The atomic mass of tungsten is 184 g/mole. The density of tungsten is 19.3 g/cm^

Question

The atomic mass of tungsten is 184 g/mole. The density of tungsten is 19.3 g/cm^3. You perform measurements with a rod of tungsten that is 2.5 m long and has a cross sectional area of 0.0225 cm^2. You attached the top of the rod to a beam along the ceiling of the lab, and hang a 415 kg mass to its lower end. You find that the tungsten rod stretches by 1.26 cm. Assuming that each tungsten atom occupies a cubic volume of side s, what is the value of s? We model the rod as a number N_p of parallel chains of atoms, each chain consisting of N_s atomic "springs" connected in series. What are the values of N_s and N_p? What is the numerical value of the effective spring constant for the entire rod? (This is easy-it doesn't involve your answers to part (b)!) What is the effective inter-atomic spring constant of a single chemical bond between adjacent atoms in your model of the rod?

Explanation / Answer

(a)

Volume of the rod used = Length*Area = 250*0.0225 = 5.625 cm3

Mass of rod = Density*Volume = 19.3 * 5.625 = 108.5625 g

Atomic mass = 184 g/mole

No. of atoms present in the rod = Mass/Atomic mass = 108.5625/184 = 0.59 mole = 0.59*6.02*1023 atoms

Volume occupied by one atom = Total volume/No. of atoms = 5.625/(0.59*6.02*1023)

Assuming the volume occupied by single atom as s3, we get :

s3 = 5.625/(0.59*6.02*1023)

Solving we get :
s = (15.837*10-24)1/3 = 2.51*10-8 cm

(b)

Np*Ns = Total number of atoms = 0.59*6.02*1023

Assuming the length occupied by each atom as 's' calculated in part (a),

Ns * s = Rod length = 250 cm

Putting values, Ns = 250/(2.51*10-8) = 99.6*108 atoms

Thus, Np = 0.59*6.02*1023 / 99.6*108 = 0.0356*1015 atoms

(c)

Using the definition of Young's modulus for defining the spring constant,

F/S = keq*(dL/L)

Here, F = applied force in Newtons, S = Cross sectional area, dL = change in length, L = original length, keq = equivalent spring constant

Thus,

keq = (415*9.8/(0.0255*10-4))*(2.5/(1.26*10-2)) = 0.316 N/m2

(d)

Assuming spring constant of one bond as 'k', we have Ns atoms connected in series along the length of the rod

Thus,

1/keq = Ns/k , because for springs in series we have 1/keq = 1/k1 + 1/k2 + 1/k3 + ... and here all k1 = k2 = ... = k

So, k = Ns*keq = 99.6*108*0.316 = 31.47*108 N/m2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.