CHM 3410 – Problem Set 4
Due date: Wednesday, October 6th
Do all of the following problems. Show your work.
1) The phase diagram for helium is given on page 142 of Atkins, and may be used to answer the following questions.
a) For each pair of phases, identify the phase that has the higher density.
liquid He-I or liquid He-II
solid hcp He or liquid He I
liquid He-I or gas phase He
b) Based on the phase diagram estimate the value for DH°vap for helium.
2) The following data are for argon, and may be of use in answering the questions below.
Normal melting point T = 83.80 K
Normal boiling point T = 87.29 K
Triple point T = 83.81 K, p = 0.680 atm.
DHsub = 7.74 kJ/mol
a) Which has a higher density, Ar(s) or Ar()? Justify your answer.
b) What is the value for DHvap, the enthalpy of vaporization, for argon.
c) Based on your answer to b and the data above, estimate the value for DHfus, the enthalpy of fusion, for argon.
d) Trouton’s rule (Atkins, page 105) is the observation that for nonpolar or slightly polar substances the entropy of vaporization is approximately equal to 85. J/mol.K. Trouton’s rule and the normal boiling point can be used to estimate the value for DHvap for a substance. Use Trouton’s rule and the data above to estimate the value for DHvap for argon, and compare the result to the value found in part b of this problem.
e) Give a sketch of the phase diagram for argon in the vicinity of the triple point. Give the phase diagram for pressures in the range 0 – 3 atm, and temperatures in the range 70. K – 100 K. (Hint – For the solid-gas and liquid-gas phase boundaries use the Clausius-Clapeyron equation to find values of p and T at several different temperatures).
3) The thermodynamic data in Table 2.8 of Atkins can be used to estimate the values for vapor pressure for solids. In cases where the vapor pressures are too small to measure experimentally, this represents one practical method for determining their values.
a) Estimate the value for the vapor pressure for each of the following substances at T = 298. K.
Cd (cadmium)
I2 (iodine)
Li (lithium)
Na (sodium)
b) For lithium, use your answer to find the equilibrium concentration of lithium atoms in the vapor phase in equilibrium with the metal. Give your answer in unis of atoms/cm3.
Exercises
4.4 b What is the maximum number of phases that can be in mutual equilibrium in a four-component system?
4.16b Calculate the melting point of ice under a pressure of 10. MPa. Assume that the density of ice under these conditions is approximately 0.915 g/cm3, and that of liquid water is 0.998 g/cm3.
5.2 b At 20 °C, the density of a 20 % by mass ethanol-water solution is 968.7 kg/m3. Given that the partial molar volume of ethanol in the solution is 52.2 cm3/mol, calculate the partial molar volume of water.
Problems
4.12 In a study of the vapor pressure of chloromethane, A. Bah and N. Dupont-Pavlovsky (J.Chem.Eng.Data 40, 869 (1995)) presented data for the vapor pressure over solid chloromethane at low temperatures. Some of that data is shown below
T(K) 145.94 147.96 149.93 151.94 153.97 154.94
p(Pa) 13.07 18.49 25.99 36.76 50.86 59.56
Estimate the standard enthalpy of sublimation of chloromethane at T = 150.0 K. (Take the molar volume of the vapor to be that of a perfect gas, and that of the solid to be negligible).
5.4 The densities of aqueous solutions of copper II sulfate at 20 °C were measured as set out below. Determine and plot the partial molar volume of CuSO4 in the range of the measurements (m(CuSO4) is the mass of CuSO4 dissolved in 100.0 g of solution).
m(CuSO4)(g) 5.0 10.0 15.0 20.0
r(g/cm3) 1.051 1.107 1.167 1.230
NOTE: The density of pure water at T = 20 °C is r = 0.997 g/cm3.
Solutions.
