Registration Dossier

Administrative data

Endpoint:
vapour pressure
Type of information:
experimental study
Adequacy of study:
key study
Study period:
between 10 July 2008 and 08 August 2008.
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 relevant results.

Data source

Reference
Reference Type:
study report
Title:
Unnamed
Year:
2008

Materials and methods

Test guideline
Qualifier:
according to
Guideline:
EU Method A.4 (Vapour Pressure)
Deviations:
no
GLP compliance:
yes (incl. certificate)
Remarks:
With the exception of the explosive properties and oxidising properties predictions, the work described was performed in compliance with UK GLP standards (Schedule 1, Good Laboratory Practice Regulations 1999 (SI 1999/3106 as amended by SI 2004/0994)).
Type of method:
effusion method: vapour pressure balance

Test material

Reference
Name:
Unnamed
Type:
Constituent

Results and discussion

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

Any other information on results incl. tables

Results:

The results from runs 1 to 5 are shown in the tables below

Run 1:

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10 Vp

181

454.15

0.002201916

189.01

1.890E-07

0.262394304

-0.581045597

182

455.15

0.002197078

195.84

1.958E-07

0.271876094

-0.565628978

183

456.15

0.002192261

209.27

2.093E-07

0.290520375

-0.536823404

184

457.15

0.002187466

217.08

2.171E-07

0.301362656

-0.520910566

185

458.15

0.002182691

228.31

2.283E-07

0.316952773

-0.499005445

186

459.15

0.002177937

243.61

2.436E-07

0.338193093

-0.470835267

187

460.15

0.002173204

246.70

2.467E-07

0.342482804

-0.465361229

188

461.15

0.002168492

269.32

2.693E-07

0.373885159

-0.427261773

189

462.15

0.002163800

266.79

2.668E-07

0.370372871

-0.431360832

190

463.15

0.002159128

284.37

2.844E-07

0.394778415

-0.403646601

A plot of Log10 (vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 1 (Figure 6.2) gives the following statistical data using an unweighted least squares treatment.

Slope                                          -4181.233
Standard deviation in slope             172.540

Intercept                                             8.627
Standard deviation in intercept            0.376

The results obtained indicate the following vapour pressure relationship:

Log10 (Vp (Pa)) = -4181.233/temp(K) + 8.627.

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -5.397.

Run 2:

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10 Vp

181

454.15

0.002201916

181.12

1.811E-07

0.251440963

-0.599563969

182

455.15

0.002197078

181.52

1.815E-07

0.251996265

-0.598605896

183

456.15

0.002192261

204.31

2.043E-07

0.283634624

-0.547240755

184

457.15

0.002187466

203.66

2.037E-07

0.282732257

-0.548624639

185

458.15

0.002182691

220.66

2.207E-07

0.306332613

-0.513806765

186

459.15

0.002177937

246.86

2.469E-07

0.342704925

-0.465079654

187

460.15

0.002173204

236.53

2.365E-07

0.328364239

-0.483644147

188

461.15

0.002168492

241.65

2.417E-07

0.335472110

-0.474343579

189

462.15

0.002163800

262.56

2.626E-07

0.364500547

-0.438301815

190

463.15

0.002159128

281.77

2.818E-07

0.391168949

-0.407635627

A plot of Log10 (vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 2 (Figure 6.3) gives the following statistical data using an unweighted least squares treatment.

Slope                                          -4408.658
Standard deviation in slope             372.785

Intercept                                             9.105
Standard deviation in intercept            0.813

The results obtained indicate the following vapour pressure relationship:

Log10 (Vp (Pa)) = -4.408.658/temp(K) + 9.105.

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -5.682.

