Registration Dossier

Physical & Chemical properties

Vapour pressure

Currently viewing:

Administrative data

Link to relevant study record(s)

Reference
Endpoint:
vapour pressure
Type of information:
experimental study
Adequacy of study:
key study
Study period:
09 June 2011 and 18 July 2011
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
guideline study
Remarks:
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. The study report was conclusive, done to a valid guideline and the study was conducted under GLP conditions.
Qualifier:
according to
Guideline:
EU Method A.4 (Vapour Pressure)
Deviations:
no
Qualifier:
according to
Guideline:
OECD Guideline 104 (Vapour Pressure Curve)
Deviations:
no
GLP compliance:
yes (incl. certificate)
Type of method:
effusion method: vapour pressure balance
Key result
Temp.:
25 °C
Vapour pressure:
0 Pa
Key result
Transition / decomposition:
no

Results

Run 6

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

135

408.15

0.002450080

19.31

1.931E-08

0.026807227

-1.571748105

136

409.15

0.002444091

18.84

1.884E-08

0.026154747

-1.582449480

137

410.15

0.002438132

18.84

1.884E-08

0.026154747

-1.582449480

138

411.15

0.002432202

19.86

1.986E-08

0.027570768

-1.559551135

139

412.15

0.002426301

21.12

2.112E-08

0.029319971

-1.532836465

140

413.15

0.002420428

21.74

2.174E-08

0.030180690

-1.520270839

141

414.15

0.002414584

24.73

2.473E-08

0.034331576

-1.464306262

142

415.15

0.002408768

27.16

2.716E-08

0.037705038

-1.423600613

143

416.15

0.002402980

28.49

2.849E-08

0.039551419

-1.402837929

144

417.15

0.002397219

27.94

2.794E-08

0.038787878

-1.411303977

145

418.15

0.002391486

31.40

3.140E-08

0.043591245

-1.360600731

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                                      -4060.130
Standard deviation in slope    370.805

Intercept                                          8.336
Standard deviation in intercept   0.896

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -4060.130 /temp(K) + 8.336

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -5.282.

Run 6 - Graph of Log10Vapour Pressure vs Reciprocal Temperature, see attachment 3 Runs 6 to10.

Run 7

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

135

408.15

0.002450080

16.95

1.695E-08

0.023530943

-1.628360676

136

409.15

0.002444091

16.09

1.609E-08

0.022337042

-1.650974335

137

410.15

0.002438132

17.50

1.750E-08

0.024294483

-1.614492330

138

411.15

0.002432202

17.03

1.703E-08

0.023642003

-1.626315731

139

412.15

0.002426301

18.13

1.813E-08

0.025169085

-1.599132575

140

413.15

0.002420428

20.02

2.002E-08

0.027792889

-1.556066306

141

414.15

0.002414584

20.17

2.017E-08

0.028001128

-1.552824481

142

415.15

0.002408768

21.35

2.135E-08

0.029639270

-1.528132499

143

416.15

0.002402980

25.35

2.535E-08

0.035192295

-1.453552415

144

417.15

0.002397219

24.88

2.488E-08

0.034539814

-1.461680003

145

418.15

0.002391486

27.55

2.755E-08

0.038246458

-1.417408776

A plot of Log10(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 7 gives the following statistical data using an unweighted least squares treatment.

Slope                                      -3931.004
Standard deviation in slope    400.317

Intercept                                          7.962
Standard deviation in intercept   0.969

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -3931.004 /temp(K) + 7.962

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -5.223.

Run 7 - Graph of Log10Vapour Pressure vs Reciprocal Temperature, see attachment 3 Runs 6 to10.

Run 8

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

135

408.15

0.002450080

14.21

1.421E-08

0.019727121

-1.704936301

136

409.15

0.002444091

14.36

1.436E-08

0.019935359

-1.700375939

137

410.15

0.002438132

16.09

1.609E-08

0.022337042

-1.650974335

138

411.15

0.002432202

16.17

1.617E-08

0.022448103

-1.648820359

139

412.15

0.002426301

17.27

1.727E-08

0.023975185

-1.620238041

140

413.15

0.002420428

18.13

1.813E-08

0.025169085

-1.599132575

141

414.15

0.002414584

17.90

1.790E-08

0.024849786

-1.604677348

142

415.15

0.002408768

20.09

2.009E-08

0.027890067

-1.554550442

143

416.15

0.002402980

21.27

2.127E-08

0.029528209

-1.529762889

144

417.15

0.002397219

21.74

2.174E-08

0.030180690

-1.520270839

145

418.15

0.002391486

25.28

2.528E-08

0.035095117

-1.454753309

A plot of Log10(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 8 gives the following statistical data using an unweighted least squares treatment.

