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Administrative data

Endpoint:
vapour pressure
Type of information:
experimental study
Adequacy of study:
key study
Study period:
15 January 2009 to 05 February 2009
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:
2009
Report date:
2009

Materials and methods

Test guideline
Qualifier:
according to guideline
Guideline:
EU Method A.4 (Vapour Pressure)
Deviations:
no
GLP compliance:
yes (incl. QA statement)
Remarks:
Date of inspection: 19 August 2008; Date of signature: 29 January 2009
Type of method:
effusion method: vapour pressure balance

Test material

Results and discussion

Vapour pressure
Temp.:
25 °C
Vapour pressure:
0 Pa

Any other information on results incl. tables

Please see attachment 2 below for graphs.

Run1

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

128

401.15

0.002492833

6.49

6.490E-09

0.009009783

-2.045285682

129

402.15

0.002486634

7.11

7.110E-09

0.009870502

-2.005660778

130

403.15

0.002480466

8.34

8.340E-09

0.011578057

-1.936364328

131

404.15

0.002474329

9.48

9.480E-09

0.013160669

-1.880722041

132

405.15

0.002468222

10.40

1.040E-08

0.014437864

-1.840497039

133

406.15

0.002462145

11.84

1.184E-08

0.016436953

-1.784178676

134

407.15

0.002456097

13.70

1.370E-08

0.019019110

-1.720809812

135

408.15

0.002450080

15.04

1.504E-08

0.020879373

-1.680282542

 

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                                       -8724.859
Standard deviation in slope     185.348

Intercept                                        19.700
Standard deviation in intercept    0.458

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -8724.859 /temp(K) + 19.700

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -9.563.

Run 2

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

128

401.15

0.002492833

6.39

6.390E-09

0.008870957

-2.052029521

129

402.15

0.002486634

6.90

6.900E-09

0.009578968

-2.018681288

130

403.15

0.002480466

7.93

7.930E-09

0.011008872

-1.958257191

131

404.15

0.002474329

9.68

9.680E-09

0.013438320

-1.871655021

132

405.15

0.002468222

11.64

1.164E-08

0.016159302

-1.791577398

133

406.15

0.002462145

11.43

1.143E-08

0.015867768

-1.799484148

134

407.15

0.002456097

12.57

1.257E-08

0.017450380

-1.758195101

135

408.15

0.002450080

15.14

1.514E-08

0.021018199

-1.677404504

 

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                                       -8737.351
Standard deviation in slope     653.118

Intercept                                        19.727
Standard deviation in intercept    1.614

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -8737.351 /temp(K) + 19.727

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -9.578.

Run 3

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

128

401.15

0.002492833

6.18

6.180E-09

0.008579423

-2.066541904

129

402.15

0.002486634

6.39

6.390E-09

0.008870957

-2.052029521

130

403.15

0.002480466

7.11

7.110E-09

0.009870502

-2.005660778

131

404.15

0.002474329

8.03

8.030E-09

0.011147697

-1.952814833

132

405.15

0.002468222

9.89

9.890E-09

0.013729854

-1.862334087

133

406.15

0.002462145

11.74

1.174E-08

0.016298128

-1.787862282

134

407.15

0.002456097

11.95

1.195E-08

0.016589662

-1.780162473

135

408.15

0.002450080

13.49

1.349E-08

0.018727576

-1.727518429

 

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                                       -8724.455
Standard deviation in slope     610.393

Intercept                                        19.657
Standard deviation in intercept    1.509

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -8724.455 /temp(K) + 19.657

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -9.605.

Run 4

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

128

401.15

0.002492833

5.36

5.360E-09

0.007441053

-2.128365589

129

402.15

0.002486634

6.08

6.080E-09

0.008440598

-2.073626799

130

403.15

0.002480466

6.39

6.390E-09

0.008870957

-2.052029521

131

404.15

0.002474329

7.93

7.930E-09

0.011008872

-1.958257191

132

405.15

0.002468222

8.45

8.450E-09

0.011730765

-1.930673670

133

406.15

0.002462145

9.06

9.060E-09

0.012577601

-1.900402181

134

407.15

0.002456097

11.43

1.143E-08

0.015867768

-1.799484148

135

408.15

0.002450080

11.84

1.184E-08

0.016436953

-1.784178676

 

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                                       -8307.816
Standard deviation in slope     498.441

Intercept                                        18.578
Standard deviation in intercept    1.232

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -8307.816 /temp(K) + 18.578

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -9.286.

Run 5

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapour Pressure (Pa)

Log10Vp

128

401.15

0.002492833

5.36

5.360E-09

0.007441053

-2.128365589

129

402.15

0.002486634

5.97

5.970E-09

0.008287890

-2.081556048

130

403.15

0.002480466

6.08

6.080E-09

0.008440598

-2.073626799

131

404.15

0.002474329

7.00

7.000E-09

0.009717793

-2.012432339

132

405.15

0.002468222

8.65

8.650E-09

0.012008416

-1.920514271

133

406.15

0.002462145

9.17

9.170E-09

0.012730309

-1.895161043

134

407.15

0.002456097

10.61

1.061E-08

0.014729398

-1.831814995

135

408.15

0.002450080

12.26

1.226E-08

0.017020021

-1.769039909

 

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                                       -8555.661
Standard deviation in slope     528.799

Intercept                                        19.180
Standard deviation in intercept    1.307

The results obtained indicate the following vapour pressure relationship:

Log10(Vp (Pa)) = -8555.661 /temp(K) + 19.180

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -9.516.


Summary of Results

Run

Log10[Vp(25ºC)]

1

-9.563

2

-9.578

3

-9.605

4

-9.286

5

-9.516

Mean

-9.510

Vapour Pressure

3.1 x 10-10Pa


Applicant's summary and conclusion

Conclusions:
The vapour pressure of the test material has been determined to be 3.1 x 10-10 Pa at 25ºC.
Executive summary:

The vapour pressure was determined using a vapour pressure balance system withasurents being made at several temperatures and linear regression analysis used to calculate the vapour pressure at 25°C. Testing was conducted using Method A4 Vapour Pressure of Council Regulation (EC) No 440/2208 of 30 May 2008.

The vapour pressure of the test material has been determined to be 3.1 x 10-10 Pa at 25ºC.