thermodynamics workshop problems model answers 2010-2011 1
TRANSCRIPT
1
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)