{-A/T}<1\ \Rightarrow\ 0<r<4.r=5⇒ln⁡(4/r)<0,r=5\ \Rightarrow\ \ln(4/r)<0,which cannot give a positive Maxwell temperature. Such data require checking the model, detector bias or bin-width conversion.

Physical check

Omitting the v2v^2 factor would reverse the interpretation of a density that rises over the selected speed interval.