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Mechanical Properties Of Fluids

Question
CBSEENPH11020191

Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms: 


 


[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

Solution
If r is the radius of the atom, then volume of each atom = 4/3 π r
Volume of all atoms in one mole of substance = 4/3 π r3 × N = M/ρ 
Therefore,
 r = [ 3M / 4πρN]1/3
For Carbon, 
M = 12.01 × 10-3 Kg
ρ = 2.22 × 103 Kg m-3

 straight r space equals space fraction numerator 3 space straight x 12.01 straight x 10 to the power of negative 3 end exponent over denominator 4 space straight x space begin display style bevelled 22 over 7 end style straight x space left parenthesis 2.22 space straight x 10 to the power of 23 right parenthesis space straight x space left parenthesis 6.023 space straight x space 10 to the power of 23 right parenthesis end fraction space

space space space equals space 1.29 space straight x space 10 to the power of negative 10 space end exponent straight m
space space space equals space 1.29 space straight A with straight o on top 
Similarly, 
 
for gold, r = 1.59 Å 
for liquid nitrogen, = 1.77 Å 
for lithium, r = 1.73 Å 
for liquid fluorine, r = 1.88 Å