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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:
From December 13, 2018 to December 20, 2018
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
guideline study
Qualifier:
according to
Guideline:
OECD Guideline 104 (Vapour Pressure Curve)
Deviations:
no
Qualifier:
according to
Guideline:
EU Method A.4 (Vapour Pressure)
Deviations:
no
GLP compliance:
yes
Type of method:
effusion method: vapour pressure balance
Key result
Test no.:
#1
Temp.:
ca. 25 °C
Vapour pressure:
ca. 0.008 Pa

Results

Recorded temperatures, mass differences and the resulting calculated values of vapor pressure are shown in the following tables:

 

Run 1

 

Table1– Vapor Pressure Data

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

35

308.15

0.003245

4.26

4.260e-09

0.00591

-2.22841

36

309.15

0.003235

5.14

5.140e-09

0.00714

-2.14630

37

310.15

0.003224

3.21

3.210e-09

0.00446

-2.35067

38

311.15

0.003214

4.00

4.000e-09

0.00555

-2.25571

39

312.15

0.003204

4.81

4.810e-09

0.00668

-2.17522

40

313.15

0.003193

4.36

4.360e-09

0.00605

-2.21824

41

314.15

0.003183

6.36

6.360e-09

0.00883

-2.05404

42

315.15

0.003173

7.47

7.470e-09

0.01037

-1.98422

43

316.15

0.003163

8.67

8.670e-09

0.01204

-1.91937

44

317.15

0.003153

8.56

8.560e-09

0.01188

-1.92518

45

318.15

0.003143

10.07

1.007e-08

0.01398

-1.85449

A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 1 gives the following statistical data using an unweighted least squares treatment (For graph run, kindly refer to the attached background material section of the IUCLID).

Slope:

-4.17 x 103

Standard error in slope:

788

 

Intercept:

11.2

Standard error in intercept:

2.52

The results obtained indicate the following vapor pressure relationship:

Log10(Vp (Pa)) = -4.17 x 103/temp(K) + 11.2

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -2.77.

 

Run 2

 

Table2– Vapor Pressure Data

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

35

308.15

0.003245

4.79

4.790e-09

0.00665

-2.17718

36

309.15

0.003235

4.79

4.790e-09

0.00665

-2.17718

37

310.15

0.003224

4.28

4.280e-09

0.00594

-2.22621

38

311.15

0.003214

4.37

4.370e-09

0.00607

-2.21681

39

312.15

0.003204

5.82

5.820e-09

0.00808

-2.09259

40

313.15

0.003193

5.05

5.050e-09

0.00701

-2.15428

41

314.15

0.003183

6.74

6.740e-09

0.00936

-2.02872

42

315.15

0.003173

6.47

6.470e-09

0.00898

-2.04672

43

316.15

0.003163

8.11

8.110e-09

0.01126

-1.94846

44

317.15

0.003153

8.20

8.200e-09

0.01138

-1.94386

45

318.15

0.003143

9.50

9.500e-09

0.01319

-1.87976

A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 2 gives the following statistical data using an unweighted least squares treatment (For graph run, kindly refer to the attached background material section of the IUCLID).

Slope:

-3.24 x 103

Standard error in slope:

477

 

Intercept:

8.28

Standard error in intercept:

1.52

The results obtained indicate the following vapor pressure relationship:

Log10(Vp (Pa)) = -3.24 x 103/temp(K) + 8.28

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -2.60.

 

Run 3

 

Table3– Vapor Pressure Data

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

35

308.15

0.003245

4.36

4.360e-09

0.00605

-2.21824

36

309.15

0.003235

3.33

3.330e-09

0.00462

-2.33536

37

310.15

0.003224

3.41

3.410e-09

0.00473

-2.32514

38

311.15

0.003214

4.78

4.780e-09

0.00664

-2.17783

39

312.15

0.003204

5.34

5.340e-09

0.00741

-2.13018

40

313.15

0.003193

4.79

4.790e-09

0.00665

-2.17718

41

314.15

0.003183

6.44

6.440e-09

0.00894

-2.04866

42

315.15

0.003173

6.38

6.380e-09

0.00886

-2.05257

43

316.15

0.003163

8.67

8.670e-09

0.01204

-1.91937

44

317.15

0.003153

9.14

9.140e-09

0.01269

-1.89654

45

318.15

0.003143

9.66

9.660e-09

0.01341

-1.87257

A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 3 gives the following statistical data using an unweighted least squares treatment (For graph run, kindly refer to the attached background material section of the IUCLID).

