Respuesta :
Answer:
[tex]\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}[/tex]
[tex]\mathbf{M_n = 49163.56431 \ g/mol }[/tex]
Explanation:
The question can be well structured in a table format as illustrated below:
Tensile Strength (MPa) Number- Average Molecular Weight (g/mol)
82 12,700
156 28,500
The tensile strength and number-average molecular weight for two polyethylene materials given above.
Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.
SOLUTION:
We know that :
[tex]T_S = T_{S \infty} - \dfrac{A}{M_n}[/tex]
where;
[tex]T_S[/tex] = Tensile Strength
[tex]T_{S \infty}[/tex] = Tensile Strength (Infinity)
[tex]M_n[/tex] = Number- Average Molecular Weight (g/mol)
SO;
[tex]82= T_{S \infty} - \dfrac{A}{12700} ---- (1)[/tex]
[tex]156= T_{S \infty} - \dfrac{A}{28500} ---- (2)[/tex]
From equation (1) ; collecting the like terms; we have :
[tex]T_{S \infty} =82+ \dfrac{A}{12700}[/tex]
From equation (2) ; we have:
[tex]T_{S \infty} =156+ \dfrac{A}{28500}[/tex]
So; [tex]T_{S \infty} = T_{S \infty}[/tex]
Then;
[tex]T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}[/tex]
Solving by L.C.M
[tex]\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}[/tex]
[tex]\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}[/tex]
By cross multiplying ; we have:
[tex]({4446000 + A})* {12700} ={28500} *({1041400 + A})[/tex]
[tex](5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)[/tex]
Collecting like terms ; we have
[tex](5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)[/tex]
[tex]2.67843*10^{10} = 15800 \ A[/tex]
Dividing both sides by 15800:
[tex]\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}[/tex]
A = 1695208.861
From equation (1);
[tex]82= T_{S \infty} - \dfrac{A}{12700} ---- (1)[/tex]
Replacing A = 1695208.861 in the above equation; we have:
[tex]82= T_{S \infty} - \dfrac{1695208.861}{12700}[/tex]
[tex]T_{S \infty}= 82 + \dfrac{1695208.861}{12700}[/tex]
[tex]T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}[/tex]
[tex]T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}[/tex]
[tex]T_{S \infty}= \dfrac{2736608.861 }{12700}[/tex]
[tex]\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}[/tex]
From equation(2);
[tex]156= T_{S \infty} - \dfrac{A}{28500} ---- (2)[/tex]
Replacing A = 1695208.861 in the above equation; we have:
[tex]156= T_{S \infty} - \dfrac{1695208.861}{28500}[/tex]
[tex]T_{S \infty}= 156 + \dfrac{1695208.861}{28500}[/tex]
[tex]T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}[/tex]
[tex]T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}[/tex]
[tex]T_{S \infty}= \dfrac{6141208.861}{28500}[/tex]
[tex]\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}[/tex]
We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.
If the Tensile Strength (MPa) is 82 MPa
Definitely the average molecular weight will be = 12,700 g/mol
If the Tensile Strength (MPa) is 156 MPa
Definitely the average molecular weight will be = 28,500 g/mol
But;
Let us assume that the Tensile Strength (MPa) = 181 MPa for example.
Using the same formula:
[tex]T_S = T_{S \infty} - \dfrac{A}{M_n}[/tex]
Then:
[tex]181 = 215.481- \dfrac{1695208.861 }{M_n}[/tex]
Collecting like terms ; we have:
[tex]\dfrac{1695208.861 }{M_n} = 215.481- 181[/tex]
[tex]\dfrac{1695208.861 }{M_n} =34.481[/tex]
1695208.861= 34.481 [tex]M_n[/tex]
Dividing both sides by 34.481; we have:
[tex]M_n[/tex] = [tex]\dfrac{1695208.861}{34.481}[/tex]
[tex]\mathbf{M_n = 49163.56431 \ g/mol }[/tex]
