Question

A system is given by two compartments separated by a thin membrane, and solute moves from one compartment to the other by diffusion. The volume of compartment 1 is 0.03 cm^3 and compartment 2 is 0.01 cm^3. Suppose 50 M of solute is dumped into compartment 2, and the concentration response for compartment 1 is c1(t)= 1250(1- e^(-0.01t) u(t)  M/cm^3 . (a) Find the transfer rate. (b) Solve for the concentration in compartment 2.

A system is given by two compartments separated by a thin membrane, and solute moves from one compartment to the other by diffusion. The volume of compartment 1 is 0.03 cm3 and compartment 2 is 0.01 cm3. Suppose 50 M of solute is dumped into compartment 2, and the concentration response for compartment 1 is c(t) 1250 (1(t (a) Find the transfer rate. (b) Solve for the concentration in compartment 2.

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Answer #1

Here the transfer rate is net Number of moles passing the meberane in a second.This net number of moles passing can be found by measuring the change in the number of moles of solute in any of the compartments.As we are given the concentration function of compartment 1,

Concentration (M/cm3)x volume of compartment 1(cm3) =( Ci x 0.03 )M gives the number of moles.So the change in number of moles in this compartment with time an be found by differentiating the above with respect to time.

(cit x 0.03)0.03 x 1250 x [(1 - e-0011 x u(t) dt x ult dt

0.03 × 1250 × [a(t) ×- lt (1-e-0.01 t)) + (1-e-0.01t) ×-a(t)l lt0.03 x 1250 x [u(t) x (10.0le 00it) + (1 - e001t)u(t)) -0.01t

Transfer Rate (Num. of moles per second)0.01t(t)M/second 37.5t) (1 0.01e 001t) + (1 - e  

b)For finding c2(t),

We know that sum of number of moles of solute in both compartments will always be equal to 50M[As there is no removal or addition of further solute to the system]

Number of moles in compartment 1 = c1t x volume of compartment 1 = c1(t) x 0.03 M

Number of moles in compartment 2 = c2t x volume of compartment 2 = c2​​​​(t) x 0.01 M

Sum of both number of moles =50M

o(t) × 0.03 + c2(t) × 0.01 = 50

1250(1-e -0.01t )u(t) × 0.034 c2(t) × 0.01 = 50

50 1250(1 - e-0olt)0.03 0.01 M/cm

0.03 0.0î (1-e-0.01 t)M/crn3 g(t) 5000-1250

0.01t c2(t)- 5000 - 3750(1 - e)M/cm

At time t=0, concentration in compartment 2 = No of moles / volume = 50/0.01 =5000M/cm3

We can verify c2(t),by substitutuing t= 0, and we get 5000 .So verified.

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