Slope:

-4.47 x 103

Standard error in slope:

573

 

Intercept:

12.2

Standard error in intercept:

1.83

The results obtained indicate the following vapor pressure relationship:

Log10(Vp (Pa)) = -4.47 x 103/temp(K) + 12.2

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -2.82.

Run 4

 

Table4– Vapor Pressure Data

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

35

308.15

0.003245

3.18

3.180e-09

0.00441

-2.35556

36

309.15

0.003235

4.29

4.290e-09

0.00596

-2.22475

37

310.15

0.003224

3.84

3.840e-09

0.00533

-2.27327

38

311.15

0.003214

4.25

4.250e-09

0.00590

-2.22915

39

312.15

0.003204

4.87

4.870e-09

0.00676

-2.17005

40

313.15

0.003193

5.63

5.630e-09

0.00782

-2.10679

41

314.15

0.003183

6.04

6.040e-09

0.00839

-2.07624

42

315.15

0.003173

6.13

6.130e-09

0.00851

-2.07007

43

316.15

0.003163

7.59

7.590e-09

0.01054

-1.97716

44

317.15

0.003153

8.67

8.670e-09

0.01204

-1.91937

45

318.15

0.003143

8.99

8.990e-09

0.01248

-1.90379

A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 4 gives the following statistical data using an unweighted least squares treatment (For graph run, kindly refer to the attached background material section of the IUCLID).

Slope:

-4.25 x 103

Standard error in slope:

294

 

Intercept:

11.5

Standard error in intercept:

0.939

The results obtained indicate the following vapor pressure relationship:

Log10(Vp (Pa)) = -4.25 x 103/temp(K) + 11.5

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -2.80.

 

Run 5

Table5– Vapor Pressure Data

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

35

308.15

0.003245

3.71

3.710e-09

0.00515

-2.28819

36

309.15

0.003235

4.06

4.060e-09

0.00564

-2.24872

37

310.15

0.003224

3.98

3.980e-09

0.00553

-2.25727

38

311.15

0.003214

4.03

4.030e-09

0.00559

-2.25259

39

312.15

0.003204

3.99

3.990e-09

0.00554

-2.25649

40

313.15

0.003193

5.49

5.490e-09

0.00762

-2.11805

41

314.15

0.003183

5.11

5.110e-09

0.00709

-2.14935

42

315.15

0.003173

6.46

6.460e-09

0.00897

-2.04721

43

316.15

0.003163

7.71

7.710e-09

0.01070

-1.97062

44

317.15

0.003153

7.58

7.580e-09

0.01052

-1.97798

45

318.15

0.003143

9.13

9.130e-09

0.01267

-1.89722

A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 5 gives the following statistical data using an unweighted least squares treatment (For graph run, kindly refer to the attached background material section of the IUCLID).

Slope:

-3.92 x 103

Standard error in slope:

434

 

Intercept:

10.4

Standard error in intercept:

1.39

The results obtained indicate the following vapor pressure relationship:

Log10(Vp (Pa)) = -3.92 x 103/temp(K) + 10.4

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -2.76.

