Difference in chemical potential(mu) problem

[jamalm55]

Homework Statement

Given: 2 cups of hot coffee, each 250 mL, T=50 degrees celsius, 323.15K
1 cup is sweetened 1tsp (5cm^3) of sucrose (342.3 g/mol)
the other cup is sweetened with 1g saccharin (183.17 g/mol)
assume coffee is very weak so the sweetener is the only solute
density of H2O at 323.15K = 988.07 kb/m^3
density of sucrose = 1505.5 kg/m^3
what is the difference between the chemical potential of the hot water
in the two cups?

Homework Equations

chemical potential (mu) of any solution: μ=RTlna_i
a_i = γ_iχ_i
lna_A = γ_Aχ_A = -χ_B
A= solvent, B= solute

The Attempt at a Solution

So since its asking for Δμ, I put Δμ_sucrose-Δμ_saccharin which simplifies to Δμ=RT(lna_sucrose-lna_saccharin).

Then I found moles of each substance from the given info: n_H2O=13.71 mol
n_sucrose= 0.022 mol
n_saccharin= 5.46 mol

Then I found their mole fractions, χ_B,
χ_B=χ_sucrose= n(sucrose)/(n(H2O)+n(sucrose))= 1.60 x 10^-3
χ_B=χ_saccharin= n(saccharin)/(n(H2O)+n(saccharin)) = 0.285

And plugged into Δμ=RT(lna_sucrose-lna_saccharin).
but I'm not sure if I have to sub in ln(χ_suc) - ln(χ_saccharin) or just
-χ_sucrose - (-χ_saccharin)

I figured since lna_A = γ_Aχ_A = -χ_B, I do the latter.

Δμ=(8.314 J/molxK)(323.15K)((-1.60x10^-3)-(-0.285)) = 761.4 J/mol

I can't find the solution anywhere to check.

 

[nate808]

Thank you for your interesting question. I would like to provide a detailed explanation of the solution to your problem.

Firstly, let's define the chemical potential (μ) of a solution. It is the change in Gibbs free energy (ΔG) per mole of substance, and it is related to the concentration of the solute (χB) and the temperature (T) by the following equation:

μ = RTlnχB

where R is the gas constant (8.314 J/molxK).

In your problem, you are asked to find the difference in chemical potential between two cups of hot coffee, one sweetened with sucrose and the other with saccharin. This can be calculated by taking the difference between the chemical potentials of the two solutions:

Δμ = μsucrose - μsaccharin

To find μsucrose and μsaccharin, we need to calculate the mole fraction of each solute in the solution, using the given information about the moles of solute and solvent:

χsucrose = nsucrose / (nH2O + nsucrose) = 0.022 / (13.71 + 0.022) = 1.60 x 10^-3

χsaccharin = nsaccharin / (nH2O + nsaccharin) = 5.46 / (13.71 + 5.46) = 0.285

Now we can calculate the chemical potentials of each solution:

μsucrose = RTlnχsucrose = (8.314 J/molxK)(323.15K)ln(1.60 x 10^-3) = -4.81 J/mol

μsaccharin = RTlnχsaccharin = (8.314 J/molxK)(323.15K)ln(0.285) = -15.25 J/mol

Finally, we can find the difference between the chemical potentials:

Δμ = μsucrose - μsaccharin = (-4.81 J/mol) - (-15.25 J/mol) = 10.44 J/mol

Therefore, the difference in chemical potential between the two cups of coffee is 10.44 J/mol. This means that the cup of coffee sweetened with saccharin has a higher chemical potential than