Chem 420/523 Chemical Thermodynamics

Homework Assignment # 4

1. *The density of aqueous solutions of acetic acid varies with the mass fraction w2 of acetic acid at 298 Kas follows:

ρ (g cm−3) w2 ρ (g cm−3) w2 ρ (g cm−3) w20.9982 0.0 1.0488 0.4 1.0700 0.81.0125 0.1 1.0575 0.5 1.0661 0.91.0263 0.2 1.0642 0.6 1.0498 1.001.0384 0.3 1.0685 0.7

Plot an appropriate volume unit vs. x2, the mole fraction of acetic acid. Determine partial molar volumesof water and acetic acid at x2 = 0, 0.25, 0.50, 0.75, 1.00. Plot V 1 and V 2 as functions of x2.

Answer

We first calculate the volume of acetic acid solutions containing 1000 g of water. The mass fraction ofacetic acid is then given by (using the letter µ for mass because we want to reserve m for molality)

w2=µ2

µ1 + µ2

=µ2

1000 + µ2

The mass of acetic acid in a solution with a weight fraction w2 of acetic acid is given by

µ2 =1000w21−w2

and the total mass of the solution is, therefore, (µ2+1000) g. From this mass and the given density, wecan calculate the molality of acetic acid m2.

ρ w2 µ2 µsol x2 m = n2 V(g cm−3) g g mol kg−1 (cm3)0.9982 0.00 0.00 1000.00 0.0000 0.0000 1001.801.0125 0.10 111.11 1111.11 0.0323 1.8502 1097.391.0263 0.20 250.00 1250.00 0.0698 4.1629 1217.971.0384 0.30 428.57 1428.57 0.1139 7.1364 1375.741.0488 0.40 666.67 1666.67 0.1667 11.1011 1589.121.0575 0.50 1000.00 2000.00 0.2308 16.6517 1891.251.0642 0.60 1500.00 2500.00 0.3103 24.9775 2349.181.0685 0.70 2333.33 3333.33 0.4118 38.8539 3119.641.0700 0.80 4000.00 5000.00 0.5455 66.6067 4672.901.0661 0.90 9000.00 10000.00 0.7297 149.8651 9379.981.0498 1.00 -

1

Plotting the volume as a function of molefraction, we get the figure below.

Volume vs. mole fraction of acetic acid

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

x2

0

2000

4000

6000

8000

10000

V (c

m3 )

To evaluate the partial molar volume, it is better to plot volume vs. molality:

Volume vs. molality of acetic acid

0

2000

4000

6000

8000

10000

V (c

m3 )

0 50 100 150

m2 (mol kg�1)

We fit this data to a second order polynomial (even though the line is almost linear, we will allow forsmall variations of slope with molality). This results in

V = 990.373 + 54.433m2 + 0.0104m22.

From this relationship, we get

V 2=

µ∂V

∂n2

¶T,p,n1

=

µ∂V

∂m2

¶T,p,n1

=54.433 + 0.0208m2. (1)

2

Since the molality relationship does not apply when w2 = 1, this result is not valid for x2 = 1. However,since x2 = 1 corresponds to pure acetic acid, the partial molar volume at this mole fraction is simply themolar volume of pure acetic acid, given by

M2

ρ(w2 = 1)=60.054 g mol−1

1.0498 g cm−3 = 57.2052 cm3 mol−1

To find the partial molar volume of water, we calculate the volume of solutions containing 1.00 kg ofacetic acid (i.e., we treat the acetic acid as solvent and water as the solute).

ρ w1 µ1 µsol x2 m = n1 V(g cm−3) g g mol kg−1 (cm3)0.9982 1.00 - - - - -1.0125 0.90 9000.00 10000.00 0.0323 499.56 9876.541.0263 0.80 4000.00 5000.00 0.0698 222.02 4871.871.0384 0.70 2333.33 3333.33 0.1139 129.52 3210.071.0488 0.60 1500.00 2500.00 0.1667 83.26 2383.681.0575 0.50 1000.00 2000.00 0.2308 55.51 1891.251.0642 0.40 666.67 1666.67 0.3103 37.00 1566.121.0685 0.30 428.57 1428.57 0.4118 23.79 1336.991.0700 0.20 250.00 1250.00 0.5455 13.88 1168.221.0661 0.10 111.11 1111.11 0.7297 6.17 1042.221.0498 0.00 - 1000.00 1.0000 - 952.56

We plot volume vs. molality of water to evaluate the partial molar volume of water:

Volume vs. molality of water

0

2000

4000

6000

8000

10000

V(cm

3 )

0 100 200 300 400 500

m1 (mol kg�1)

Analysis similar to that above yields

V = 927.865 + 17.523m1 + 0.0008m21.

