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staceybiomed
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I am not really good at using the specials characters, etc so please forgive me!
a) Calculate the work required to concentrate a dilute sucrose solution. Assume that we begin with 10 wt % solution of sucrose in water and we want to end with a 30 wt % sucrose solution. Pick a basis of 1 kg of 10 wt % solution and express answer in Joules.
b) Compare your to answer to that which would be obtained if you assumed the sucrose-water solution to be ideal (i.e. where the activities are set equal to the mole fraction)
T of separation = 20C
MW of sucrose = 342.3 g/mol
For 10 wt% sucrose solution: molality (moles sucrose/kg H2O) = 0.3249, mole fraction H2O = 0.99419, Activity of water = 0.99393
For 30 wt% sucrose solution: molality = 1.253, mole fraction H2O = 0.99794, activity of water = 0.97596
non-ideal:
G = \Sigma(x<sub>i</sub> * \gamma)
where \gamma is activity coefficient
ideal:
G = \Sigma(x<sub>i</sub>) + RT * \Sigma(x<sub>i</sub>*ln x<sub>i</sub>)
I know that if we start with 1 kg of 10 wt % solution, I start with 0.100 kg of sucrose (also known as 0.292 moles) and the mass/moles of sucrose remains constant during concentration. It is just the amount of water that reduces (goes from 0.900 kg to 0.233 kg in 30 wt % solution). Also, the solution mass goes from 1 kg to 0.333 kg.
I think that if I can get Gibbs F.E. for non-ideal by just adding the (mole fraction * activity coefficients) at each concentration but not sure if that is the right way to start. Also, I'm not sure how to connect work to Gibbs free energy.
Please help!
P.S. I have 2 other problems that I need help on so stay tuned... :-)
Homework Statement
a) Calculate the work required to concentrate a dilute sucrose solution. Assume that we begin with 10 wt % solution of sucrose in water and we want to end with a 30 wt % sucrose solution. Pick a basis of 1 kg of 10 wt % solution and express answer in Joules.
b) Compare your to answer to that which would be obtained if you assumed the sucrose-water solution to be ideal (i.e. where the activities are set equal to the mole fraction)
T of separation = 20C
MW of sucrose = 342.3 g/mol
For 10 wt% sucrose solution: molality (moles sucrose/kg H2O) = 0.3249, mole fraction H2O = 0.99419, Activity of water = 0.99393
For 30 wt% sucrose solution: molality = 1.253, mole fraction H2O = 0.99794, activity of water = 0.97596
Homework Equations
non-ideal:
G = \Sigma(x<sub>i</sub> * \gamma)
where \gamma is activity coefficient
ideal:
G = \Sigma(x<sub>i</sub>) + RT * \Sigma(x<sub>i</sub>*ln x<sub>i</sub>)
The Attempt at a Solution
I know that if we start with 1 kg of 10 wt % solution, I start with 0.100 kg of sucrose (also known as 0.292 moles) and the mass/moles of sucrose remains constant during concentration. It is just the amount of water that reduces (goes from 0.900 kg to 0.233 kg in 30 wt % solution). Also, the solution mass goes from 1 kg to 0.333 kg.
I think that if I can get Gibbs F.E. for non-ideal by just adding the (mole fraction * activity coefficients) at each concentration but not sure if that is the right way to start. Also, I'm not sure how to connect work to Gibbs free energy.
Please help!
P.S. I have 2 other problems that I need help on so stay tuned... :-)