Dilute Solutions Flashcards
When is a solution considered dilute?
If solute molecules are much less abundant than the solvent molecules. (Solutes never directly interact with other solutes)
Gibbs Free Energy of Pure Solvent
G = Nauo(T, P)
What parts of the change in Gibbs energy depends on N?
dG = dU + PdV - TdS
dU and PdV do not depend on N but depend only on how the B molecule added to solvent interacts with its members.
TdS depends on where the B molecule is placed. (number of choices proportional or N thus increasing multiplicity by factor proportional to N)
dS = kln(N) + terms independent of N
thus dG = f(T, P) - kTlnN (for adding one B molecule)
How does the change in Gibbs Free Energy change when adding two identical molecules?
There is a greater change in entropy since two molecules are identical thus interchanging does not result in distinct state. Thus we must divide multiplicity by 2 or subtract kln2 from the entropy.
dG = 2f(T, P) - 2kTlnN + kTln2
Gibbs Free Energy of adding B Molecules
G = Nau(T, P) + Nbf(T, P) - NbkTlnNa + NbkTlnNb- NbkT
valid for limit Nb «_space;Na (dilute)
NbkTlnNb- NbkT accounts for interchangeability of particle B.
dS = -kTlnA since we have freedom of choosing where to put particle B.
Solvent and Solute Chemical Potentials
Solvent:
ua = (dG/dN)T, P, Nb = uo(T, P) - NbkT/Na
Solute:
ub = (dG/dN)T, P, Na = f(T, P) + kTln(Nb/Na)
Adding more solute reduces chemical potential of A and increase chemical potential of B.
Molality
Number of moles of solute per kg of solvent.
Constant times ratio Nb/Na.
m = moles of solute / kilograms of solvent
New Solute Chemical Potential:
ub = uo(T, P) + kTlnmB
uo = chemical potential under standard condition mB = 1
Osmosis
Chemical potential of the solvent interacts the position is less than that of the pure solvent. Particles flow towards lower chemical potential so solvent molecules will flow from the pure solvent to solution.
Solution in equilibrium with liquid.
How can you stop osmotic flow?
Chemical potential of solvent must be the same on both sides of membrane.
uo(T, P1) = uo(T, P2) - NbkT/Na
P1 is pressure of pure solvent
P2 is pressure of solution
If not different, can approximate:
uo(T, P2) ~ uo(T, P1) + (P2 - P1) duo/dP
Then we can obtain:
(P2 - P1)duo/dP = NbkT/Na
Osmotic Pressure
the amount of force/pressure applied to a solution that prevents solvent from moving across a semipermeable membrane
(P2 - P1)duo/dP = NbkT/Na
duo/dP = V/N
Thus
(P2 - P1) V/Na = NbkT/Na
(P2 - P1) = NbkT/V = nBRT/V (Van’t Hoff’s Formula)
Boiling Point
Dilute solution in equilibrium with its gas phase. Gas contains no solute!
Equilibrium condition for solvent:
uA liq(T, P) = uA gas(T, P)
Rewrite as:
uo(T, P) - NbkT/Na = u gas(T, P)
uo is chemical potential of pure solvent
Want pure solvent to in equilibrium.
uo(T, Po) = u gas(T, Po)
- uo(T, P) - NbkT/Na = u gas(T, P) becomes
uo(T, Po) + (P - Po)duo/dP - NbkT/Na = u gas(T, Po) + (P- Po)dU gas / dP
- So then
(P- Po) (V/N)liq - NbkT/Na = (P-Po)(V/N)gas
Volume per particle in gas phase is kT/Po while volume per particle in liquid is negligible.
P - Po = -Nb/Na Po or P/Po = 1- Nb/Na (Raoult’s Law)
Temperature needed to maintain equilibrium in the presence of solute. (Pressure fixed)
uo(To, P) = u gas(To, P)
u(To, P) + (T - To)duo/dT - NbkT/Na = ugas(To, P) + (T - To)du gas/dT
dU/dT is minus the entropy of particle for that phase. (dG/dT = -S)
-(T - To) (S/N)liq- NbkT/Na = -(T-To)(S/N)gas
T - To = NbkTo^2/L = nBRTo^2/L
