Maxwell's speed distribution law
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(See attached file for full problem description with diagram and symbols)
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1) Find the fractional number of molecules of a gas having speeds between vm and 1.05vm where vm = Sq.root(2kT/m) and is denoted as the most probable speed. [vm denotes v subscript m]
2) At what radial "distance" v from the origin of velocity space is the "density" of representative points one-half as great as at the origin.
3) How many oxygen molecules in one mole have speeds greater than 103 m/s at a temperature of 1000 K?
4) The oven in the figure (see attachment) contains bismuth at a temperature of 827 K. The drum is 10 cm in diameter and rotates at 6000 rpm. Find the displacement on the glass plate G, measured from a point directly opposite the slit, of the points of impact of the molecules Bi and Bi2. Assume that all the molecules of each species travel with the root mean square speed appropriate to that species.
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Solution Summary
Problems have been solved with the application of Maxwell's speed distribution law. How many oxygen molecules are in one mole are determined.
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SOLUTION
Basic theory
1) As per Maxwell speed distribution law :
a) The number of molecules per unit volume having speed v is given by :
f(v) = n (m/2ΠkT)3/2exp( -mv2/2kT) where n = total number of molecules, m = mass of a molecule, k = Boltzmann constant, T = Absolute temp. ...............(1a)
Or f(v) = n (M/2ΠRT)3/2exp( -Mv2/2RT) where n = total number of molecules, M = molecular mass of the gas, R = Universal gas constant, T = Absolute temp. ...(1b)
b) The fraction of molecules of a gas having speeds between v and dv is given by :
F(v) dv = 4Π (m/2ΠkT)3/2 v2 exp (-mv2/2kT) dv ............(2a)
Or F(v) dv = 4Π (M/2ΠRT)3/2 v2 exp (-Mv2/2RT) dv ............(2b)
1) Most probable speed
To find the fraction of molecules having velocities between vm and 1.05vm we use equation (2a) above : F(v) dv = 4Π (m/2ΠkT)3/2 v2 exp (-mv2/2kT) dv
We take v in the above equation as the mid point between the range of velocities i.e.
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v = (vm +1.05vm)/2 = 1.025vm = 1.025 √(2kT/m)
dv = 1.05vm - vm = 0.05 vm
mv2/2kT = (m/2kT)(1.025)2(2kT/m) = 1.05
Putting values in the above equation we get,
F(v) dv = 4Π (m/2ΠkT)3/2 v2 exp (-mv2/2kT) dv = 4Π ...
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