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Physikalisch-chemisches Praktikum, Departement Chemie, Universität Basel Experiment: Vapor pressure PC Praktikum SS / ×WS …………. Date of experiment (to be filled in by the assistant): ……..…………………. Chemistry Biology Nanotechnology × Pharmaceutical Sciences Last Name, First Group number Email address Points (to be filled out by the assistant) /6

Vapor pressure PC Praktikum SS

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Page 1: Vapor pressure PC Praktikum SS

Physikalisch-chemisches Praktikum,

Departement Chemie, Universität Basel

Experiment:

Vapor pressure

PC Praktikum □ SS / ×WS ………….

Date of experiment (to be filled in by the assistant): ……..………………….

□ Chemistry

□ Biology

□ Nanotechnology

× Pharmaceutical Sciences

Last Name, First

Group number Email address

Points (to be filled out by the

assistant)

/6

Page 2: Vapor pressure PC Praktikum SS

Date

2

Vapor pressure

Gruppe XXX

Vorname Name (1)

Vorname Name (2)

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Index 1.) Introduction .......................................................................................................................... 4

2) Experimental part ................................................................................................................... 6

3.) Results ................................................................................................................................... 7

3.1) Measurements of the orange liquid ............................................................................... 7

3.2) Measurements of the green liquid ................................................................................. 8

4.) Calculations ........................................................................................................................... 9

4.1) Calculation of with Clausius-Clapeyron equation .............................................. 9

4.1.1) of orange liquid....................................................................................... 10

4.1.2) of green liquid ......................................................................................... 10

4.2) Error calculation ............................................................................................................ 11

4.2.1) Orange liquid ........................................................................................................ 11

4.2.2) Green liquid .......................................................................................................... 12

5) Discussion ............................................................................................................................. 13

6) Literature .............................................................................................................................. 14

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1.) Introduction

Every liquid has its own specific vapor pressure as its basic. It is possible to identify an unknown solution by his vapour pressure. Therefore it can be exploit that the molecules in a closed system change from gas to liquid phase and the other way round to establish an equilibrium between the different phases. During a progressive evaporation process a lot more molecules change to gas phase aground of vaporisation and also the following condensation rate between gas to liq-uid rises. After a while a state of equilibrium forms. Further the vapor pressure is defined as pressure above the liquid, because gaseous molecules need more space, the pressure rises. Va-por pressure depends on the substance and is proportional to the temperature.

These conditions can be shown by the following illustration of the phase diagram, which is char-acteristical for each substance and with it, it is possible to recognize under which term a mole-cule is in gas, liquid or solid state. For the experiment the curve between the triple point and the critical point is of interest. On the triple point the 3 different states are in equilibrium with each other, there exist a triple of phases. The curve between the gas phase and fluidity is called the vapour pressure curve, which includes the critical point with its critical temperature. That means that the vapour pressure is there equal to the critical pressure, also the velocity of evapo-ration is equal to the velocity of condensation and it is the last point a molecule can be clearly defined as liquid or gaseous above it a differentiation from each other is not possible anymore.

Image 1: illustration of a phase diagram with the temperature as x-axis and pressure as y-axis

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The aim of the experiment was to identify two liquids through measuring the vapour pressure by known temperature. With the experimental values it was possible to calculate the intermo-lecular forces in a system, like ΔHvap and ∆S respective their errors. Afterwards the results were compared with literature values to find out which solvent has been measured.

For the equilibrium between two phases applies the Clausius-Clapeyron equitation, given by the entropy of gaseous state subtracted by the entropy of the liquid state, resultant the vaporisation entropy. Divided by the molecular volume of the vaporisation, consists out of subtraction of the volume of the gas phase by the volume of liquid phase:

(1)

Sm = molar entropy Vm = molar volume 1,2 = Indices fort phases in equilibrium

For ΔS another formula can be used then in equilibrium applies that the Gibbs free enthalpy is equal to zero and can be calculated by the enthalpy minus the temperature multiplied by the entropy.

ΔG= ΔH-TΔS=0 or ΔS=

(2)

By insert equation (2) in equation (1) result:

(3)

With the assumption that the ideal gas theory applies that the mole volume of the liquid coun-terpart of the gas is really small and the enthalpy of vaporization depends on the tem-perature, than the following formula can be used:

(4)

Image 2: the formula of the ideal gas theory

Through the formula, it can be shown that the vapour pressure and the temperature have a pro-portional relation. Beside the vapour pressure and the value of are proportional to each other, too.

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2) Experimental part

Material

The following solvents: -Methanol -Ethanol -Isopropanol -Cyclohexane

For the procedure two of these solvents were selected (orange and green). The aim of the ex-periment was to find out which solvents were taken. During the measurements the values of were obtained.

