Registration Dossier

Administrative data

Endpoint:
vapour pressure
Type of information:
experimental study
Adequacy of study:
key study
Study period:
15th January 2010 to 23rd February 2010.
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
other: Study conducted in compliance with agreed protocols, with no or minor deviations from standard test guidelines and/or minor methodological deficiencies, which do not affect the quality of the relevant results.

Data source

Reference
Reference Type:
study report
Title:
Unnamed
Year:
2010

Materials and methods

Test guideline
Qualifier:
according to
Guideline:
EU Method A.4 (Vapour Pressure)
Deviations:
no
GLP compliance:
yes (incl. certificate)
Remarks:
Date of inspection: 15/09/09 Date of Signature: 26/11/09
Type of method:
effusion method: vapour pressure balance

Test material

Reference
Name:
Unnamed
Type:
Constituent
Details on test material:
Sponsor's identification : DVS005u (aka Weston 705)
Description : clear colourless viscous liquid
Batch number : MW9F23A901
Date received : 05 November 2009
Expiry date : 15 June 2010
Storage conditions : approximately 4°C, under nitrogen, in the dark

Results and discussion

Vapour pressure
Temp.:
25 °C
Vapour pressure:
0 Pa
Transition / decomposition
Transition / decomposition:
no

Any other information on results incl. tables

Run1

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

175

448.15

0.002231396

380.86

3.809E-07

0.528731256

-0.276765016

176

449.15

0.002226428

400.72

4.007E-07

0.556302024

-0.254689360

177

450.15

0.002221482

431.72

4.317E-07

0.599337966

-0.222328210

178

451.15

0.002216558

468.06

4.681E-07

0.649787196

-0.187228850

179

452.15

0.002211655

491.85

4.919E-07

0.682813811

-0.165697703

180

453.15

0.002206775

528.82

5.288E-07

0.734137643

-0.134222507

181

454.15

0.002201916

567.99

5.680E-07

0.788515638

-0.103189689

182

455.15

0.002197078

611.47

6.115E-07

0.848877018

-0.071155224

183

456.15

0.002192261

667.44

6.674E-07

0.926577717

-0.033118148

184

457.15

0.002187466

736.91

7.369E-07

1.023019875

0.009884071

185

458.15

0.002182691

804.02

8.040E-07

1.116185749

0.047736473

A plot of Log10(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 1 gives the following statistical data using an unweighted least squares treatment.

Slope                                       -6610.581
Standard deviation in slope     177.616

Intercept                                        14.462
Standard deviation in intercept    0.392

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -6610.581/temp(K) + 14.462

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -7.710.

Run 1 - Graph of Log10Vapour Pressure vs Reciprocal Temperature see attached Run 1 to 9 - Graph of Log10 Vapour Pressure vs Reciprocal Temperature.

Run 2

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

175

448.15

0.002231396

383.13

3.831E-07

0.531882597

-0.274184219

176

449.15

0.002226428

405.19

4.052E-07

0.562507529

-0.249871660

177

450.15

0.002221482

438.63

4.386E-07

0.608930816

-0.215432047

178

451.15

0.002216558

472.30

4.723E-07

0.655673403

-0.183312433

179

452.15

0.002211655

509.04

5.090E-07

0.706677935

-0.150778469

180

453.15

0.002206775

551.98

5.520E-07

0.766289656

-0.115607037

181

454.15

0.002201916

598.44

5.984E-07

0.830788039

-0.080509764

182

455.15

0.002197078

643.42

6.434E-07

0.893231803

-0.049035822

183

456.15

0.002192261

698.68

6.987E-07

0.969946841

-0.013252067

184

457.15

0.002187466

754.73

7.547E-07

1.047758601

0.020261235

185

458.15

0.002182691

805.20

8.052E-07

1.117823891

0.048373387

A plot of Log10(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 2 gives the following statistical data using an unweighted least squares treatment.

Slope                                       -6791.343
Standard deviation in slope       64.728

Intercept                                        14.873
Standard deviation in intercept    0.143

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -6791.343/temp(K) + 14.873

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -7.905.

Run 2 - Graph of Log10Vapour Pressure vs Reciprocal Temperature see attached Run 1 to 9 - Graph of Log10 Vapour Pressure vs Reciprocal Temperature.

Run 3

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

175

448.15

0.002231396

382.98

3.830E-07

0.531674359

-0.274354284

176

449.15

0.002226428

401.03

4.010E-07

0.556732383

-0.254353516

177

450.15

0.002221482

434.39

4.344E-07

0.603044610

-0.219650560

178

451.15

0.002216558

468.30

4.683E-07

0.650120378

-0.187006221

179

452.15

0.002211655

505.04

5.050E-07

0.701124911

-0.154204602

180

453.15

0.002206775

543.73

5.437E-07

0.754836543

-0.122147083

181

454.15

0.002201916

577.17

5.772E-07

0.801259830

-0.096226629

182

455.15

0.002197078

627.49

6.275E-07

0.871116882

-0.059923570

183

456.15

0.002192261

682.83

6.828E-07

0.947942980

-0.023217785

184

457.15

0.002187466

740.52

7.405E-07

1.028031480

0.012006414

185

458.15

0.002182691

782.83

7.828E-07

1.086768600

0.036137082

A plot of Log10 (vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 3 gives the following statistical data using an unweighted least squares treatment.

