a) You start with your modeling predictions. You assume that the polymer does no

Question

a) You start with your modeling predictions. You assume that the polymer does not degrade. Rather, you assume that the dissolution rate of the polymer is proportional to a constant k, and the surface area of the polymer, A.

dissolution rate = kA
k = 2 g/(hr x cm2)
A = 10 cm2

How long will it take for the 100 g polymer slab to dissolve completely in water?

b) Realizing that the first assumption is too simple, you try to model the dissolution rate as

dissolution rate = kA0t1/2
k = 2 g/(hr x cm2)
A0 = 10 cm2/hr1/2

How long will it take for the 100 g polymer slab to dissolve completely in water?

c) During the first hour the degradation rate increases linearly to a value of 10 g/hr, then plateaus and is a constant at 10 g/hr. Using this model for the dissolution rate from part a, you combine the dissolution and degradation terms into one model. You begin the trial with 100 g and stop when the mass is 20 g. Using the model, how long will it take for the polymer to reduce in mass from 100 g to 20 g in water?

Explanation / Answer

(a) Dissolution rate, R = k*A

k = 2 g/(h*cm2)

A = 10 cm2

Thus, R = 20 g/hr

Thus, from the rate we see that 20 g polymer dissolver in 1 hr

Thus, 100 g will dissolve in 1*(100/20) = 5 hrs

(b)

In this case, R = k*A0*t1/2

Thus, R = -dM/dt = k*A0*t1/2

Rearranging above eqn and integrating, we get :

[M] = -k*A0*(2/3)*[t3/2]

Putting limits and solving, we get :

t = 7.52/3 = 3.83 hrs

(c)

Now, two factors contribute to the disappearance of polymer, one is dissolution and other is degradation.

Rate due to dissolution is : k*A = 20 g/hr = 1/3 g/min

During first hour, rate due to degradation is : (10/60)*t g/min = (1/6)*t g/min

Thus disappearance rate for the first hour = -dM/dt = 1/3 + (1/6)*t g/min

Solving the equation again as above, we get :

[-M] = t2/12 + t/3

Now, putting limits for disapperance of mass from 100 g to 20 g, we get :

t2 + 4t - 960 = 0

Solving for +ve value of t, we get :

t = 28.98 mins

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