Answer
$M_{n}= 27,692.31 \approx 27,692 g/mol$
Work Step by Step
Required:
Number-average molecular weight that is required to give a tensile strength of 140 MPa.
Solution:
Using Equation 15.3 and substituting the given values, two expressions with two unknowns will be expressed:
$TS = TS_{∞} - \frac{A}{M_{n}}$
$90 MPa = TS_{∞} - \frac{A}{20,000 g/mol}$
$180 MPa = TS_{∞} - \frac{A}{40,000 g/mol}$
Computing the two expressions simultaneously,
$TS_{∞} = 270 MPa, A= 3.60 \times 10^{6} MPa-g/mol$
Substituting the computed values into Equation 15.3 for $TS = 140 MPa$:
$M_{n} = \frac{A}{TS_{∞} - TS} = \frac{3.60 \times 10^{6} MPa-g/mol}{(270 MPa - 140 MPa)} = 27,692.31 \approx 27,692 g/mol$
