Question

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.

Answer 1

`\mathbf{T_(S \infty ) \ \approx 215.481 \ MPa}`

`\mathbf{M_n = 49163.56431 \ g/mol }`

Step-by-step 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 :

`T_S = T_(S \infty) - (A)/(M_n)`

where;

`T_S` = Tensile Strength

`T_(S \infty)` = Tensile Strength (Infinity)

`M_n` = Number- Average Molecular Weight (g/mol)

SO;

`82= T_(S \infty) - (A)/(12700) ---- (1)`

`156= T_(S \infty) - (A)/(28500) ---- (2)`

From equation (1) ; collecting the like terms; we have :

`T_(S \infty) =82+ (A)/(12700)`

From equation (2) ; we have:

`T_(S \infty) =156+ (A)/(28500)`

So;
`T_(S \infty) = T_(S \infty)`

Then;

`T_(S \infty) =82+ (A)/(12700) =156+ (A)/(28500)`

Solving by L.C.M

`(82(12700) + A)/(12700) =(156(28500) + A)/(28500)`

`(1041400 + A)/(12700) =(4446000 + A)/(28500)`

By cross multiplying ; we have:

`({4446000 + A})* {12700} ={28500} *({1041400 + A})`

`(5.64642*10^(10) + 12700A) =(2.96799*10^(10)+ 28500A)`

Collecting like terms ; we have

`(5.64642*10^(10) - 2.96799*10^(10) ) =( 28500A- 12700A)`

`2.67843*10^(10) = 15800 \ A`

Dividing both sides by 15800:

`( 2.67843*10^(10) )/(15800) =(15800 \ A)/(15800)`

A = 1695208.861

From equation (1);

`82= T_(S \infty) - (A)/(12700) ---- (1)`

Replacing A = 1695208.861 in the above equation; we have:

`82= T_(S \infty) - (1695208.861)/(12700)`

`T_(S \infty)= 82 + (1695208.861)/(12700)`

`T_(S \infty)= (82(12700) +1695208.861 )/(12700)`

`T_(S \infty)= (1041400 +1695208.861 )/(12700)`

`T_(S \infty)= (2736608.861 )/(12700)`

`\mathbf{T_(S \infty ) \ \approx 215.481 \ MPa}`

From equation(2);

`156= T_(S \infty) - (A)/(28500) ---- (2)`

Replacing A = 1695208.861 in the above equation; we have:

`156= T_(S \infty) - (1695208.861)/(28500)`

`T_(S \infty)= 156 + (1695208.861)/(28500)`

`T_(S \infty)= (156(28500) +1695208.861 )/(28500)`

`T_(S \infty)= (4446000 +1695208.861 )/(28500)`

`T_(S \infty)= (6141208.861)/(28500)`

`\mathbf{T_(S \infty ) \ \approx 215.481 \ MPa}`

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:

`T_S = T_(S \infty) - (A)/(M_n)`

Then:

`181 = 215.481- (1695208.861 )/(M_n)`

Collecting like terms ; we have:

`(1695208.861 )/(M_n) = 215.481- 181`

`(1695208.861 )/(M_n) =34.481`

1695208.861= 34.481
`M_n`

Dividing both sides by 34.481; we have:

`M_n` =
`(1695208.861)/(34.481)`

`\mathbf{M_n = 49163.56431 \ g/mol }`