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 = kA₀t^1/2
k = 2 g/(hr x cm²)
A₀ = 10 cm²/hr^1/2

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?

Answer 1

(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*t^1/2

Thus, R = -dM/dt = k*A₀*t^1/2

Rearranging above eqn and integrating, we get :

[M] = -k*A₀*(2/3)*[t^3/2]

Putting limits and solving, we get :

t = 7.5^2/3 = 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