FLORIDA INTERNATIONAL UNIVERSITY



CHM 3410 – Problem Set 5

Due date: Wednesday, October 5th

Do all of the following problems. Show your work.

"If things were simple word would have gotten around." - Jacques Derrida

1) Based on the Clapeyron equation (equn 4.6 of Atkins) there is a general relationship between the slope of the boundary between two phases (in a p vs T phase diagram) and the relative density of the phases.

If dp/dT > 0 then the lower temperature phase has a larger density (smaller molar volume) than the higher temperature phase

If dp/dT < 0 then the lower temperature phase has a smaller density (larger molar volume) than the higher temperature phase.

Use this relationship to identify which of the following two phases in water has a higher density. Fig 4.9 of Atkins gives a phase diagram for water.

a) Ice I (the normal solid form of water) or liquid water

b) Ice VII or liquid water

c) Ice VII or Ice VIII

2) If we assume (Hvap is independent of temperature the following equation can be derived from the Clausius-Clapeyron equation

ln(p2/p1) = - ((H(vap/R) [ (1/T2) - (1/T1) ] (2.1)

where p1 is the vapor pressure of the liquid at T1 and p2 is the vapor pressure of the liquid at T2.

a) Use equn 2.1 and the data given below to estimate the value for the vapor pressure of water (H2O) at T = 50.0 (C. Give your final answer in units of torr.

b) Equation 2.1 is derived assuming that the value for (Hvap is independent of temperature. A better equation may be obtained by including a term to account for the dependence of (Hvap on temperature. Let

(Hvap = a + b (T - T() (2.2)

where a = (H(vap, the enthalpy of vaporization at T(, the normal boiling point of the liquid, and b = (Cp,m = Cp,m(g) - Cp,m(() is the difference between the constant pressure molar heat capacities of the gas and liquid.

Using equn 2.2 and the Clausius-Clapeyron equation

d ln (p) = (Hvap (2.3)

dT RT2

find an improved expression for ln(p2/p1) that takes into account the temperature dependence of (Hvap.

c) Use your answer in b and the data below to estimate the value for the vapor pressure of water at T = 50.0 (C. Give your final answer in units of torr.

d) Compare your estimated values for the vapor pressure of water found in part a and part c of this problem to the experimental value at T = 50.0 (C, pvap = 92.6 torr.

(H(vap(H2O) = 40656. J/mol T((H2O) = 100.00 (C (Cp,m(H2O) = - 41.71 J/mol.K

3) The following data are for argon.

Normal melting point T = 83.80 K

Normal boiling point T = 87.29 K

Triple point T = 83.81 K, p = 0.680 atm.

(Hsub = 7.74 kJ/mol (Hvap = 6.74 kJ/mol

Give 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).

4) "Synthetic air" is a mixture of nitrogen (N2) and oxygen (O2) with XN2 = 0.80. Find (Gmix, (Hmix, and (Smix when 1.000 mol of synthetic air is produced by mixing an appropriate amount of pure nitrogen and pure oxygen at p = 1.000 atm and T = 25.0 (C. Assume ideal behavior.

Also do the following from Atkins:

Exercises.

4.8 b The molar volume of a certain solid is 142.0 cm3/mol at 1.00 atm and 427.15 K, its melting temperature. The molar volume of the liquid at this temperature and pressure is 152.6 cm3/mol. At 1.2 MPa the melting temperature changes to 429.26 K. Find the enthalpy and entropy of fusion for the solid.

Problems.

5.2 The volume of an aqueous solution of NaCl at T = 25.0 (C was measured at a series of molalities b, and it was found that its volume fit the expression

V = 1003. + 16.62 x + 1.77 x3/2 + 0.12 x2

where V is the volume of a solution formed from 1.000 kg of water and added sodium chloride, and x = b/b(. Find the partial molar volume of water and sodium chloride for a solution where b = 0.100 mol/kg.

Solutions.

1) a) Boundary slopes left, so dp/dT < 0. Therefore the lower temperature phase (ice I) has a lower density than liquid water (the reason ice I floats in liquid water).

b) Boundary slopes right, so dp/dT > 0. Therefore the lower temperature phase (ice VII) has a higher density than liquid water.

c) Boundary slopes left, so dp/dT < 0. Therefore the lower temperature phase (ice VIII) has a lower density than ice VII.

2) a) If we assume (H(vap is independent of temperature, then

ln(p2/p1) = - ((H(vap/R) [ (1/T2) - (1/T1) ]

Let T1 = 100.0 (C = 373.15 K T2 = 50.0 (C = 323.15 K

p1 = 760. torr (H(vap = 40656. J/mol R = 8.3145 J/mol.K

Substitution into the above equation gives

ln(p2/p1) = - [(40656. J/mol)/(8.3145 J/mol.K)] [ (1/323.15 K) - (1/373.15 K) ] = - 2.0275

p2/p1 = exp(-2.0275) = 0.13166

p2 = (0.13166) p1 = (0.13166) (760 torr) = 100.1 torr

b) To take the temperature dependence of (H(vap into account we say

(H(vap = a + b (T - T() where T( = 100.0 (C = 373.15 K

Substitution into the Clausius-Clapeyron equation gives

d ln (p) = (Hvap = [ a + b (T - T() ] multiply both sides by dT and rearrange

dT RT2 RT2

d ln (p) = [ (a - bT() + bT ] dT integrate from p1, T1 to p2, T2

RT2

(p1p2 d ln (p) = (T1T2 [ (a - bT()/RT2 + b/RT ] dT

ln (p2/p1) = - [ (a - bT()/R] [ (1/T2) - (1/T1) ] + (b/R) ln(T2/T1)

c) Substituting the information from part a of the problem, and a = 40656. J/mol ; b = - 41.71 J/mol.K

ln(p2/p1) = - [ (40656. - (- 41.71) (373.15) ] J/mol [ (1/323.15 K) - (1/373.15 K) ]

