Work required to concentrate a dilute solution

[staceybiomed]

I am not really good at using the specials characters, etc so please forgive me!

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 = [tex]\Sigma(x_i * $$\gamma$$)[/tex]
where $$\gamma$$ is activity coefficient

ideal:
G = [tex]\Sigma(x_i)[/tex] + RT * [tex]\Sigma(x_i*ln x_i)[/tex]

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... :-)

 

[staceybiomed]

I am still looking for help on this question...please help me!

 

[Mene]

Hello,

Thank you for your question. I am happy to help you with this problem.

To begin, let's define some terms and equations that will be useful in solving this problem. The work required to concentrate a solution can be calculated using the following equation:

W = ΔG = ΔH - TΔS

Where W is the work, ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is the temperature, and ΔS is the change in entropy.

In this case, we are dealing with a non-ideal solution, so we will need to use the following equation to calculate Gibbs free energy:

ΔG = Σ(xi * γ)

Where xi is the mole fraction of each component and γ is the activity coefficient.

Now, let's apply this to the problem at hand. We are starting with 1 kg of a 10 wt % sucrose solution, which means we have 0.100 kg of sucrose and 0.900 kg of water. We want to end up with a 30 wt % sucrose solution, which means we will have 0.300 kg of sucrose and 0.700 kg of water.

Using the given information, we can calculate the mole fraction of sucrose and water in both the initial and final solutions:

Initial solution:
Mole fraction of sucrose = 0.100 kg / 0.3249 mol = 0.308
Mole fraction of water = 0.900 kg / 0.3249 mol = 2.769

Final solution:
Mole fraction of sucrose = 0.300 kg / 1.253 mol = 0.239
Mole fraction of water = 0.700 kg / 1.253 mol = 0.560

Now, we need to calculate the activity coefficients for each component at both concentrations. This can be done using the following equations:

ln γ = -A ln(molality) + B molality + C
Where A, B, and C are constants that can be found in tables for different substances.

For sucrose:
Initial concentration:
ln γ = -1.044 ln(0.3249) + 0.006 molality + 0.048 = 0.021
Final concentration:
ln γ = -1.044 ln(1.253) + 0.