V 1 =

µ∂V

∂n1

¶T,p,n2

=

µ∂V

∂m1

¶T,p,n2

= 17.523 + 0.0016m. (2)

3

Again, for reasons mentioned above, this expression is not valid at x2 = 0. However, since this corre-sponds to pure water, V 1 = V ∗m,1 = 18.016 g mol−1/0.9982 g cm−3 = 18.049 cm3 mol−1.

Therefore, we tabulate the partial molar volumes calculated from Eqs. (1) and (2) below. Note that m2is calculated for solutions containing 1.00 kg of water (55.51 mol H2O) and m1 is calculated for solutionscontaining 1.00 kg of acetic acid (16.65 mol HAc).

x2 m1 m2 V 1(cm3 mol−1) V 2(cm3 mol−1)0 - 0.0 18.05 54.43

0.25 49.96 18.50 17.60 54.820.50 16.65 55.51 17.55 55.590.75 5.55 166.52 17.53 57.901.00 0.0 - 17.52 57.21

A plot of the partial molar volumes vs. mole fractions of acetic acid is given below.

Partial molar volumes vs. mole fraction of HAc

10

20

30

40

50

60

Parti

al m

olar

vol

umes

0.00 0.20 0.40 0.60 0.80 1.00

x2

Note: It should be clear to you by now that partial molar volumes are extremely sensitive to the methodof analysis used. You could easily justify using a linear fit for the V vs. mi curves in which case thepartial molar volumes would have no concentration dependence at all. Alternately, one could have useda model like

V = a+ bm1/2 + cm+ dm3/2 + em2

and obtained somewhat different results from those reported above. The main reason I avoided the modelwith the m1/2 term was to avoid the singularity in the derivative when m = 0.

It must also be mentioned that one of the authors of the text book, Dr. Boerio-Goates, suggests that westart with Eqs. (5.33) and (5.34), use the Redlich-Kister equation given in Eq. (5.40), and find the partialmolar volumes. We should expect those results to be qualitatively similar, but quantitatively different,from those obtained above. I will try to solve this problem using this method and provide you with theresults.

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2. At 298.15 K, the total volume of a solution formed from 1.00 kg of water mixed with m moles of MgSO4is given by

V/(cm3) = 1001.21 + 34.69(m− 0.07)2.The expression applies up to a molality of approximately m = 1.10 mol kg−1. What are the partialmolar volumes of (a) the MgSO4, (b) the water when m = 0.050 mol kg−1?

Answer

(a) The partial molar volume of MgSO4 is given by 2× 34.69(0.05− 0.07) = −1.3876

V 2 =

µ∂V

∂m

¶T,p,n1

= 2× 34.69(m− 0.07)

= −1.3876 cm3 mol−1

(b) The partial molar volume of water is determined from the relationship (with m = 0.05 mol kg−1 andn1 = 1000 g/18.016 g mol−1)

V=n1V 1 + n2V 2

V 1=(V − n2V 2)/n1=1

n1

h1001.21 + 34.69(m− 0.07)2 − 2× 34.69(m− 0.07)

i=

1

55.51

h1001.21 + 34.69(0.05− 0.07)2 − 2× 34.69(0.05− 0.07)

i=18.062 cm3 mol−1.

3. The apparent heat capacity of sucrose (2) in water (1) is given as a function of molality, m by theexpression

φCp/(J K−1 mol−1) = 633.88 + 4.728m− 0.1948m2

with C∗p,1 = 75.40 J K−1 mol−1. (a) Derive expressions for Cp,1 and Cp,2. (b) Calculate the heat capacityof a solution consisting of 0.30 moles of sucrose and 600 g of water.

Answer

(a) Using Eq. (5.44), we get

Cp,2 =

µ∂Cp∂m

¶T,p,n1

= φCp +m

µ∂φCp∂m

¶T,p,n1

= 633.88 + 4.728m− 0.1948m2 +m (4.728− 2× 0.1948m)= 633.88 + 9.456m− 0.5844m2

From the Gibbs-Duhem equation we have

n1dCp,1 + n2dCp,2 = 0,

dCp,1 = −n2n1dCp,2

= −mn1(9.456− 2× 0.5844m)dm.

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Simplifying and integrating between the limits of 0 molality of solute (pure solvent) and the molality m,we get

Z Cp,1

C∗p,1dCp,1 = − 1

n1

Z m

0(9.456m− 1.1688m2)dm

Cp,1 −C∗p,1 = −1

n1

Ã9.456

m2

2− 1.1688m

3

3

!

from which we get

Cp,1 = C∗p,1 −

1

n1

³4.728m2 − 0.3896m3

´(b) The molality of the solution is given by

m=0.30 mol600 g

× 1000 g kg−1

=0.50 mol kg−1

The heat capacity of the solution is calculated as (with n1 = 55.51 mol)

Cp=mφCp + n1C∗p,1

=636.2 + 55.51× 75.40=4821.7 J K−1

4. The following data are given for aqueous solutions of calcium chloride solutions at 20◦C:

CaCl2 Density CaCl2 Density% by weight (g cm−3) % by weight (g cm−3)