1) a) To answer this question we use the fact that if dp/dT > the phase on the left has the higher density, while if dp/dT < 0 the phase on the right has the higher density.
liquid He-I and liquid He-II – dp/dT < 0, and so liquid He-I is more dense
solid hcp He and liquid He-I – dp/dT > 0, and so solid hcp He is more dense
liquid He-I and gas phase He – dp/dT > 0, and so liquid He-I is more dense
b) By the Clausius-Clapeyron equation
d(ln p)/d(1/T) = - DHvap/R, and so
DH°vap = - R [ d(ln p)/d(1/T) ] (This is true of we find the slope at p = 1 atm)
I obtain the slope from the following two points on the phase diagram
T(K) p(bar) 1/T (K-1) ln p
4.3 1.00 0.2325 0.000
2.5 0.10 0.4000 - 2.303
So d(ln p)/d(1/T) = (2.303)/(-0.1675 K-1) = - 13.75 K
and so DH°vap = - (8.314 J/mol.K) (- 13.75 K) = 110. J/mol
2) a) If we take the pressure and temperature at the triple point and normal fusion point, then
dp/dT = (1.00 atm – 0.680 atm)/(83.80 K – 83.81 K) < 0.
Since the slope is negative, that means that liquid argon is more dense than solid argon.
(Note: While the above is the correct conclusion for the data given, solid argon is almost certainly more dense than liquid argon. Unlike water, there is no good physical reason to expect the liquid phase to be more dense. Further, my (limited) checking of density data for argon found the solid has a density of ~ 1.6 g/cm3 and the liquid a density of ~ 1.4 g/cm3. The erroneous sign in the temperature difference above is likely due to obtaining data at the triple point and normal fusion point from different sources).
b) We can use the same procedure as in problem 1b above
T(K) p(atm) 1/T (K-1) ln p
83.81 0.680 0.011932 - 0.3856
87.29 1.000 0.011456 0.0000
So d(ln p)/d(1/T) = (- 0.3856)/(0.000476 K-1) = - 810.1 K
and so DH°vap = - (8.314 J/mol.K) (- 810.8 K) = 6735. J/mol
c) If we assume that DHfus + DHvap = DHsub, then
DHfus = DHsub - DHvap = 7740. J/mol – 6735 J/mol = 1005. J/mol
d) Using Trouton’s rule we would estimate DHvap @ (85. J/mol.K) (87.29 K) = 7420. J/mol. This is about 10% higher than the actual value.
e) To do the sketch of the phase diagram we will begin at the triple point. We can assume the solid-liquid boundary is a straight vertical line (since the solid and liquid density are approximately the same). For the solid-gas and liquid-gas boundaries we can use a form of the Clausius-Clapeyron equation
ln(p2/p1) = - (DHpt/R) { (1/T2) – (1/T1) }
where DHpt is either DHsub or DHvap. We will pick p1 and T1 to be the pressure and temperature at the triple point.
solid – gas (T < 83.81 K)
ln(p/0.680) = - (931.0 K-1) { (1/T2) – (1/83.81 K) }
liquid – gas (T > 83.81 K)
ln(p/0.680) = - (810.1 K-1) { (1/T2) – (1/83.81 K) }
A table of values for p and T, found using the above equations, is given below
T(K) p(atm) T(K) p(atm)
70.0 0.076 84.0 0.695
72.0 0.110 86.0 0.870
74.0 0.156 88.0 1.077
76.0 0.217 90.0 1.322
78.0 0.297 92.0 1.608
80.0 0.401 94.0 1.939
82.0 0.532 95.0 2.123
96.0 2.320
97.0 2.531
98.0 2.756
99.0 2.996
3) a) For the solid to gas phase transition
M(s) ® M(g) ln p = - DG°rxn/RT
where DG°rxn = DG°f(M(g)) - DG°f(M(s)), and p is in units of torr.
substance DG°rxn (kJ/mol) ln p p (bar)
Cd 77.41 - 31.24 2.7 x 10-14
I2 19.33 - 7.80 4.1 x 10-4
Li 126.66 - 51.12 6.3 x 10-23
Na 76.76 - 30.98 3.5 x 10-14
b) pV = nRT, and so
N/V = pNA/RT = (6.3 x 10-23 bar) (6.022 x 1023 mol-1) = 0.0015 atom/cm3. (about 1.5 atoms per liter of volume).