 

Run 6

 

Table6– Vapor Pressure Data

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

35

308.15

0.003245

3.64

3.640e-09

0.00505

-2.29671

36

309.15

0.003235

4.09

4.090e-09

0.00568

-2.24565

37

310.15

0.003224

3.66

3.660e-09

0.00508

-2.29414

38

311.15

0.003214

4.46

4.460e-09

0.00619

-2.20831

39

312.15

0.003204

4.11

4.110e-09

0.00571

-2.24336

40

313.15

0.003193

5.42

5.420e-09

0.00752

-2.12378

41

314.15

0.003183

5.73

5.730e-09

0.00795

-2.09963

42

315.15

0.003173

5.74

5.740e-09

0.00797

-2.09854

43

316.15

0.003163

5.97

5.970e-09

0.00829

-2.08145

44

317.15

0.003153

8.17

8.170e-09

0.01134

-1.94539

45

318.15

0.003143

8.89

8.890e-09

0.01234

-1.90868

A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 6 gives the following statistical data using an unweighted least squares treatment (For graph run, kindly refer to the attached background material section of the IUCLID).

Slope:

-3.68 x 103

Standard error in slope:

414

 

Intercept:

9.61

Standard error in intercept:

1.32

 

The results obtained indicate the following vapor pressure relationship:

Log10(Vp (Pa)) = -3.68 x 103/temp(K) + 9.61

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -2.73.

 

Run 7

 

Table7– Vapor Pressure Data

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

35

308.15

0.003245

3.97

3.970e-09

0.00551

-2.25885

36

309.15

0.003235

4.03

4.030e-09

0.00559

-2.25259

37

310.15

0.003224

3.74

3.740e-09

0.00519

-2.28483

38

311.15

0.003214

4.33

4.330e-09

0.00601

-2.22113

39

312.15

0.003204

4.14

4.140e-09

0.00575

-2.24033

40

313.15

0.003193

4.91

4.910e-09

0.00682

-2.16622

41

314.15

0.003183

5.29

5.290e-09

0.00734

-2.13430

42

315.15

0.003173

6.69

6.690e-09

0.00929

-2.03198

43

316.15

0.003163

7.01

7.010e-09

0.00973

-2.01189

44

317.15

0.003153

8.59

8.590e-09

0.01193

-1.92336

45

318.15

0.003143

7.14

7.140e-09

0.00991

-2.00393

A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 7 gives the following statistical data using an unweighted least squares treatment (For graph run, kindly refer to the attached background material section of the IUCLID).

Slope:

-3.46 x 103

Standard error in slope:

452

 

Intercept:

8.91

Standard error in intercept:

1.45

 

The results obtained indicate the following vapor pressure relationship:

Log10(Vp (Pa)) = -3.46 x 103/temp(K) + 8.91

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -2.69.

The values of vapor pressure at 25 °C, extrapolated from each graph, are summarized in the following table:

Table 8 Summary of Vapor Pressure Data

Run

Log10[Vp(25 ºC)]

1

-2.07

2

-2.09

3

-2.08

4

-2.11

5

-2.11

6

-2.12

7

-2.12

Mean

-2.10

Vapor Pressure

7.93 x 10-3Pa

 

Discussion

The test substance did not change in appearance under the conditions used in the determination. A total of 7 runs were completed for the main sequence. All 7 runs were used as equilibrium had been reached.The results may represent rounded values obtained by calculations based on the exact raw data.

 

Conclusion

The vapor pressure of the test substance has been determined to be 7.9 x 10-3 Pa at 25 ºC.

Conclusions:
Under the study conditions, the vapoure pressure of the test substance was determined to be 7.9 x 10-3 Pa at 25°C.
Executive summary:

A study was conducted to determine the vapour pressure of the test substance using the vapoure pressure balance method, according to OECD Guideline 104 and EU Method A4, in compliance with GLP. A total of 7 runs were completed for the main sequence and all 7 runs were used as equilibrium had been reached. Under the study conditions, the vapoure pressure of the test substance was determined to be 7.9 x 10-3 Pa at 25°C (Ford, 2019).

Description of key information

The vapoure pressure of the test substance was determined using the vapoure pressure balance method, according to OECD Guideline 104 and EU Method A.4 (Ford, 2019).

Key value for chemical safety assessment

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

Additional information