EXPERII,IENT
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II{STROCTOR'S
Tips
Equipment
Special
\
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A beaker tall enough to cover the inverted 10-mL graduatedcylinder with water is required. A l-liter
beaker is the best choice. Other size beakersmay work, depending on the height of the l0-ml graduated
cylinderused.
i
andTechniques
Procedures
Laboratory
t
The laboratory procedures are simple and straightforward involving heating a beaker and recording the
volume of air trapped in an inverted 10-mL graduatedcylinder as the temperaturechanges.The calculations
may be confusing, and students may need help in understandingthem. Studentsshould be able to recognize
the purposeof each of the steps in the calculationsand graph.
I
Disposal
t
Only. tap water is used; it can be poured down the drain.
OataandGalculations
Sample
r('c)
?(K)
V(mL)
Corrected
V(mL)
(mmHg)
Put.
P*"t".
lD Pr"t",
(mmHe)
i
ur
(K-l)
80.0
353
9.38
9.18
439
310
5.74
2.83x10-3
75.0
348
8.39
8.19
485
264
5.58
2.87x7O-3
.l'
i
70.0
343
7.59
7.39
530
2r9
5.39
2.92xl}-3
65.0
338
6.98
6.78
s69
180
5.19
2.96xl}-1
60.0
JJJ
6.50
6.30
604
145
4.98
3.00x10-3
55.0
328
6.15
5.95
629
t20
4.79
3.05x10-3
50.0
323
5.80
5.60
659
90
4.5
3.10x10-3
2.O
275
4.40
4.20
747
2
o.7
3.63x10-3
rj i l
:
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itt-
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Barometricpressure:744 mmHg
Differencein heightbetweenHrO in cylinderandHrO in beaker:63 mm
l. Correctvolumemeasurements
9.38mL -0.2m[- = 9.18mL
in thecylinder
2. Calculatethetotalpressure
Pcylinde,= Put,ro.ph"r,+ h(mmHrO)
x
= 744 mmHg + 63 mmHrO x
Pcyrinder
1.00mmHg
13.6mmHrO
1.00mmHg
13.6mmHrO
= 749 mmHg
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218
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3. Calculate the moles of trapped air by using the volume of air presentnear OoC
PV
nair=
i=
4.
749 mmHg x 0.00420L
l'83xloamoles
ffi=
For each temperaturebetween 50'C and 80"C, calculate the partial pressureof air in the cylinder:
Pr. at 80"C:
jn-,.R T
1.83x 104 mol x62.4 rnmHg.[./6ol.K x 353 K
_
=
= 439 mmHg
P^,-=
a'!r 4- v
0.00919L
5. Calculate the vapor pressureof water at eachtemperature:
P*r,".4t 80'C:
Pui, = 749 mmllg -439 mmHg = 310 mmHg
P*",", = Pcylinder'
6. Plot ln P*",", on the vertical axis versus llT on the horizontal axis. Draw the best fitting straight line
through the points. Determine the slope of the line, and calculate the value of Alluro of water.
Compare to the reported value for the enthalpyof vaporization of water.
Water
NaturalLogarithm
of VaporPrcssure(mmHg)versusReciprocal
of AbsoluteTempenture
s.80
5.63
/
\\-
\
\
5.45
\
5.28
ln
Pressur€ 5 . 1 0
mmHg
(
4.93
4.7s
4.58
4.40
2.80 2.85 2.90 2.95 3.00 3.05 3.10 3.1s
Reciprocalol AbsoluteTemperaturex l Og
K-r
4.52- 5.69
v^- Jv,t
=-=-4700K
S
---Elooe='L
xn- x,
3.10x 10-3- 2.85x 10-3
Mn^o- -(slope xR) = -(-4700 K x 8.314J/mol'K) = 39000 J/mol = 39 kJ/mol
Accepted value = 41 kJ/mol
Percent Error =
Accepted value - Calculatedvalue
4l -39
4l
Accepted value
\Y
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219
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EXPERIT.IEITT
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to lfiscussion
Answers
Questions
1. What is vaporpressureandwhy doesit changewith temperature.
Vaporpressureis the pressureexertedby a liquid in equilibriumwith its vapor.As the temperature
andmoreof themaremovingrapidly.It is possiblefor
risis thi kinetic energyof moleculesincreases
more moleculesto break away from the liquid phaseinto the vaporphase,so the pressureof the
vaporphaseincreases.
r,{t_
tj
t,,
What is enthalpyof vaPorization?
to changefrom the liqEnthalpyof vaporizationis the heatrequiredto causeonemole of a substance
uid to thevaporphase.
3 . The assumptionwas madethat the vapor pressureof water is negligibleat a temperatureclose to
andcommenton the validity of
zero.Find the actualvaporpressureof waterat your low temperature
theassumption.
f-
I
The low temperature obtained was 2oC. The vapor pressureof water at this temperature is 5 mmHg.
This is a small percentageof most of the values recorded, so the assumptionis fairly good.
1
4 . The assumption was also made that the slight changes in "h", the depth under the surface of the
water, will not significantly change the total pressure in the graduatedcylinder. Comment on the
validity of this assumption.
t
It requires a depth of 13.6 mm of water to change the pressure of the gas by 1 mmHg. Since the
atmoipheric pressure was 744 mmHg, the relative effect on the atmospheric pressure is extremely
small.
5 . Were your data values close to a straight line graph?
The values were close to a straight line.
6 . Write out the long "two-point" form of the Clausius-Clapeyron equation. Why does the graphical
method of analysis give a better value for the enthalpy of vaporization than does this form of the
equation using two temperature-vaporpressurevalues?
_li
r
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nE=+(+-')
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The graphical analysis averagesa larger number of values to give a more accurateanswer.
LabAssignment
to Preliminary
ouestions
Answers
l.
A graduatedcylinder containingsomeair is immersedin waterasshownin Figure 1 in the lab directions. The trelgtrtbetweenthe watersurfaceandthe top of the waterinsidethe graduatedcylinder is
106mm. Calculatethe correctionthatmustbe addedto the barometricpressureto find the total pressureof the gasesin the cYlinder.
106mmH,ox
I S= z.8ommHg
iFmmH"o
t
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2. The following experimental values are for ethanol..Graph the data as shown in the experimental
directions. Label the axes; draw the best straight line plot. Calculate thb value of Al/vapfor ethanol
from the slope.
Temperature
UT
("c)
Vapor Pressure
(mmHg)
Temperature
(K)
10.0
23.8
283
0.00353
3.170
15.0
32.0
288
0.00347
3.466
20.o
44.1
293
0.00341
3.786
25.O
59.0
298
0.00336
4.078
30.0
78.6
303
0.00330
4.364
Ethanol
NaturalLogarithmof VaporPressurc(mmHg)versusReciprocal
of Ab solute Temperature
3.60
e(E
\
3.50
\(.-
ln
Pressure
mmHg
\
3.45
I
3.40
\a
3.35
3.30
3.25
3.20
3.00 3.25 3.50
Reciprocal of Absolute Temperature x 1d
K-t
Slope:Ly I Lx = (4.36-3.17)l(3.30x 10-3-3.53x 10-3)= -5200K
Calculateheatof vaporization:
M,^o- -(slopex R) = {-5200K x 8.314J/mol.K) = 43000J/mol = 43 kJ/mol
{'..-!
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221
In P""p
(K-l)