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Physical & Chemical properties

Vapour pressure

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Reference
Endpoint:
vapour pressure
Type of information:
experimental study
Adequacy of study:
key study
Study period:
2010-10-04 - 2010-10-20
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
comparable to guideline study
Qualifier:
equivalent or similar to guideline
Guideline:
EU Method A.4 (Vapour Pressure)
GLP compliance:
yes (incl. QA statement)
Type of method:
effusion method: vapour pressure balance
Specific details on test material used for the study:
Sponsor's identification: Amine C8
Description: dark brown liquid
Chemical name: Ethanol, 2, 2-oxybis-, reaction products with ammonia, morpholine derivs. residues
Batch number: T7-271109
Date received: 21 September 2010
Expiry date: 31 December 2011
Storage conditions: room temperature in the dark
Key result
Temp.:
25 °C
Vapour pressure:
0.55 Pa
Transition / decomposition:
no

Run1

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

21

294.15

0.003399626

248.31

2.483E-07

0.344717897

-0.462536169

22

295.15

0.003388108

279.46

2.795E-07

0.387962077

-0.411210724

23

296.15

0.003376667

312.71

3.127E-07

0.434121596

-0.362388609

24

297.15

0.003365304

350.35

3.504E-07

0.486375559

-0.313028257

25

298.15

0.003354016

390.59

3.906E-07

0.542238989

-0.265809259

26

299.15

0.003342805

437.31

4.373E-07

0.607098318

-0.216740970

27

300.15

0.003331667

488.13

4.881E-07

0.677649498

-0.168994879

28

301.15

0.003320604

543.65

5.437E-07

0.754725482

-0.122210986

29

302.15

0.003309614

605.45

6.055E-07

0.840519715

-0.075452095

30

303.15

0.003298697

675.64

6.756E-07

0.937961418

-0.027815025

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                                       -4294.026
Standard deviation in slope          6.934

Intercept                                        14.137
Standard deviation in intercept    0.023

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -4294.026/temp(K) + 14.137

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.265.

Run 1 - Graph of Log10Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log10Vapour Pressure vs Reciprocal Temperature.

Run 2

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

20

293.15

0.003411223

237.63

2.376E-07

0.329891320

-0.481629111

21

294.15

0.003399626

261.59

2.616E-07

0.363153939

-0.439909241

22

295.15

0.003388108

291.54

2.915E-07

0.404732212

-0.392832229

23

296.15

0.003376667

322.69

3.227E-07

0.447976393

-0.348744872

24

297.15

0.003365304

357.24

3.572E-07

0.495940644

-0.304570298

25

298.15

0.003354016

398.18

3.982E-07

0.552775853

-0.257450936

26

299.15

0.003342805

443.20

4.432E-07

0.615275147

-0.210930627

27

300.15

0.003331667

494.42

4.944E-07

0.686381630

-0.163434348

28

301.15

0.003320604

547.14

5.471E-07

0.759570496

-0.119431913

29

302.15

0.003309614

609.74

6.097E-07

0.846475335

-0.072385692

30

303.15

0.003298697

679.33

6.793E-07

0.943084083

-0.025449585

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                                       -4067.764
Standard deviation in slope       26.030

Intercept                                        13.389
Standard deviation in intercept    0.087

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -4067.764/temp(K) + 13.389

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.255.

Run 2 - Graph of Log10Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log10Vapour Pressure vs Reciprocal Temperature.

Run 3

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

20

293.15

0.003411223

236.43

2.364E-07

0.328225413

-0.483827797

21

294.15

0.003399626

261.09

2.611E-07

0.362459811

-0.440740140

22

295.15

0.003388108

290.74

2.907E-07

0.403621607

-0.394025593

23

296.15

0.003376667

321.80

3.218E-07

0.446740845

-0.349944339

24

297.15

0.003365304

358.64

3.586E-07

0.497884203

-0.302871653

25

298.15

0.003354016

397.08

3.971E-07

0.551248771

-0.258652366

26

299.15

0.003342805

442.51

4.425E-07

0.614317250

-0.211607289

27

300.15

0.003331667

492.13

4.921E-07

0.683202523

-0.165450539

28

301.15

0.003320604

546.64

5.466E-07

0.758876368

-0.119828971

29

302.15

0.003309614

605.95

6.060E-07

0.841213844

-0.075093589

30

303.15

0.003298697

678.44

6.784E-07

0.941848535

-0.026018933

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                                       -4067.078
Standard deviation in slope       21.916

Intercept                                        13.385
Standard deviation in intercept    0.074

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -4067.078/temp(K) + 13.385

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.256.

Run 3 - Graph of Log10Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log10Vapour Pressure vs Reciprocal Temperature.