Run 3:

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10 Vp

181

454.15

0.002201916

168.99

1.690E-07

0.234601415

-0.629669373

182

455.15

0.002197078

172.00

1.720E-07

0.238780066

-0.622001932

183

456.15

0.002192261

180.14

1.801E-07

0.250080472

-0.601920220

184

457.15

0.002187466

202.60

2.026E-07

0.281260706

-0.550890938

185

458.15

0.002182691

207.15

2.072E-07

0.287577272

-0.541245441

186

459.15

0.002177937

215.78

2.158E-07

0.299557923

-0.523519190

187

460.15

0.002173204

218.95

2.190E-07

0.303958695

-0.517185429

188

461.15

0.002168492

230.59

2.306E-07

0.320117997

-0.494689909

189

462.15

0.002163800

243.04

2.430E-07

0.337401786

-0.471852622

A plot of Log10 (vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 3 (Figure 6.4) gives the following statistical data using an unweighted least squares treatment.

Slope                                          -4234.395
Standard deviation in slope             294.509

Intercept                                             8.692
Standard deviation in intercept            0.643

The results obtained indicate the following vapour pressure relationship:

Log10 (Vp (Pa)) = -4234.395/temp(K) + 8.692.

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -5.510.

Run 4:

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10 Vp

181

454.15

0.002201916

153.94

1.539E-07

0.213708159

-0.670178897

182

455.15

0.002197078

159.72

1.597E-07

0.221732280

-0.654171077

183

456.15

0.002192261

166.63

1.666E-07

0.231325130

-0.635777184

184

457.15

0.002187466

173.39

1.734E-07

0.240709742

-0.618506332

185

458.15

0.002182691

180.14

1.801E-07

0.250080472

-0.601920220

186

459.15

0.002177937

188.44

1.884E-07

0.261602998

-0.582357283

187

460.15

0.002173204

192.43

1.924E-07

0.267142140

-0.573257599

188

461.15

0.002168492

206.99

2.070E-07

0.287355151

-0.541581014

A plot of Log10 (vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 4 (Figure 6.5) gives the following statistical data using an unweighted least squares treatment.

Slope                                          -3693.440
Standard deviation in slope             136.406

Intercept                                             7.461
Standard deviation in intercept            0.298

The results obtained indicate the following vapour pressure relationship:

Log10 (Vp (Pa)) = -3693.440/temp(K) + 7.461.

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -4.927.

Run 5:

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10 Vp

181

454.15

0.002201916

143.61

1.436E-07

0.199367473

-0.700345697

182

455.15

0.002197078

152.72

1.527E-07

0.212014487

-0.673634463

183

456.15

0.002192261

163.14

1.631E-07

0.226480116

-0.644969921

184

457.15

0.002187466

174.36

1.744E-07

0.242056351

-0.616083518

185

458.15

0.002182691

162.97

1.630E-07

0.226244113

-0.645422713

186

459.15

0.002177937

185.10

1.851E-07

0.256966222

-0.590123960

187

460.15

0.002173204

191.86

1.919E-07

0.266350834

-0.574545939

188

461.15

0.002168492

209.60

2.096E-07

0.290978499

-0.536139100

189

462.15

0.002163800

215.62

2.156E-07

0.299335802

-0.523841337

190

463.15

0.002159128

221.39

2.214E-07

0.307346040

-0.512372378

A plot of Log10 (vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 5 (Figure 6.6) gives the following statistical data using an unweighted least squares treatment.

Slope                                          -4416.055
Standard deviation in slope             341.863

Intercept                                             9.027
Standard deviation in intercept            0.745

The results obtained indicate the following vapour pressure relationship:

Log10 (Vp (Pa)) = -4416.055/temp(K) + 9.027.

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -5.785.

Summary of Results:

Run

Log10 [Vp(25ºC)]

1

-5.397

2

-5.682

3

-5.510

4

-4.927

5

-5.785

Mean

-5.460

Vapour Pressure

3.5 x 10-6 Pa

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

Applicant's summary and conclusion

Conclusions:

The vapour pressure of the test item has been determined to be 3.5 x 10-6 Pa at 25ºC.
Executive summary:

The vapour pressure of SLI(76)stripped was determined in study number 2517/0001 using a vapour pressure balance with measurements being made at several temperatures and linear regression analysis used to calculate the vapour pressure at 25°C.  Testing was conducted using Method A4 of Commission Directive 92/69/EEC (which constitutes Annex V of Council Directive 67/548/EEC). The vapour pressure of the test item has been determined to be 3.5 x 10-6 Pa at 25ºC.