CHM 3400 Fundamentals of Physical Chemistry

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 ______. . 23 R = 0.08206 L atm/mole K NA = 6.022 x 10 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 pi = 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(ℓ)), 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 XA (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 TB (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 pi = 10.00 atm to pf = 1.00 atm is such a process.

f Ssyst = i (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

f Ssyst = (1/T) i (dq)rev = qrev/T

For an isothermal reversible expansion of an ideal gas, qrev = 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) - ( 8.314 J/mol  K) ln(400/100) = 32.57 kJ/mol [ (1/T2) - (1/T1) ] [ (1/334.0 K) – (1/298.7 K) ]

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

-3 -1 And so T2 = Tvap = (1/3.158 x 10 K ) = 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

-3 npoly = Xpoly nH2O = (0.002295) (25.00 g/18.02 g/mol) = 3.183 x 10 mol polymer

-3 Mpoly = mpoly/npoly = (1.28 g/3.183 x 10 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

pA = XA pA*

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 = pA = pA = 264.0 torr = 0.5427 pA + pB ptotal 486.5 torr

5) 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 ZA = 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

nℓ YA – ZA (0.828 – 0.600)

nℓ = ng/0.3465 = 2.886 ng

ntotal = nℓ + ng = 2.886 ng + ng = 3.886 ng = 1.000 mol

ng = (1.000 mol)/3.886 = 0.2573 mol in the gas phase.