Run 4

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

20

293.15

0.003411223

236.23

2.362E-07

0.327947762

-0.484195329

21

294.15

0.003399626

261.09

2.611E-07

0.362459811

-0.440740140

22

295.15

0.003388108

289.55

2.896E-07

0.401969582

-0.395806809

23

296.15

0.003376667

321.30

3.213E-07

0.446046717

-0.350619653

24

297.15

0.003365304

357.04

3.570E-07

0.495662993

-0.304813505

25

298.15

0.003354016

396.38

3.964E-07

0.550276992

-0.259418645

26

299.15

0.003342805

440.21

4.402E-07

0.611124261

-0.213870475

27

300.15

0.003331667

489.63

4.896E-07

0.679731882

-0.167662359

28

301.15

0.003320604

543.45

5.435E-07

0.754447831

-0.122370786

29

302.15

0.003309614

605.45

6.055E-07

0.840519715

-0.075452095

30

303.15

0.003298697

674.74

6.747E-07

0.936711987

-0.028393922

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                                       -4053.160
Standard deviation in slope       22.971

Intercept                                        13.337
Standard deviation in intercept    0.077

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -4053.160/temp(K) + 13.337

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.257.

Run 4 - Graph of Log10Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log10Vapour Pressure vs Reciprocal Temperature.

Run 5

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

20

293.15

0.003411223

233.93

2.339E-07

0.324754773

-0.488444458

21

294.15

0.003399626

261.09

2.611E-07

0.362459811

-0.440740140

22

295.15

0.003388108

287.95

2.880E-07

0.399748372

-0.398213296

23

296.15

0.003376667

320.10

3.201E-07

0.444380809

-0.352244705

24

297.15

0.003365304

355.94

3.559E-07

0.494135911

-0.306153583

25

298.15

0.003354016

395.68

3.957E-07

0.549305213

-0.260186280

26

299.15

0.003342805

439.31

4.393E-07

0.609874831

-0.214759290

27

300.15

0.003331667

489.33

4.893E-07

0.679315406

-0.167928536

28

301.15

0.003320604

542.35

5.424E-07

0.752920749

-0.123250734

29

302.15

0.003309614

608.35

6.084E-07

0.844545658

-0.073376866

30

303.15

0.003298697

671.35

6.714E-07

0.932005799

-0.030581385

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                                       -4074.568
Standard deviation in slope       21.543

Intercept                                        13.408
Standard deviation in intercept    0.072

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -4074.568/temp(K) + 13.408

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.258.

Run 5 -Graph of Log10Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log10Vapour Pressure vs Reciprocal Temperature.

Run 6

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

20

293.15

0.003411223

234.13

2.341E-07

0.325032424

-0.488073314

21

294.15

0.003399626

258.00

2.580E-07

0.358170099

-0.445910673

22

295.15

0.003388108

280.46

2.805E-07

0.389350333

-0.409659449

23

296.15

0.003376667

318.90

3.189E-07

0.442714902

-0.353875859

24

297.15

0.003365304

353.15

3.532E-07

0.490262677

-0.309571168

25

298.15

0.003354016

393.78

3.938E-07

0.546667526

-0.262276724

26

299.15

0.003342805

438.01

4.380E-07

0.608070098

-0.216046353

27

300.15

0.003331667

487.34

4.873E-07

0.676552776

-0.169698320

28

301.15

0.003320604

541.45

5.415E-07

0.751671319

-0.123972021

29

302.15

0.003309614

604.05

6.041E-07

0.838576157

-0.076457490

30

303.15

0.003298697

669.65

6.697E-07

0.929645763

-0.031682506

A plot of Log10(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 6 gives the following statistical data using an unweighted least squares treatment.

Slope                                       -4102.766
Standard deviation in slope       35.676

Intercept                                        13.500
Standard deviation in intercept    0.120

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -4102.766/temp(K) + 13.500

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.261.

Run 6 - Graph of Log10Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log10Vapour Pressure vs Reciprocal Temperature.

Conclusions:
The vapour pressure of the test item has been determined to be 0.55 Pa at at 25 ºC.
Executive summary:

INTRODUCTION

Hazardous physico-chemical properties of the test item have been determined. Methods employed were designed to be compatible with Commission Regulation (EC) No 440/2008 of 30 May 2008, Part A: Methods for the determination of physico-chemical properties

Method

The vapour pressure was determined 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 a method designed to be compatible with Method A4 Vapour Pressure of Commission Regulation (EC) No 440/2008 of 30 May 2008.

Summary of Results

Run

Log10[Vp(25ºC)]

1

-0.265

2

-0.255

3

-0.256

4

-0.257

5

-0.258

6

-0.261

Mean

-0.259

Vapour Pressure

5.5 x 10-1Pa

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

Discussion

The unshaded point on run 1 has been omitted from the data analysis, because its position in relation to the other 10 points clearly indicates it is erroneous.

Conclusion

The vapour pressure of the test item has been determined to be 5.5 x 10-1Pa at 25 ºC.

Description of key information

The vapour pressure of the test substance was determined to be 0.55 Pa at 25 °C using the effusion method (vapour pressure balance) following a GLP study compatible with EC test method A.4 (Atwal et al., 2010).

Key value for chemical safety assessment

Vapour pressure:
0.55 Pa
at the temperature of:
25 °C

Additional information