CHM 3400 – Fundamentals of Physical Chemistry
Second Hour Exam
March 6, 2013
There are five problems on the exam. Do all of the problems. Show your work
______________________________________________________________________________________
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
______________________________________________________________________________________
1. (20 points) The pressure of 1.000 mol of an ideal gas is changed from an initial value p i = 10.00 atm to
a final value pf = 1.000 atm, at a constant temperature T = 320.0 K. The constant pressure molar heat
capacity of the gas is Cp,m = 20.79 J/molK. q for the process is 2400.0 J.
What are w, U, H, Ssyst, Ssurr, Suniv, and G for the above process?
2. (25 points) The vapor pressure of cyclohexane (C6H12()), M = 84.16 g/mol) is p = 100.0 torr at T =
25.5 C, and p = 400.0 torr at T = 60.8 C. Based on this information find Hvap (the enthalpy of
vaporization), Svap, and Tvap (the normal boiling point) for cyclohexane.
3. (15 points) A solution of a small polymer molecule is formed by dissolving 1.28 g of the polymer in
25.00 g of water (H2O, M = 18.02 g/mol). The vapor pressure of pure water at T = 37.0 C is p*(H2O) =
47.067 torr. The vapor pressure of the polymer solution, measured at the same temperature, is lower than
the vapor pressure of pure water by 0.108 torr. Based on this information what is the molecular mass of the
polymer molecule?
4. (15 points) Two volatile liquids A and B form an ideal solution. The total pressure above a solution of
A and B, measured at T = 60.0 C, is ptotal = 486.5 torr. The vapor pressure of the pure liquids, measured at
the same temperature, are pA* = 617.4 torr and pB* = 388.7 torr. Based on this information find X A (mole
fraction of A in the liquid) and YA (mole fraction of A in the vapor above the liquid).
5. (25 points) The temperature vs mole fraction A phase diagram for two miscible liquids A and B is
given below and may be used to answer the following questions.
a) What are TA (the normal boiling point of pure A) and T B (the normal boiling point of pure B)?
b) Do A and B form any azeotropes? If your answer is yes, give the temperature and mole fraction
of A of each azeotrope that forms.
c) A solution is prepared at room temperature (T = 25. C) by mixing together 0.600 moles of A
and 0.400 moles of B. This solution is slowly heated in a closed system at a pressure 1.00 atm. At what
temperature does the solution first begin to boil? At what temperature is boiling complete?
d) For the solution in part c, at T = 70. C, the mole fraction of A at points A, B, and C are 0.521
at point A, 0.600 at point B, and 0.828 at point C. Based on this information, what is the total number of
moles of vapor present in the system?
Solutions.
1)
The process is isothermal and the gas is ideal, and so U = H = 0.
From the first law, U = q + w. Since U = 0, w = -q = - 2400. J
To find Ssyst, we need a reversible process with the same initial and final state as the unspecified
process in the problem. A reversible isothermal expansion from p i = 10.00 atm to pf = 1.00 atm is such a
process.
Ssyst = if (dq)rev/T. The process being used to find Ssyst is isothermal and reversible, and so we
can take T outside the integral, to get
Ssyst = (1/T) if (dq)rev = qrev/T
For an isothermal reversible expansion of an ideal gas, q rev = nRT ln(pi/pf)
And so Ssyst = (1/T) [ nRT ln(pi/pf) ] = nR ln(pi/pf) = (1.00 mol) (8.314 J/molK) ln(10.0/1.00)
= 19.14 J/K
Ssurr = - qsyst/T = - 2400.J/320.0 K = - 7.50 J/K (Note that moving 2400. J of heat from the
system to the surroundings is reversible from the point of view of the surroundings, even if the process
taking place is irreversible from the point of view of the system.)
Suniv = Ssyst + Ssurr = 19.14 J/K = ( - 7.50 J/K) = 11.64 J/K
Finally, for an isothermal process
G = H - T S = 0. - (320.0 K)(19.14 J/K) = - 6126. J
2)
a) We may find the enthalpy of vaporization using
ln(p2/p1) = - (Hvap/R) [ (1/T2) - (1/T1) ]
Hvap = -
R ln(p2/p1)
[ (1/T2) - (1/T1) ]
- (8.314 J/molK) ln(400/100)
[ (1/334.0 K) – (1/298.7 K) ]
= 32.57 kJ/mol
b) We can start with the same equation as before. Now we know Hvap. We can use one of the
previous data points for p1 and T1, and make p2 = 760. torr, the pressure at the normal boiling point. If we
solve for T2, we get
(1/T2) = (1/T1) - (R/Hvap) ln(p2/p1)
= (1/334.0 K ) - [ (8.314 J/molK)/(32570. J/mol) ] ln(760/400)
= 3.158 x 10-3 K-1
And so T2 = Tvap = (1/3.158 x 10-3 K-1) = 353.3 K = 80.1 C.
Since for a phase transition Gvap = Hvap - Tvap Svap = 0
Svap = Hvap/Tvap = (32570. J/mol)/(353.3 K) = 92.2 J/molK
3) For vapor pressure lowering
p = Xpoly p*H2O
So
Xpoly = p/p*H2O = (0.108)/(47.067) = 0.002295
But Xpoly = npoly/(npoly + nH2O). Since Xpoly << 1, we may say
Xpoly  npoly/nH2O
npoly = Xpoly nH2O = (0.002295) (25.00 g/18.02 g/mol) = 3.183 x 10 -3 mol polymer
Mpoly = mpoly/npoly = (1.28 g/3.183 x 10-3 mol) = 402. g/mol
4) For an ideal solution of two volatile liquids both components will obey Raoult's law, and so the partial
pressures of A and B are
p A = X A p A*
pB = XB pB* = (1 - XA) pB*
where we have written pB in terms of XA (using XA + XB = 1)
So
ptotal = pA + pB = XA pA* + (1 - XA) pB* = XA (pA* - pB*) + pB*
and so XA = ptotal - pB* = (486.5 – 388.7) = 0.4276
pA* - pB*
(617.4 – 388.7)
Then, from Raoult’s law
pA = XA pA* = (0.4276)(617.4 torr) = 264.0 torr
From Dalton's law, the mole fraction of A in the vapor phase is
YA =
5)
pA
= pA
pA + pB
ptotal
= 264.0 torr = 0.5427
486.5 torr
a) From the phase diagram,
TA = 50. C (normal boiling point of A)
TB = 60. C (normal boiling point of A)
b) There is one azeotrope, at T = 90. C, ZA = 0.39
c) Based on the composition of the system, ZA = 0.60. If we look at the line corresponding to Z A =
0.60, we see that boiling begins at T = 64. C and ends at T = 83. C (these are the points in the phase
diagram where the line corresponding to ZA = 0.600 crosses the boundaries between the one phase and two
phase regions).
d) Using the lever rule
ng = ZA - XA = (0.600 – 0.521) = 0.3465
n
YA – Z A
(0.828 – 0.600)
n = ng/0.3465 = 2.886 ng
ntotal = n + ng = 2.886 ng + ng = 3.886 ng = 1.000 mol
ng = (1.000 mol)/3.886 = 0.2573 mol in the gas phase.