A hanging titanium wire with diameter 2.0 mm (2.0 × 10-3 m) is initially 2.5 m long. When a 9 kg mass is hung from it, the wire stretches an amount 0.605 mm. A mole of titanium has a mass of 48 grams, and its density is 4.51 g/cm3. Find the approximate value of the effective spring stiffness of the interatomic bond.

The approximate value of the effective spring stiffness of the interatomic bond of titanium is 494.391 kN/m per mole.

Explanation:

Spring stiffness is given by

F = k × e

That is

Force = mass × acceleration

= 9×9.81 = 88.29 N

Extension = 0.605 mm = 6.05×10^(-4) m

Spring stiffness = k = F/e = (88.29 N)/(6.05×10^(-4) m) = 145.934 kN/m

Molar mass = 48 g

Density = 4.51 g/cm^3

Volume of one mole = Mass/ Density = 48/4.51 = 10.64 cm^3 = 1.06×10^(-5)m^3

Therefore a cross section of 2.0×10^(-3)m with an area = (pi×D^2)/4 = 3.14×10^(-6) m^2 contains approximately 1/(3.14×10 (-6)) moles or 0.295 moles

Therefore 0.295 moles has a stiffness value of 145.934 kN/m

Therefore, one mole will have an attractive force of 494.391 kN/(m×mole)

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