Procedure: One of the two liquids was filled in the round bottom flask with an injection via a thin tube until 2/3. The vacuum flask was connected, through a u-tube named dilatometer, to a manometer and a vacuum pump for the regulation of the pressure. The exact definition for this equipment is isoteniscop, it was located in a heat-controlled water bath and it measured the vapor pressure in the system. For the correct meter-reading it was important that not too much liquid was in the dilatometer, otherwise it would have been impossible to determine in which moment the equi-librium state was achieved. Afterwards the cooling system was started, the required tempera-ture was set and a vacuum was produced by a manual vacuum pump. Through this the liquid in the glass flask started boiling and condensed in the dilatometer. For the measurement the ventil of the pump were open slowly and an equilibrium in the u-tubes consisted, in case of the two meniscus were on the same height. In this situation both pressures the one in the inner-/ and outside of the system were equal. The pressure was indicated by the digital display and was noted. The measurement started at a temperature of 65 degrees and went down in steps of 5 degrees until 30 degrees. At every temperature the procedure had to be repeated three times.

The experimental setup is shown by the illustration on image 3.

A) round bottom flaskB) dilatometerC) water bathD) cooling systemE) vacuum pumpF) vacuum flaskG) ventilH) digital manometer

.

Image.3 This illustration shows what the experimental “machine” looks like

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3.) Results

3.1) Measurements of the orange liquid

Temperature T [K]

1/T [K]

Pressure p1

[mbar]

Pressure p2

[mbar]

Pressure p3

[mbar]

Mean value [mbar]

ln(p) [Pa]

338,15 0,00295727 581 580 579 580 10,9681983 333,15 0,00300165 463 463 464 463,333333 10,7436169 328,15 0,00304739 373 373 372 372,666667 10,5258546 323,15 0,00309454 298 300 299 299 10,3056138 318,15 0,00314317 238 238 237 237,666667 10,0760393 313,15 0,00319336 185 185 184 184,666667 9,82372258 308,15 0,00324517 138 139 139 138,666667 9,53724316 303,15 0,0032987 112 111 113 112 9,32366906

Table 1: relevant datas of the orange liquid measurements

Image 4: Temperatur T vs. ln(p) of the orange liquid

y = -4855.6x + 25.326 R² = 0.9994

9.2

9.4

9.6

9.8

10

10.2

10.4

10.6

10.8

11

11.2

0.0029 0.00295 0.003 0.00305 0.0031 0.00315 0.0032 0.00325 0.0033 0.00335

ln(p

) [Pa

]

1/T [K]

Temperature T vs. ln(p) of orange liquid

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3.2) Measurements of the green liquid

Temperature T

[K]

1/T [K]

Pressure p1

[mbar]

Pressure p2

[mbar]

Pressure p3

[mbar]

Mean value

Standard devia-tion Sx

338,15 0,00295727 454 453 453 453,333333 0,57735027 333,15 0,00300165 363 362 362 362,333333 1,15470054 328,15 0,00304739 285 285 285 285 2,081666 323,15 0,00309454 224 224 225 224,333333 1,52752523 318,15 0,00314317 178 178 177 177,666667 0,57735027 313,15 0,00319336 136 135 134 135 0,57735027 308,15 0,00324517 105 103 104 104 0,57735027 303,15 0,0032987 79 79 79 79 0,57735027

Image 5: Temperatur T vs. ln(p) of the orange liquid

Due to low errors the error bars are not recognisable.

y = -4520.5x + 24.142 R² = 0.9953

0

2

4

6

8

10

12

0.0029 0.00295 0.003 0.00305 0.0031 0.00315 0.0032 0.00325 0.0033 0.00335

ln(p

) [Pa

]

1/T [K]

1/T vs. ln(p) of green liquid

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4.) Calculations

4.1) Calculation of with Clausius-Clapeyron equation

By integrating the following equation and separating it is possible to calculate with the help of linear regression:

The derivation of this equation is shown in the introduction.

The formula of the linear regression is:

m is defined as the slope of the curve. The received curve is dependent on the pressure and temperature. So the slope can be seen as:

Due to a negative slope:

This means for

With R = gas constant = 8,314

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4.1.1) of orange liquid

To calculate the boiling temperature, the two equations can be combinded:

pqmT

mRH

pqRHq

TRHp

boil ln1

andln1T 1ln boil

��� �

� '

��

' ���¸

¹·

¨©§ '�

As pressure, the standard pressure of p = 1 atm = 101325 Pa is used.

C78.71K 9.351526.11326.25

1 65.4855 q �

� KTboil

With the results of the enthalpy and the boiling temperature, it is now possible to calculate the entropy ∆S:

KmolJ

KmolJ

THSboil �

'

' 72.114 9.351

40369.87

4.1.2) of green liquid

CKKpq

mTboil q �

� �

�� 94.83 09.357526.11854.25

1 414.5116ln1

KmolJ

KmolJ

THSboil �

'

' 124.119 09.357

87.42537

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4.2) Error calculation

4.2.1) Orange liquid

The error of ∆∆Hvap can be calculated with the Gauss’ error propagation:

� �

� � � �22

22

mRH

mmH

H

vap

vapvap

'�� ''

'�¸̧¹

·¨̈©

§w

'w ''

The error oft he slope can be calculated in excel with this equation:

� �

Km

xn

yxcyxmynnm

ii

iiiii

433.49

2 22

2

'

