MATSCI 204 Thermodynamics and Phase Equilibria Winter 2012 ...
MATSCI 204 Thermodynamics and Phase Equilibria Winter 2012 Problem Set #4 Due: Monday Feb. 11th Problem 1 (40 points) d (#gmix ) for an ideal binary solution. Sketch the behavior of dx Δgmx near the axis (i.e. x " 0 ) and comment on the ability of a material to stay perfectly pure. 2. Using this understanding, let's think about how this practically affects 'single' ! component phase diagrams. Assume we have a non-volatile (not evaporating) ! not crystallize out with the solvent upon freezing, so it is only solute which does present in the liquid phase. Explain how the presence of the solute affects the freezing and boiling points of the solvent phase? 3. Qualitatively show these effects by drawing a µ vs. T diagram for the solid, the pure liquid, the liquid solvent when in solution, and the vapor phases based on the information provided above. The graph should have 4 different functions for on the same plot. Label the arbitrary melting and boiling points for clarity. 4. Qualitatively draw how the unary (P,T) phase diagram would change upon the presence of such a solute which exists only in the liquid phase. What happens to the triple point? Your graph should have a phase diagram in the case of no solute present as well as a phase diagram with a solute present in the liquid phase only.
1. Calculate limx "0
Answer
!
1- From the notes, "gmix = RT [ x ln( x ) + (1# x ) ln(1# x )] therefore, $ x ' d ("gmix ) d (#gmix ) = RT ln& = $% . Δgmix decreases ) and limx "0 %1# x ( dx dx near the axis ! with an infinitely steep slope. Therefore, the purer a material, the stronger its tendency to dissolve impurities in solution. Because of the infinite slope of Δgmix, it is impossible to ! keep a material perfectly pure.
2- The presence of the solute decreases the chemical potential of the solvent because of the Δgmix being negative. In practical terms, this means that the liquid solvent is “stabilized” by the solute. Because the solvent freezes or boils as a pure phase, if the liquid is “stabilized” by being in solution (i.e. its chemical potential is decreased), this means that its temperature window where it is stable is wider: the melting temperature decreases and the boiling temperature increases. 3
Since we have a non-volatile solute, we assume it is only present in the liquid phase. Thus, the chemical potentials of the solid and vapor are not affected by the presence of a solute, only the chemical potential of the liquid phase is affected as shown in the plot.
4-
Problem 2 (25 points) I have a very large bottle of a water-ethanol mixture (I am expecting my group over for dinner) on my deck. The mixture is 15 mol% ethanol (I will not comment on my students’ drinking habits as this corresponds to ~44% v/v or ~77 proof…). I accidentally spill one mole of water in the mixture (the container is so large that its composition does not change appreciably). My students won’t be pleased about my being so clumsy but I want you to calculate (show me how you did it or you won’t get credit): 1- the volume change of the mixture (15 points) 2- the heat exchanged with the surroundings (magnitude and sign) (10 points) Data: molar volume of ethanol: 58 ml; molar volume of water: 18 ml
Answer The volume increase will be the partial molar volume of H2O at the solution composition stated in the problem (15% ethanol) and the heat exchanged will be the partial molar enthalpy of mixing of water in the solution at the same composition. Partial molar quantities are found at the intercepts of the tangents to the mixing quantities for xH2O=1. The excess volume is ~ -0.35 cm3 therefore the volume change is 18-0.35=17.65 cm3 Heat is released upon mixing (negative enthalpy) and it’s ~-775 J.
Problem 3 (25 points) Below is the unit cell volume of LixNO2 as a function of x. What is the partial molar volume of Li in this material at x=1?
Answer
Problem 4 (10 points) The vapor pressure of pure B at 1000K is 3x10-5 atm. B forms an ideal solution with A at 1000K. Its vapor pressure in solution will be: a. unchanged (3x10-5 atm) b. higher than 3x10-5 atm c. lower than 3x10-5 atm d. can’t say, it will depend on concentration Answer c
1. Calculate limx "0
Answer
!
1- From the notes, "gmix = RT [ x ln( x ) + (1# x ) ln(1# x )] therefore, $ x ' d ("gmix ) d (#gmix ) = RT ln& = $% . Δgmix decreases ) and limx "0 %1# x ( dx dx near the axis ! with an infinitely steep slope. Therefore, the purer a material, the stronger its tendency to dissolve impurities in solution. Because of the infinite slope of Δgmix, it is impossible to ! keep a material perfectly pure.
2- The presence of the solute decreases the chemical potential of the solvent because of the Δgmix being negative. In practical terms, this means that the liquid solvent is “stabilized” by the solute. Because the solvent freezes or boils as a pure phase, if the liquid is “stabilized” by being in solution (i.e. its chemical potential is decreased), this means that its temperature window where it is stable is wider: the melting temperature decreases and the boiling temperature increases. 3
Since we have a non-volatile solute, we assume it is only present in the liquid phase. Thus, the chemical potentials of the solid and vapor are not affected by the presence of a solute, only the chemical potential of the liquid phase is affected as shown in the plot.
4-
Problem 2 (25 points) I have a very large bottle of a water-ethanol mixture (I am expecting my group over for dinner) on my deck. The mixture is 15 mol% ethanol (I will not comment on my students’ drinking habits as this corresponds to ~44% v/v or ~77 proof…). I accidentally spill one mole of water in the mixture (the container is so large that its composition does not change appreciably). My students won’t be pleased about my being so clumsy but I want you to calculate (show me how you did it or you won’t get credit): 1- the volume change of the mixture (15 points) 2- the heat exchanged with the surroundings (magnitude and sign) (10 points) Data: molar volume of ethanol: 58 ml; molar volume of water: 18 ml
Answer The volume increase will be the partial molar volume of H2O at the solution composition stated in the problem (15% ethanol) and the heat exchanged will be the partial molar enthalpy of mixing of water in the solution at the same composition. Partial molar quantities are found at the intercepts of the tangents to the mixing quantities for xH2O=1. The excess volume is ~ -0.35 cm3 therefore the volume change is 18-0.35=17.65 cm3 Heat is released upon mixing (negative enthalpy) and it’s ~-775 J.
Problem 3 (25 points) Below is the unit cell volume of LixNO2 as a function of x. What is the partial molar volume of Li in this material at x=1?
Answer
Problem 4 (10 points) The vapor pressure of pure B at 1000K is 3x10-5 atm. B forms an ideal solution with A at 1000K. Its vapor pressure in solution will be: a. unchanged (3x10-5 atm) b. higher than 3x10-5 atm c. lower than 3x10-5 atm d. can’t say, it will depend on concentration Answer c
