A Maxwell-Boltzmann distribution implies that a given molecule (mass m ) will have a speed between vand v + dv with probability equal to f(v)dv where ƒ(v) ∞ v² e-mv²/2kBI and the proportionality sign is used because a normalization constant has f(v)dv.) For this distribution, calculate the been omitted. (You can correct for this by dividing any averages you work out by mean speed (v) and the mean inverse speed (1/v). Show that (v)(1/v) = ¼/ .
A Maxwell-Boltzmann distribution implies that a given molecule (mass m ) will have a speed between vand v + dv with probability equal to f(v)dv where ƒ(v) ∞ v² e-mv²/2kBI and the proportionality sign is used because a normalization constant has f(v)dv.) For this distribution, calculate the been omitted. (You can correct for this by dividing any averages you work out by mean speed (v) and the mean inverse speed (1/v). Show that (v)(1/v) = ¼/ .
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Transcribed Image Text:A Maxwell-Boltzmann distribution implies that a given molecule (mass m ) will have a speed between 1 and v + dv with
probability equal to f(v)dv where ƒ(v) ∞ v² e-mv²/2k³I and the proportionality sign is used because a normalization constant has
f(v)dv.) For this distribution, calculate the
been omitted. (You can correct for this by dividing any averages you work out by
mean speed (v) and the mean inverse speed (1/v). Show that (v)(1/v) = ¼/ .
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