thermodynamics workshop problems model answers 2010-2011 1

9
1 Thermodynamics Workshop Problems Model Answers 1. The partial molar volumes of acetone and chloroform in a mixture in which the mole fraction of CHCl 3 is 0.4693 are 74.166 cm 3 mol -1 and 80.235 cm 3 mol -1 , respectively. Show that the volume of a solution of mass 1.000 kg is 886.8 cm 3 . Let A = acetone, B = chloroform. Molar masses: M A = 58.077 g mol -1 ; M B = 119.368 g mol -1 . The total volume is given by, V = V A n A + V B n B (1) where V A , V B are the partial molar volumes, and , A B n n are the number of moles of each. The total mass is, T A A B B m Mn Mn (2) Equations (1) and (2) need to be rewritten in terms of mole fractions, ; A B A B A B A B n n x x n n n n Dividing both sides of (1) and (2) by A B n n , B B A A B A x V x V n n V (3) T A A B B A B m Mx Mx n n (4) Finally dividing (3) by (4), we can eliminate the number of moles, which we don’t know, 3 cm 8 . 886 B B A A B B A A T x M x M x V x V m V 2. At 25°C, the density of a 50 per cent by mass ethanol / water solution is 0.914 g cm -3 . Given that the partial molar volume of water in the solution is 17.4 cm 3 mol -1 , show that the partial molar volume of the ethanol is 56.3 cm 3 mol -1 . Let A = H 2 O; B = C 2 H 5 OH. Molar masses: M A = 18.016 g mol -1 , M B = 46.067 g mol -1 For convenience, let V = 1 cm 3 (since the partial molar volumes, by definition, do not depend on the total volume, we are free to choose any value. The mass of the mixture is 0.914 g. Hence the mass of A = mass of B = 0.457 g. The number of moles of each is,

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Thermodynamics Workshop Problems Model Answers

1. The partial molar volumes of acetone and chloroform in a mixture in which the mole

fraction of CHCl3 is 0.4693 are 74.166 cm3

mol-1

and 80.235 cm3

mol-1

, respectively.

Show that the volume of a solution of mass 1.000 kg is 886.8 cm3.

Let A = acetone, B = chloroform. Molar masses: MA = 58.077 g mol-1

; MB = 119.368 g mol-1

.

The total volume is given by,

V = VAnA + VBnB (1)

where VA, VB are the partial molar volumes, and ,A Bn n are the number of moles of each. The

total mass is,

T A A B Bm M n M n (2)

Equations (1) and (2) need to be rewritten in terms of mole fractions,

;A BA B

A B A B

n nx x

n n n n

Dividing both sides of (1) and (2) by A Bn n ,

BBAA

BA

xVxVnn

V

(3)

TA A B B

A B

mM x M x

n n

(4)

Finally dividing (3) by (4), we can eliminate the number of moles, which we don’t know,

3cm8.886

BBAA

BBAAT

xMxM

xVxVmV

2. At 25°C, the density of a 50 per cent by mass ethanol / water solution is 0.914 g cm-3

.

Given that the partial molar volume of water in the solution is 17.4 cm3 mol

-1, show that

the partial molar volume of the ethanol is 56.3 cm3 mol

-1.

Let A = H2O; B = C2H5OH. Molar masses: MA = 18.016 g mol-1

, MB = 46.067 g mol-1

For convenience, let V = 1 cm3 (since the partial molar volumes, by definition, do not depend

on the total volume, we are free to choose any value. The mass of the mixture is 0.914 g.

Hence the mass of A = mass of B = 0.457 g.

The number of moles of each is,

2

-1

-1

0.457 g0.02537 moles

18.016 gmol

0.457 g0.00992 moles

46.067 gmol

A

B

n

n

Using, BBAA nVnVV ,

13133

molcm3.56mole00992.0

)mole02537.0molcm4.17(cm1

B

AAB

n

nVVV

3. A piston-cylinder arrangement contains a mixture of NO2(g) and N2O4(g), which are in

chemical equilibrium at 298.15 K. If the piston is adjusted so that the total pressure

inside the cylinder is 2 bar, determine the partial pressures of NO2(g) and N2O4(g)

inside the cylinder, assuming both gases behave ideally. The equilibrium constant at T

= 298.15 K for the reaction 2 NO2(g) → N2O4(g) is Kp = 6.739.

The equilibrium constant at T = 298.15 K for the reaction

2 2 42NO (g) N O (g)

is Kp = 6.739. In terms of partial pressures,

739.6)( 2

2

42

NO

ONp

p

pK

Since the sum of the partial pressures is the total pressure,

422422 ONtotalNOONNOtotal pppppp

Hence,

739.6)( 2

42

42

ONtotal

ONp

pp

pK

Rearranging into the form of a quadratic equation,

0)12(

)2(

)(

22

22

2

4242

424242

4242

ONpONtotalptotalp

ONONONtotaltotalp

ONONtotalp

pKppKpK

pppppK

pppK

This is a quadratic equation of the form, 2 0ax bx c , where,

2

);12(;;42 totalptotalppON pKcpKbKapx

The solution is

2 42.624bar or 1.524bar

2

b b acx

a

3

Only the second of these is physically meaningful, since the partial pressure must be less than

the total pressure; hence,

bar476.0

bar524.1

2

42

NO

ON

p

p

4. Addition of 5.00 g of a compound to 250 g of naphthalene lowers the freezing point by

0.780 K. Calculate the molar mass of the compound. Note: it is necessary to find

additional data in order to answer this question.