Slope                                      -3936.817
Standard deviation in slope    266.472

Intercept                                          7.930
Standard deviation in intercept    0.645

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -3936.817 /temp(K) + 7.930

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -5.274.

Run 8 - Graph of Log10Vapour Pressure vs Reciprocal Temperature, see attachment 3 Runs 6 to10.

Run 9

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

135

408.15

0.002450080

13.34

1.334E-08

0.018519338

-1.732374549

136

409.15

0.002444091

13.19

1.319E-08

0.018311099

-1.737285583

137

410.15

0.002438132

14.05

1.405E-08

0.019505000

-1.709854054

138

411.15

0.002432202

13.50

1.350E-08

0.018741459

-1.727196610

139

412.15

0.002426301

14.84

1.484E-08

0.020601722

-1.686096478

140

413.15

0.002420428

14.91

1.491E-08

0.020698900

-1.684052735

141

414.15

0.002414584

15.54

1.554E-08

0.021573501

-1.666079364

142

415.15

0.002408768

17.11

1.711E-08

0.023753064

-1.624280369

143

416.15

0.002402980

17.90

1.790E-08

0.024849786

-1.604677348

144

417.15

0.002397219

19.07

1.907E-08

0.026474046

-1.577179686

145

418.15

0.002391486

20.96

2.096E-08

0.029097850

-1.536139100

A plot of Log10(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 9 gives the following statistical data using an unweighted least squares treatment.

Slope                                      -3350.205
Standard deviation in slope    325.576

Intercept                                          6.447
Standard deviation in intercept   0.788

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -3350.205 /temp(K) + 6.447

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -4.790.

Run 9 - Graph of Log10Vapour Pressure vs Reciprocal Temperature, see attachment 3 Runs 6 to10.

Run10

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

135

408.15

0.002450080

11.77

1.177E-08

0.016339775

-1.786753916

136

409.15

0.002444091

12.09

1.209E-08

0.016784017

-1.775104078

137

410.15

0.002438132

12.56

1.256E-08

0.017436498

-1.758540739

138

411.15

0.002432202

11.93

1.193E-08

0.016561896

-1.780889935

139

412.15

0.002426301

12.72

1.272E-08

0.017658619

-1.753043267

140

413.15

0.002420428

14.60

1.460E-08

0.020268540

-1.693177523

141

414.15

0.002414584

15.86

1.586E-08

0.022017743

-1.657227196

142

415.15

0.002408768

15.93

1.593E-08

0.022114921

-1.655314603

143

416.15

0.002402980

15.23

1.523E-08

0.021143142

-1.674830475

144

417.15

0.002397219

16.80

1.680E-08

0.023322704

-1.632221097

145

418.15

0.002391486

19.31

1.931E-08

0.026807227

-1.571748105

A plot of Log10(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 10 gives the following statistical data using an unweighted least squares treatment.

Slope                                      -3479.425
Standard deviation in slope    407.520

Intercept                                          6.719
Standard deviation in intercept   0.986

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -3479.425 /temp(K) + 6.719

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -4.951.

Run 10 - Graph of Log10Vapour Pressure vs Reciprocal Temperature, see attachment 3 Runs 6 to10 (attached background material)

Summary of Results

Run

Log10[Vp(25°C)]

6

-5.282

7

-5.223

8

-5.274

9

-4.790

10

-4.951

Mean

-5.104

Vapour pressure

7.9 x 10-6Pa

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

Conclusions:
The vapour pressure of the test item has been determined to be 7.9 x 10-6Pa at 25ºC.
Executive summary:

The vapour pressure was determined using a vapour pressure balance with measurents being made at several temperatures and linear regression analysis used to calculate the vapour pressure at 25°C. Testing was conducted using a procedure designed to be compatible with Method A4 Vapour Pressure of Commission Regulation (EC) No 440/2008 of 30 May 2008, and Method 104 specified in the OECD Guidelines for Testing of Chemicals, 23 March 2006.

Summary of Results:

Run

Log10[Vp(25ºC)]

6

-5.282

7

-5.223

8

-5.274

9

-4.790

10

-4.951

Mean

-5.104

Vapour Pressure

7.9 x 10-6Pa

Conclusion

The vapour pressure of the test item has been determined to be 7.9 x 10-6 Pa at 25ºC.

Description of key information

The vapour pressure of the test item has been determined to be 7.9 x 10-6 Pa at 25°C.

Key value for chemical safety assessment

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

Additional information

The vapour pressure was determined using a vapour pressure balance with measurents being made at several temperatures and linear regression analysis used to calculate the vapour pressure at 25°C. Testing was conducted using a procedure designed to be compatible with Method A4 Vapour Pressure of Commission Regulation (EC) No 440/2008 of 30 May 2008, and Method 104 specified in the OECD Guidelines for Testing of Chemicals, 23 March 2006.