��

�����

� '

¦ ¦¦ ¦¦

So it follows:

� � � � � �

molkJH

molJK

KmolJmRH

vap

vap

)41.0 37.40(

99.410 433.49314.8 22

22

r 'o

��

'�� ''

For the error of the entropy , the same procedure is used:

22

22

22

22

2

22

22

22

22

2

ln1

ln1

1

qpq

mmpq

qqTm

mTT

TH

TTHH

TT

TSH

HSS

boilboilboil

vapboil

boilboilboil

boil

'�¸̧¹

·¨̈©

§�

��'�¸̧¹

·¨̈©

§�

� '�¸̧¹

·¨̈©

§w'w

�'�¸¸¹

·¨¨©

§

w'w

'

'' '�¸̧

¹

·¨̈©

§ '��''�¸̧

¹

·¨̈©

§ '�¸̧

¹

·¨̈©

§w'w

�''�¸¹·

¨©§'w'w

''

In excel, the error of ∆q is calculated which gives the result: 155.0 'q

KKTboil 28.14155.053.11)326.25(

1) 65.4855(433.4953.11326.25

1 22

22

2

�¸̧¹

·¨̈©

§�

����¸¹·

¨©§

�� '

KmolJ

TH

S vap

'' �¸

¸¹

·¨¨©

§���¸

¹·

¨©§ '' 17.1

9.35199.41028.14

9.35187.4036999.410

9.3511 2

2

22

2

This leads to the final entropy of :

KmolJS�

r ' )17.172.114(

Date

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4.2.2) Green liquid The calculation with excel of the error of the slope gives:

48.19 'm

� � � � � �

molkJH

molJK

KmolJmRH

vap

vap

)16.154.42(

96.161 48.19314.8 22

22

r 'o

��

'�� ''

In excel, the error of ∆q is calculated which gives the result: 061.0 'q

KKTboil 2.08 061.053.11)854.25(

1) 414.5116(48.1953.11854.25

1 22

22

2

�¸̧¹

·¨̈©

§�

����¸¹·

¨©§

�� '

KmolJ

TH

S vap

'' �¸

¸¹

·¨¨©

§���¸

¹·

¨©§ '' 0.454

357.09161.9608.2

09.35787.4253796.161

09.3571 2

2

22

2

KmolJS�

r ' )45.012.119(

Date

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Date

13

5) Discussion

The goal of the experiment was to identify two solvents with an application of the vapour pres-sure. For it the data of the vaporization enthalpy, vaporization entropy and the boiling point of the solutions were given. Afterwards the results were compared to the literature values.

Literature Values

Methanol Isopropanol Cyclohexane Ethanol Std entropy change of vaporization ∆vap S [J/(mol.K)]

9.55.109 r 2.115.126 r 7.52.93 r 7.55.119 r

Std enthalpy change of vaporization ∆vap H [kJ/mol]

0.20.37 r 0.40.45 r 0.20.33 r 0.20.42 r

Boiling point [°C]

64.7 82.6 80.74 78.35

Values are taken from http://webbook.nist.gov/ (21.12.2015)

Omitted Values

Orange Green Std entropy change of vaporization ∆vap S [J/(mol.K)]

17.172.114 r 45.012.119 r

Std enthalpy change of vaporization ∆vap H [kJ/mol]

41.037.40 r 16.154.42 r

Boiling point [°C] 71.78 94.83

In conclusion we can say that the results make sense after all. The data were all in a reasonable area, so large errors can be excluded. According to our experimental results the orange liquid has to be Ethanol and the green liquid Isopropanol.

Than during the procedure of the experiment a lot of error sources existed, the following errors were not fully excludable.

x After the measurement of the first solvent, the solution was changed through evacuate the liquid and evaporate the whole flask, but it could not be excluded that a little con-tamination of the second solvent occurred.

x It could be possible that the isoteniscop was not completely tight, that can be a reason for some deviations, and the other equipments themselves can be reasonable for errors.

x The error of the analyser, is fixed at the last seen digit. x The used formula was based on the ideal gas theory, so there is not a hundred percent

accordance with the real environment. x It was not easy to recognize, in which moment the meniscus in the u-tube were exactly

on the same level by the notation of the value, because it changed quickly. x One time, during the measurement of the orange solution, there was an overpressure

and we had to repeat a part of the measurement, this could be the reason for some higher error.

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6) Literature

Image 1 : http://www.chemieseite.de/anorganisch/node5.php(17.12.15, 7 :30Uhr) Image 2: http://www.calctool.org/CALC/chem/c_thermo/ideal_gas (17.12.15, 7:40Uhr) Image 3: http://www.chemie.unibas.ch/~pcpraktikum/Dampfdruck (18.12.15, 21.00 Uhr) Image 4/5 : self-made on excel

Literature values: http://webbook.nist.gov/ (21.12.2015, 15.10 Uhr)

- Atkins, Peter W. et al (2008): Kurzlehrbuch Physikalische Chemie. WILEY-VCH. Weinheim,

- Mortimer, Charles E. et al (2010): Chemie. Thieme Verlag

Date