Use the equation for the melting point depression to determine the mole fraction of

naphthalene,

0.0144)05.353(314.8

19090780.0780.0

780.0

22*

2*

RT

Hx

xH

RTT

fusB

Bfus

The mole fraction may be written as,

0144.0/517.128/250

/5

B

B

BA

BB

M

M

nn

nx

Rearranging gives,

1molg4.175250

17.128)55(

55

17.128

250

55

17.128

250

B

BB

BBB

B

BB

BB

x

xM

xMM

x

Mx

Mx

5. Vapour pressures of CCl4(l) are given below, show that the enthalpy of vaporisation is

vapH = 33.26 kJ mol-1

.

This question requires use of the equation relating pressure to temperature on the coexistence

curve between a liquid and a vapour,

4

*

11ln

* TTR

H

p

p vap

For p* and T* we choose one of the T/p data points (e.g. 253.55 K and 10 torr)

Hence in a plot of

*ln

p

pagainst

*

11

TT, the slope will be

R

Hvap .

From the graph, slope ~ -4000 K, hence, 1molkJ3.33 Hvap .

6. The solubility limit of anthracene (molar mass 178.2 g / mol) in 100 g of toluene (molar

mass 92.1 g / mol) is shown below, calculate the enthalpy of fusion of anthracene.

This question requires use of the equation relating the maximum mole fraction of solute that

can be dissolved in a solvent (i.e., the limit of solubility) to temperature,

TR

H

RT

H

TTR

Hx fusfusfus

A

111ln

**

First you need to calculate mole fractions from,

A AA

A A B B

m Mx

m M m M

where mA, mB are the mass of anthracene and toluene (mB = 100 g), and MA, MB are the molar

masses.

A plot of ln xA against 1/T will have a slopeR

Hfus . From the graph, slope ~ -3480 K,

hence, 1molkJ9.28 Hfus .

5

7. One kilogram of benzene is mixed with 0.5 kg of toluene at 298 K. Assuming ideal

behaviour

a.) Calculate mixG, mixS, mixH.

b.) What is the minimum value for the Gibbs free energy of mixing per mole of a two

component mixture that is behaving ideally, and what masses of benzene and toluene

would be required for such a mixture?

c.) What additional information would be required to work out mixG, mixS, mixH,

if the mixture did not behave ideally?

a.) ( .ln .ln )mix A A B BG RT n x n x

= -27.5 kJ

mixmix

GS

T

= 92.3 J K-1

0mixH

b.) minimum in free energy

for 0.5A Bx x

-11.72 kJ molmixG

benzene = 39.06 g

toluene = 46.07 g

c.) activities required

8. The following partial pressures were obtained for a mixture of benzene (B) and acetic

acid (A) at 50oC.

a.) Plot the vapour pressure curves for the individual components and the mixture and

sketch the lines for Raoult's Law and Henry's Law for each component.

b.) Deduce the activity coefficients for benzene at each composition.

c.) What is the physical origin of the deviation from Raoult's Law demonstrated in

these results?

Raoult’s Law: *B

BB

p

px (ideal), but

**BB

BB

B

BBB

px

p

p

px (non-ideal)

*Bp = 267.2 torr for benzene if we assume that the measurement at xB = 0.984 obeys Raoult’s

law.

6

Note that for both Raoult’s law and Henry’s law the pressure must equal zero when the mole

fraction of the respective component is zero.

Ax /A

p Torr /B

p Torr /totp Torr Bx B

0 0 267.2 267.2 1 1

0.0160 3.63 262.9 266.53 0.9840 1

0.0439 7.25 257.2 264.45 0.9561 1.007

0.0835 11.51 249.6 261.11 0.9165 1.019

0.1714 18.4 231.8 250.2 0.8286 1.047

0.3696 28.7 195.6 224.3 0.6304 1.161

0.6604 40.2 135.1 175.3 0.3396 1.489

0.8437 50.7 75.3 126 0.1563 1.803

0.9931 54.7 3.5 58.2 0.0069 1.899

1 55.1 0 55.1 0 n/a

The mixture exhibits positive deviations from Raoult’s Law, which suggests that the

molecules want to escape into the vapour more than Raoult’s Law predicts. This is an

indication that A-B interactions are less favourable than A-A, B-B interactions.