0 0.9982 20 1.17754 1.01316 25 1.22848 1.0659 30 1.281612 1.1015 35 1.337316 1.1386 40 1.3957

(a) Calculate the volume of the solutions containing 1.00 kg of water.(b) Use a graphical procedure to determine partial molar volume of CaCl2 at molalities corresponding to 12,

20, 25, and 35 percent weight.(c) * Using a curve-fitting program (Mathcad or the freely available CurveExpert), fit the volume vs. molality

data found in part (b) to an expression like

V = a+ bm1/2 + cm+ dm3/2 + em2

and report the values and units of the constants a, b, c, d, and e.(d) * Calculate the partial molar volumes at the same compositions as in part (b) and compare them.(e) * Derive an expression for the partial molar volume of water from the results of part (c).

Answer

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(a) The volume and molality data for solutions containing 1.00 kg of water are tabulated below. Molar massof CaCl2 is 110.978 g mol−1.

CaCl2 Density Mass of Solute Mass of solution m V% by weight (g cm−3) (g) (g) (mol kg−1) (cm3)

0 0.9982 0.00 1000.00 0.0000 1001.804 1.01316 41.67 1041.67 0.3755 1028.148 1.0659 86.96 1086.96 0.7835 1019.7612 1.1015 136.36 1136.36 1.3387 1031.6516 1.1386 190.48 1190.48 1.7163 1045.5620 1.1775 250.00 1250.00 2.2527 1061.5725 1.2284 333.33 1333.33 3.0036 1085.4230 1.2816 428.57 1428.57 3.8618 1114.6835 1.3373 538.46 1538.46 4.8520 1150.4240 1.3957 666.67 1666.67 6.0072 1194.14

(b) A plot of the volume vs. molality is given below.

0 1 2 3 4 5 6 7

m (mol kg�1)

Question 4 (b), (c)

1000

1050

1100

1150

1200

V (c

m3 )

The line is almost perfectly linear beyond the third data point, which corresponds to a molality of 1.2287mol kg−1, or 12 % by weight CaCl2. Therefore, we estimate the derivatives by central difference:

V 2 =

µ∂V

∂mi

¶T,p,n1

=1

2

µVi − Vi−1mi −mi−1 +

Vi+1 − Vimi+1 −mi

¶.

The partial molar volume of water is found from

V = n1V 1 + n2V 2.

The results are tabulated below. The number of moles of solute, n1, is 55.51 in all cases, correspondingto 1000.00 g of water.

Weight % m V 2 V 112 1.2287 27.62 17.9720 2.2527 30.81 17.8725 3.0036 32.93 17.7735 4.8250 36.97 17.49

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(c) * The curve shown in the figure above is the result of fitting the volume-molality data tabulated in part(a) to the model

V = a+ bm1/2 + cm+ dm3/2 + em2.

The result of the fit are a = 1002.312, b = 87.29322, c = −139.753, d = 95.91172, e = −16.5161.(d) * The partial molar volumes of the solute, V 2, are found by differentiation of the model above. The

partial molar volume of the solvent is found as in part (b). The results are tabulated below.

Weight % m V 2 V 112 1.2287 18.51 18.1820 2.2527 30.85 17.8725 3.0036 35.55 17.6335 4.8250 36.69 17.52

Except for the case of the 12 % by weight solution, the numerical derivatives found in part (b) and theanalytical derivatives above are in good agreement. The model is clearly deviating from the data in thevicinity of the first concentration considered and, therefore, we would expect the model�s prediction tobe inaccurate at that value.

(e) The expression for the partial molar volume of the solute is obtained directly from differentiating themodel with respect to molality:

V 2 =

µ∂V

∂m

¶T,p,n1

=87.29322

2m−1/2 − 139.753 + 3

2× 95.91172m1/2 − 2× 16.5161m

= 43.647m−1/2 − 139.753 + 143.87m1/2 − 33.032m.

From the Gibbs-Duhem equation, n1dV 1 + n2dV 2 = 0, and n1 = 55.51,we get

dV 1 = −n2n1dV 2

= −mn1

µ−12× 43.647m−3/2 + 1

2× 143.87m−1/2 − 33.032

¶= 0.39315m−1/2 + 1.2959m1/2 − 0.59506m

Integrating this expression from molality = 0 (V 1 = V ∗m,1) to m, and n1 = 55.51, we get

Z V 1

V ∗m,1dV 1 =

1

n1

Z m

0

³0.39315m−1/2 + 1.2959m1/2 − 0.59506m

´dm

V 1 = V∗m,1 −

0.3932

1/2m1/2 +

1.2959

3/2m3/2 − 0.59506

2m2

= 18.048− 0.7864m1/2 + 0.8639m3/2 − 0.29753m2.

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