Slope                                       -6568.269
Standard deviation in slope       92.203

Intercept                                        14.373
Standard deviation in intercept    0.203

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -6568.269/temp(K) + 14.373

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -7.657.

Run 3 - Graph of Log10Vapour Pressure vs Reciprocal Temperature see attached Run 1 to 9 - Graph of Log10 Vapour Pressure vs Reciprocal Temperature.

Run 4

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

175

448.15

0.002231396

382.11

3.821E-07

0.530466576

-0.275341975

176

449.15

0.002226428

401.97

4.020E-07

0.558037344

-0.253336737

177

450.15

0.002221482

432.82

4.328E-07

0.600865048

-0.221223058

178

451.15

0.002216558

456.13

4.561E-07

0.633225300

-0.198441742

179

452.15

0.002211655

484.55

4.846E-07

0.672679541

-0.172191781

180

453.15

0.002206775

525.68

5.257E-07

0.729778518

-0.136808925

181

454.15

0.002201916

584.00

5.840E-07

0.810741620

-0.091117532

182

455.15

0.002197078

639.73

6.397E-07

0.888109138

-0.051533661

183

456.15

0.002192261

688.01

6.880E-07

0.955134147

-0.019935628

184

457.15

0.002187466

736.52

7.365E-07

1.022478455

0.009654166

185

458.15

0.002182691

797.59

7.976E-07

1.107259261

0.044249322

A plot of Log10(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 4 gives the following statistical data using an unweighted least squares treatment.

Slope                                        -6770.783
Standard deviation in slope     187.810

Intercept                                        14.818
Standard deviation in intercept    0.414

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -6770.783/temp(K) + 14.818

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -7.891.

Run 4 - Graph of Log10Vapour Pressure vs Reciprocal Temperature see attached Run 1 to 9 - Graph of Log10 Vapour Pressure vs Reciprocal Temperature.

Run 5

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

175

448.15

0.002231396

377.01

3.770E-07

0.523386469

-0.281177509

176

449.15

0.002226428

401.03

4.010E-07

0.556732383

-0.254353516

177

450.15

0.002221482

426.85

4.269E-07

0.592577158

-0.227255093

178

451.15

0.002216558

460.14

4.601E-07

0.638792207

-0.194640391

179

452.15

0.002211655

499.78

4.998E-07

0.693822683

-0.158751506

180

453.15

0.002206775

535.88

5.359E-07

0.743938732

-0.128462830

181

454.15

0.002201916

581.80

5.818E-07

0.807687456

-0.092756662

182

455.15

0.002197078

625.52

6.255E-07

0.868382017

-0.061289179

183

456.15

0.002192261

676.94

6.769E-07

0.939766151

-0.026980202

184

457.15

0.002187466

733.14

7.331E-07

1.017786149

0.007656536

185

458.15

0.002182691

788.40

7.884E-07

1.094501187

0.039216237

A plot of Log10(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 5 gives the following statistical data using an unweighted least squares treatment.

Slope                                       -6687.856
Standard deviation in slope       78.723

Intercept                                        14.634
Standard deviation in intercept    0.174

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -6687.856/temp(K) + 14.634

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -7.797.

Run 5 - Graph of Log10Vapour Pressure vs Reciprocal Temperature see attached Run 1 to 9 - Graph of Log10 Vapour Pressure vs Reciprocal Temperature.

Please note we have included runs 6,7 and 8 in the overall remarks section, please see these below:

Applicant's summary and conclusion

Conclusions:
The vapour pressure of the test material has been determined to be 1.6 x 10-8 Pa at 25ºC.
Executive summary:

Method: The vapour pressure was determined using a vapour pressure balance with measurments being made at several temperatures and linear regression analysis used to calculate the vapour pressure at 25°C. Testing was conducted using Method 104 specified in the OECD Guidelines for Testing of Chemicals, 23 March 2006.

Summary of Results:

Run

Log10[Vp(25ºC)]

1

-7.710

2

-7.905

3

-7.657

4

-7.891

5

-7.797

6

-7.746

7

-7.623

8

-7.856

9

-7.876

Mean

-7.785

Vapour Pressure

1.6 x 10-8Pa

The test material did not change in appearance under the conditions used in the determination.

Conclusion:

The vapour pressure of the test material has been determined to be 1.6 x 10-8 Pa at 25ºC.