CHM 3410 - Physical Chemistry 1
Second Hour Exam
October 19, 2012
There are five problems on the exam. Do all of the problems. Show your work.
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R = 0.08206 L.atm/mole.K NA = 6.022 x 1023
R = 0.08314 L.bar/mole.K 1 L.atm = 101.3 J
R = 8.314 J/mole.K 1 atm = 1.013 bar = 1.013 x 105 N/m2
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1. (16 points) The following question involves the one of the Maxwell relationships.
a) Beginning with the relationship
(¶V/¶T)p = - (¶S/¶p)T (1.1)
show that the change in entropy for a reversible isothermal process where pressure is changed from an initial value pi to a final value pf is given by the expression
DS = - òif (¶V/¶T)p dp (1.2)
b) A particular substance obeys the following equation of state
V = a + bT – cp (1.3)
T
where a, b, and c are constants.
Find an expression for DS when the pressure the above substance is changed isothermally and reversibly from an initial value pi to a final value pf.
2. (12 points) The Clausius-Clapeyron equation can be modified to take into account the temperature dependence of DHvap, the enthalpy of vaporization. One equation that is often used is
ln(p) = a - (b/T) - c ln(T) (2.1)
where a, b, and c are constants determined by fitting equn 2.1 to experimental data.
Starting with equn 2.1 find an expression for DHvap.
3. (20 points) Consider a solution of a nonvolatile solute dissolved in water, with a vapor phase above the solution.
a) How many degrees of freedom does such a system possess? Justify your answer.
b) A solution is prepared by dissolving 7.548 g of a nonvolatile solute in 250.0 g of water. The vapor pressure of this solution, measured at T = 20.0 °C, is p = 2.3018 kPa. The vapor pressure of pure water at this temperature is pH2O* = 2.3149 kPa. What is the molecular mass of the solute?
4. (16 points) Chloroform (C = CHCl3) and ethanol (E = C2H5OH) are miscible liquids.
a) A solution is prepared by mixing together 1.000 moles of chloroform and 0.250 moles of ethanol at T = 35.0 °C. What is DGmix for this process? Assume ideal mixing.
b) Chloroform and ethanol do not in fact form an ideal solution. The experimental values for the activities of chloroform and ethanol in the above solution are aC = 0.894 and aE = 0.408, where Raoult’s law defines ideal behavior for both liquids. Based on this information, find the actual value for DGmix for the above mixing process, and gC, the activity coefficient for chloroform.
5. (36 points) Consider the mixture of two volatile liquids A and B, whose phase diagram is given below (at a constant pressure p = 1.000 atm). Based on the phase diagram and other information in the problem, answer the following questions.
a) What is the normal boiling point for liquid A?
b) Describe what happens when the temperature of 1.00 mole of a solution whose mole fraction A in the system is ZA = 0.650 (vertical dashed line) is increased from an initial value Ti = 50. °C to a final value Tf = 100. °C in a closed system. Clearly indicate what phases and how many phases are present, and when these values change.
c) Consider a system with 1.000 total number of moles of substance at a temperature T = 68.5 °C (horizontal dashed line). Points A, B, and C are located at 0.368, 0.650, and 0.733 respectively. Find the total number of moles of liquid in the system for these conditions.
d) At T = 68.5 °C the vapor pressure of pure A is pA* = 538. torr. What are the numerical values for the activity (aA) and activity coefficient (gA) for A in the solution phase in part c of this problem?
e) What is DH°vap(A), the enthalpy of vaporization for liquid A?
Solutions.
1) a) Since (¶V/¶T)p = - (¶S/¶p)T , then for an isothermal process
- dS = (¶V/¶T)p
dp
Multiplying both sides by – dp, we get
dS = - (¶V/¶T)p dp
If we integrate from initial state i to final state f for an isothermal process, we get
òif dS = - òif (¶V/¶T)p dp
or, finally
DS = - òif (¶V/¶T)p dp
b) Since V = a + bT – (cp/T)
then (¶V/¶T)p = ¶/¶T)p [ a + bT – (cp/T) ] = b + (cp/T2)
DS = - òif [ b + (cp/T2) ]dp = - [ bp + (cp2/2T2) ]if
= - [ b(pf – pi) + (c/2T2) (pf2 – pi2) ]
2) From the Clausius-Clapeyron equation
d ln(p) = DHvap
dT RT2
and so DHvap = RT2 (d ln(p)/dT)
DHvap = RT2 d/dT [a – (b/T) - c ln(T) ]
= RT2 [ (b/T2) - (c/T) ] = R (b – cT)
3) a) C = 2, P = 2, and so F = 2 + 2 – 2 = 2
b) The formula for vapor pressure lowering is
Dp = pA* - pA = XBpA*
and so XB = Dp = (2.3149 kPa – 2.3018 kPa) = 0.00566
pA* (2.3149 kPa)
Since XB = nB @ nB
(nA + nB) nA
and so nB @ XB nA
But nA = 250.0 g 1 mol = 13.881 mol
18.01 g
and so nB = (0.00566)(13.881 mol) = 0.07857 mol
MB = 7.548 g = 96.1 g/mol
0.07857 mol
4) a) For ideal mixing
DGmix = nRT [ XC ln(XC) + XE ln(XE) ]
n = nC + nE = (1.000 mol) + (0.250 mol) = 1.250 mol
XC = nC = 1.000 mol = 0.800
n 1.250 mol
XE – 1 – XC = 1 – 0.800 = 0.200
DGmix = (1.250 mol) (8.3145 J/mol.K) (308.15 K) [ (0.800) ln(0.800) + (0.200) ln(0.200) ]
= - 1603. J
b) For a real solution, the expression for DGmix is
DGmix = nRT [ XC ln(aC) + XE ln(aE) ]
Since we are given the activities, then
DGmix = (1.250 mol) (8.3145 J/mol.K) (308.15 K) [ (0.800) ln(0.894) + (0.200) ln(0.408) ]
= - 861. J
Also, since aC = gC XC
then gC = aC/XC = 0.894/0.800 = 1.118
5) a) The normal boiling point for liquid A is 78. °C.
b) Initially there is a single liquid phase present in the system. At T = 67. °C, the liquid begins to boil, and there are two phases, a liquid phase and a gas phase. We remain in the two phase region until T = 73. °C, at which point all of the liquid has been converted into vapor. Above 73. °C there is only one phase present in the system, a vapor phase.
c) By the lever rule
ng = XA – ZA = (0.733 – 0.650) = 0.294
n ZA – YA (0.650 – 0.368)
So ng = (0.294) n
But n = n + ng = n + (0.294) n = 1.294 n
n = n/(1.294) = (1.000 mol)/(1.294) = 0.773 mol
d) aA = pA/pA*
But pA = YA p = (0.368) (760. torr) = 279.7 torr
aA = (279.7 torr)/(538. torr) = 0.520
aA = gA XA
gA = aA/XA = (0.520/(0.733) = 0.709
e) We may find the enthalpy of vaporization using the Clausius-Clapeyron equation in the form
ln(p2/p1) = - (DH°vap/R) { (1/T2) – (1/T1) }
DH°vap = - R ln(p2/p1)
{ (1/T2) – (1/T1) }
We now need the vapor pressure of pure A at two different temperatures. We can take
T1 = 68.5 °C, p1 = 538. torr (vapor pressure at this temperature given in the problem)
T2 = 78. °C, p2 = 760. torr (normal boiling point, from the phase diagram)
DH°vap = - (8.3145 J/mol.K) ln(760./538.) = 36.3 kJ/mol
[ (1/351.15 K) – (1/341.65 K) ]