(8.314 J/mol.K)

+ ( - 41.71 J/mol.K) ln(323.15/373/15)

( 8.3145 J/mol.K)

= - 2.8039 + 0.7217 = - 2.0822

p2/p1 = exp(- 2.0822) = 0.12465

p2 = (0.12465) p1 = (0.12465) (760 torr) = 94.7 torr

d) The experimental value for the vapor pressure at T = 50.0 (C is p = 92.6 torr. The result in part c (2.3 % higher) is a bit better than the result from part a (5.7 % higher). I'm not sure what is responsible for the remaining difference, but I believe the biggest remaining error is in the assumption that the vapor phase behaves like an ideal gas.

3) 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 expected to be approximately the same). For the solid-gas and liquid-gas boundaries we can use a form of the Clausius-Clapeyron equation

ln(p2/p1) = - ((Hpt/R) { (1/T2) – (1/T1) }

where (Hpt is either (Hsub or (Hvap. We will pick p1 and T1 to be the pressure and temperature at the triple point.

solid – gas (T < 83.81 K) ((H(sub/R = (7740. J/mol)/(8.314 J/mol.K) = 931.0 K-1)

ln(p/0.680) = - (931.0 K-1) { (1/T2) – (1/83.81 K) }

liquid – gas (T > 83.81 K) ((H(vap/R = (6740. J/mol)/(8.314 J/mol.K) = 810.7 K-1)

ln(p/0.680) = - (810.7 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

The phase diagram is given on the next page.

4) For ideal mixing

(Gmix = nRT [XN2 lnXN2 + XO2 lnXO2] = (1.00 mol) (8.3145 J/mol.K) (298.15 K) [0.8 ln(0.8) + 0.2 ln(0.2)]

= - 1240. J

(Smix = - nR [XN2 lnXN2 + XO2 lnXO2] = - (1.00 mol) (8.3145 J/mol.K) [0.8 ln(0.8) + 0.2 ln(0.2)]

= 4.161 J/K

(Hmix = 0.

Exercise 4.8b

From the Clapeyron equn.

dp = (H(fus

dT T (Vfus

We could bring dT to the right side and then integrate. However, if the temperature change is small (as is expected for moving along a solid-liquid phase boundary) we may say

dp ( (p = (H(fus

dT (T T(Vfus

(H(fus = T ((p) ((Vfus)

((T)

T = 428.2 K (we have taken the average temperature)

(p = 1.2 x 106 Pa - 1.01 x 105 Pa = 1.10 x 106 Pa

(Vfus = V( - Vs = 152.6 cm3/mol – 142.0 cm3/mol = 10.6 cm3/mol = 10.6 x 10-6 m3/mol

(T = 429.26 K – 427.15 K = 2.11 K

So (H(fus = T ((p) ((Vfus) = (428.2 K) (1.10 x 106 Pa) (10.6 x 10-6 m3/mol) = 2.37 kJ/mol

((T) (2.11 K)

(S(fus = (H(fus = 2370. J/mol = 5.53 J/mol.K

Tfus 428.2 K

We could argue that it would be better to use T(fus rather than the average temperature in the above calculations, and that we should separate temperature and integrate the Clapeyron equation, but neither of these introduces significant error compared to the other approximatins implied in the problem (that, for example, (Hfus and (Vfus are independent of temperature).

Problem 5.2

The molr volume of sodium chloride is

Vm(NaCl) = ((V/(nNaCl)p,T,nH2O = ((V/(x)p,T,nH2O ((x/(nNaCl)p,T,nH2O

But ((x/(nNaCl)p,T,nH2O = 1 (because for molality the mass of solvent , and therefore the moles of solvent, is constant)

So Vm(NaCl) = ((V/(x)p,T,nH2O = ((/(x) (1003. + 16.62 x + 1.77 x3/2 + 0.12 x2 )

= 16.62 + (3/2) (1.77) x1/2 + (2) 0.12 x

For x = 0.100, Vm(NaCl) = 17.484 cm3/mol

Since V = Vm(H2O) nH2O + Vm(NaCl) nNaCl

Vm(H2O) = V – Vm(NaCl) nNaCl

nH2O

But nH2O = 1000.0 g 1.0 mol = 55.494 mol

18.02 g

V = 1003. + 16.62 (0.100) + 1.77 (0.100)3/2 + 0.12 (0.100)2 = 1004.72 cm3

and so Vm(H2O) = 1004.72 cm3 – (17.484 cm3/mol) (0.100 mol) = 18.074 cm3/mol

(55.494 mol)

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