(0.08314 L.bar/mol.K) (298. K)
Exercise 4.4b F = 2 + C – P, and so P = 2 + C – F. The smallest value for F is F = 0, and so
P = 2 + 4 – 0 = 6 is the maximum number of phases that can be in mutual equilibrium.
Exercise 4.16b From the Clapeyron equation
dp/dT = (DH)/T(DV)
If we assume that dp/dT @ Dp/DT, then
DT = T(Dp)(DV)/DH
T = 273.15 K Dp = 10. MPa = 1. x 107 Pa. DHfus = 6008. J/mol
DV = Vl – Vs = [ (1 cm3/0.998 g) – (1 cm3/0.915 g) ] 18.0 g 1 m3 -= - 1.64 x 10-6 m3/mol
1. mol 106 cm3
And so DT = (273.15 K) (1. x 107 Pa) (- 1.64 x 10-6 m3/mol) = - 0.75 K
(6008. J/mol)
So the new melting point temperature will be T = (273.15 + (- 0.75)) K = 272.40 K
Exercise 5.2b
V = nW VW + nE VE
Assume 1000. cm3 for the volume. Then the number of grams of solution will be m = 968.7 g. Since the solution is 20% by mass ethanol
mass ethanol = (0.20) (968.7 g) = 193.74 g nE = 193.74 g (1 mol/46.07 g) = 4.205 mol
mass water = (0.80) (968.7 g) = 774.96 g nW = 774.96 g (1 mol/18.01 g) = 43.029 mol
Since we know V, nW, nE, and VE, we can find VW.
VW = ( V – nE VE) = (1000. cm3 – (4.205 mol) (52.2 cm3/mol) ) = 18.14 cm3/mol
nW 43.029 mol
Problem 4.12
We can find the enthalpy of sublimation from the Clausius-Clapeyron equation by plotting ln p vs 1/T.
T (K) p(Pa) 1/T (K-1) ln p
145.94 13.07 0.0068521 2.5703
147.96 18.49 0.0067586 2.9172
149.93 25.99 0.0066698 3.2577
151.94 36.76 0.0065815 3.6044
153.97 50.86 0.0064948 3.9291
154.94 59.56 0.0064541 4.0870
The data are plotted below.
Linear least square fitting of the data to a line gives slope = - 3821. K
intercept = 28.75
And so DHsub = - R (slope) = - (8.314 J/mol.K) (- 3821. K) = 31.77 kJ/mol
Problem 5.4
The partial molar volume of CuSO4 (VC) is given by the expression
VC = (¶V/¶nC)p,T,nW
It is therefore best to adjust the data so that we have a constant number of moles of water. A convenient value is 100.0 g water. For this mass, nW = (100.0 g) (1 mol/18.01 g) = 5.552 mol. Note M(CuSO4) = 159.60 g/mol
mC(g) mW(g) mC’(g) mW’(g) nC(mol) nW(mol) XC V (cm3)
0.000 100.000 0.000 100.000 0.00000 5.552 0.0000 100.301
5.000 95.000 5.263 100.000 0.03298 5.552 0.0059 100.155
10.000 90.000 11.111 100.000 0.06962 5.552 0.0124 100.371
15.000 85.000 17.647 100.000 0.11057 5.552 0.0195 100.811
20.000 80.000 25.000 100.000 0.15664 5.552 0.0274 101.626
The data are plotted below.
The data were fit to a second order polynomial using the POLY program, to give
V = 100.28 cm3 – (4.99 cm3/mol) nC + 87.29 (cm3/mol2) nC2
The molar volumes can then be calculated using VC = (¶V/¶nC)p,T,nW. The values are given below
nC (mol) XC VC (cm3/mol)
0.0000 0.0000 - 5.0
0.0330 0.0059 + 0.8
0.0696 0.0124 + 7.2
0.1106 0.0195 + 14.3
0.1566 0.0274 + 22.4
Notice that in the limit nC ® 0 the molar volume of copper sulfate is negative. Also notice that we would not have been aware of that if we had not included the data point corresponding to pure water. I don’t think Atkins noticed this.