0.0 0.2 0.4 0.6 0.8 1.0

0

50

100

150

200

250

300

p(A)

p(B)

p(total)

p /

torr

xA

measured

Raoult (ideal)

Henry

7

9. The following vapour pressure results were obtained for a solution of sucrose in water

at 298 K.

molality / mol kg-1

0.01 0.1 0.2 0.4 0.75 1.25 1.75

vapour pressure / mmHg 23.751 23.712 23.668 23.579 23.414 23.156 22.873

activity, aA

freezing point depression

/ K

osmotic pressure / N m-2

height / m

mole Fraction, xB

a. Determine the activity of the solvent, aA.

b. Calculate the freezing point depression for each solution relative to pure water.

c. Calculate the osmotic pressure of each solution and the height of solution that could be

supported in an osmotic pressure experiment (usually these experiments are not done in

this way). On the basis of your results, suggest which measurements (v.p., f.p.d. or o.p.)

would be best to use for dilute solutions.

d. Plot the activity of the solvent, aA, against mole fraction, xA, and compare this with

what you would expect to find for ideal behaviour.

Data and assumptions:

vapour pressure of pure water = 23.756 mmHg,

fusH = 6010 J mol-1

, assume that each solution has a density of 1.0 g cm

-3,

assume that the activities used do not depend on temperature.

(a) Use *A

AA

p

pa and *

Ap 23.756 mmHg.

(b) Use

TTR

Ha

11)ln(

*fus

A with fusH = 6010 J mol-1

and T* = 273.15 K (0°C):

Rearrange to give: 1

fus

A*

)ln(1

H

aR

TT The freezing point depression is then: T = T* - T

(c) Use RTcB and = gh (We have not been given the osmotic virial

coefficient. Therefore, we use the van’t Hoff equation. Molality and molarity is

approximate the same for dilute aqueous solutions.)

(d) First, calculate mole fraction from

AA

A B

nx

n n.

For 1 litre: nA = 1000 g / 18.015 g mol-1

= 55.5 mol; nB = cB

The graph of Aa vs. Ax shows deviations from Aa = Ax .

molality/

mol kg-1

0.01 0.1 0.2 0.4 0.75 1.25 1.75

v.p./mm Hg 23.751 23.712 23.668 23.579 23.414 23.156 22.873

activity aA 0.99979 0.998148 0.996296 0.992549 0.985604 0.974743 0.96283

f.p.d./K 0.0217 0.1911 0.3822 0.7688 1.4854 2.6060 3.8351

o.p./N m-2

2.44E+04 2.44E+05 4.87E+05 9.75E+05 1.83E+06 3.05E+06 4.27E+06

height/m 2.48 24.85 49.69 99.38 186.35 310.58 434.81

xA 0.99982 0.99820 0.99641 0.99285 0.98668 0.97800 0.96946

8

0.96 0.97 0.98 0.99 1.000.96

0.97

0.98

0.99

1.00

aA

xA

10. EtOH and MeOH form very nearly ideal liquid mixtures. At 20C the vapour pressure

of EtOH is 44.5 mmHg and of MeOH is 88.7 mmHg. Calculate the partial pressures of

each above a solution containing 100 g of each liquid. Calculate also the vapour

composition.

For nearly ideal liquid mixtures, we can assume that Raoult’s Law is obeyed,

BAtotal ppp

*AAA pxp and

*BBB pxp

Ethanol:

mole fraction

100

no mols. ethanol46

100 100 no mols. methanol

46 32

= 0.410

Methanol:

mole fraction = 1 - 0.410 = 0.590

ethanol partial p = 0.410 44.5 mmHg

(Raoult’s Law) =18.2 mmHg

methanol partial p = 0.590 88.7 mmHg

(Raoult’s Law) = 52.3 mmHg

ptotal = 18.2 + 52.3 = 70.5 mmHg

methanol mol. fraction in vapour = 52.3

0.74270.5

ethanol mol. fraction in vapour = 18.2

0.25870.5

9

11. A new polypeptide antibiotic has been isolated and only 2 mg are available. By an

ultracentrifugal method, the molecular weight is approx 12,000 g mol-1

. It is decided

to check this by another method. Calculate the freezing point depression, lowering of

v.p. and osmotic pressure (at 20C) for a solution of 2 mg in 1 ml of water. Which

method would you advise on the basis of your calculation? Why?

Data: v.p. (water) 23.75 mm Hg, fusH = 6010 J mol-1

Use very dilute solution and assume ideality; e.g. dissolve in 1 ml of water, so that the mole

fraction of solute (xB) = 3 × 10-6

.

(a) Use Bfus

xH

RTT

2*

with with fusH = 6010 J mol-1

and T* = 273.15 K.

Hence, the freezing/melting point is depressed by 0.0003 K.

(b) BAA

A xxp

p 1

*

v.p. lowering =

mmHg101.7103)1( 5*6**** AAAAAAAA pxppxppp

(c)

K15.293molJ314.8m101

mol1067.1 1

36

7

RTV

nA

= 407 J m-3 = 407 N m-2 (1 J = 1 N m)

gh

m04.0mkg1000sm81.9

mN40732

2

h (1 N = 1 kg m s-2)

(easy to measure – definitely method of choice)