Answer
$TS = 112.5 MPa$
Work Step by Step
Required:
Tensile strength at a number-average molecular weight of 40,000 g/mol
Solution:
Using Equation 15.3 and substituting the given values, two expressions with two unknowns will be expressed:
$TS = TS_{∞} - \frac{A}{M_{n}}$
$50 MPa = TS_{∞} - \frac{A}{30,000 g/mol}$
$150 MPa = TS_{∞} - \frac{A}{50,000 g/mol}$
Computing the two simultaneously, we find:
$TS_{∞} = 300 MPa, A= 7.50 \times 10^{6} MPa-g/mol$
Substituting the computed values into Equation 15.3 for $M_{n} = 40,000 g/mol$:
$TS = TS_{∞} - \frac{A}{M_{n}} = (300 MPa) - \frac{(7.50 \times 10^{6} MPa-g/mol)}{(40,000 g/mol)} = 112.5 MPa$