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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:
The study was conducted between 06 August 2014 and 08 August 2014.
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
guideline study
Remarks:
The study is considered to be a reliability 1 as it has been conducted according to OECD Test Guideline 104 using a vapor pressure balance method and in compliance with GLP.
Qualifier:
according to guideline
Guideline:
OECD Guideline 104 (Vapour Pressure Curve)
Deviations:
no
GLP compliance:
yes (incl. QA statement)
Type of method:
effusion method: vapour pressure balance
Key result
Temp.:
25 °C
Vapour pressure:
15 Pa

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

 

Run 1

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

-10

263.15

0.003800114

176.15

1.762E-07

0.244541329

-0.611647731

-9

264.15

0.003785728

222.56

2.226E-07

0.308970300

-0.510083266

-8

265.15

0.003771450

254.43

2.544E-07

0.353214025

-0.451962061

-7

266.15

0.003757280

288.12

2.881E-07

0.399984376

-0.397956973

-6

267.15

0.003743215

320.10

3.201E-07

0.444380809

-0.352244705

-5

268.15

0.003729256

364.38

3.644E-07

0.505852794

-0.295975847

-4

269.15

0.003715400

413.27

4.133E-07

0.573724639

-0.241296499

-3

270.15

0.003701647

471.23

4.712E-07

0.654187968

-0.184297448

-2

271.15

0.003687996

517.69

5.177E-07

0.718686351

-0.143460603

-1

272.15

0.003674444

591.30

5.913E-07

0.820875890

-0.085722500

0

273.15

0.003660992

646.56

6.466E-07

0.897590928

-0.046921546

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.

 

Slope:

-3.91 x 103

Standard error in slope:

100

Intercept:

14.3

Standard error in intercept:

0.375

 

The results obtained indicate the following vapor pressure relationship:

Log10(Vp (Pa)) = -3.91 x 103/temp(K) + 14.3

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of 1.17.

Run 2

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

-10

263.15

0.003800114

183.59

1.836E-07

0.254869955

-0.593681357

-9

264.15

0.003785728

213.94

2.139E-07

0.297003531

-0.527238387

-8

265.15

0.003771450

241.73

2.417E-07

0.335583171

-0.474199827

-7

266.15

0.003757280

286.05

2.861E-07

0.397110686

-0.401088427

-6

267.15

0.003743215

318.86

3.189E-07

0.442659371

-0.353930337

-5

268.15

0.003729256

369.81

3.698E-07

0.513391025

-0.289551728

-4

269.15

0.003715400

414.97

4.150E-07

0.576084675

-0.239513678

-3

270.15

0.003701647

436.01

4.360E-07

0.605293585

-0.218033929

-2

271.15

0.003687996

508.52

5.085E-07

0.705956042

-0.151222340

-1

272.15

0.003674444

587.24

5.872E-07

0.815239570

-0.088714749

0

273.15

0.003660992

657.96

6.580E-07

0.913417048

-0.039330887

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.

 

Slope:

-3.91 x 103

Standard error in slope:

80.4

Intercept:

14.3

Standard error in intercept:

0.300

 

The results obtained indicate the following vapor pressure relationship:

 

Log10(Vp (Pa)) = -3.91 x 103/temp(K) + 14.3

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of 1.16.

Run 3

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

-10

263.15

0.003800114

191.04

1.910E-07

0.265212464

-0.576406069

-9

264.15

0.003785728

212.57

2.126E-07

0.295101620

-0.530028406

-8

265.15

0.003771450

246.60

2.466E-07

0.342343979

-0.465537306

-7

266.15

0.003757280

275.92

2.759E-07

0.383047650

-0.416747197

-6

267.15

0.003743215

306.76

3.068E-07

0.425861471

-0.370731650

-5

268.15

0.003729256

365.66

3.657E-07

0.507629762

-0.294452924

-4

269.15

0.003715400

409.86

4.099E-07

0.568990686

-0.244894843

-3

270.15

0.003701647

428.91

4.289E-07

0.595436966

-0.225164207

-2

271.15

0.003687996

522.74

5.227E-07

0.725697045

-0.139244645

-1

272.15

0.003674444

559.30

5.593E-07

0.776451692

-0.109885559

0

273.15

0.003660992

659.97

6.600E-07

0.916207443

-0.038006184

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.

 

Slope:

-3.83 x 103

Standard error in slope:

81.8

Intercept:

14.0

Standard error in intercept:

0.305

 

The results obtained indicate the following vapor pressure relationship:

 

Log10(Vp (Pa)) = -3.83 x 103/temp(K) + 14.0

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of 1.13.

Run 4

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

-10

263.15

0.003800114

190.77

1.908E-07

0.264837635

-0.577020299

-9

264.15

0.003785728

198.09

1.981E-07

0.274999670

-0.560667827

-8

265.15

0.003771450

244.26

2.443E-07

0.339095459

-0.469678026

-7

266.15

0.003757280

295.63

2.956E-07

0.410410180

-0.386781875

-6

267.15

0.003743215

300.12

3.001E-07

0.416643450

-0.380235441

-5

268.15

0.003729256

353.44

3.534E-07

0.490665271

-0.309214680

-4

269.15

0.003715400

411.67

4.117E-07

0.571503429

-0.242981159

-3

270.15

0.003701647

457.92

4.579E-07

0.635710278

-0.196740767

-2

271.15

0.003687996

512.68

5.127E-07

0.711731188

-0.147684003

-1

272.15

0.003674444

581.73

5.817E-07

0.807590278

-0.092808918

0

273.15

0.003660992

660.11

6.601E-07

0.916401799

-0.037914067

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.

 

Slope:

-3.95 x 103

Standard error in slope:

113

Intercept:

14.4

Standard error in intercept:

0.421

 

The results obtained indicate the following vapor pressure relationship:

 

Log10(Vp (Pa)) = -3.95 x 103/temp(K) + 14.4

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of 1.18.

Run 5

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

-10

263.15

0.003800114

196.99

1.970E-07

0.273472589

-0.563086199

-9

264.15

0.003785728

196.92

1.969E-07

0.273375411

-0.563240552

-8

265.15

0.003771450

240.41

2.404E-07

0.333750673

-0.476577850

-7

266.15

0.003757280

288.66

2.887E-07

0.400734034

-0.397143771

-6

267.15

0.003743215

322.87

3.229E-07

0.448226279

-0.348502685

-5

268.15

0.003729256

368.04

3.680E-07

0.510933811

-0.291635357

-4

269.15

0.003715400

412.45

4.125E-07

0.572586269

-0.242159071

-3

270.15

0.003701647

463.51

4.635E-07

0.643470631

-0.191471270

-2

271.15

0.003687996

532.29

5.323E-07

0.738954892

-0.131382071

-1

272.15

0.003674444

587.03

5.870E-07

0.814948036

-0.088870082

0

273.15

0.003660992

665.47

6.655E-07

0.923842852

-0.034401897

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.

 

Slope:

-3.98 x 103

Standard error in slope:

111

Intercept:

14.6

Standard error in intercept:

0.413

 

The results obtained indicate the following vapor pressure relationship:

 

Log10(Vp (Pa)) = -3.98 x 103/temp(K) + 14.6

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of 1.19.

Run 6

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

-10

263.15

0.003800114

194.56

1.946E-07

0.270099126

-0.568476821

-9

264.15

0.003785728

204.88

2.049E-07

0.284425930

-0.546030813

-8

265.15

0.003771450

238.97

2.390E-07

0.331751584

-0.479186995

-7

266.15

0.003757280

286.60

2.866E-07

0.397874227

-0.400254193

-6

267.15

0.003743215

304.99

3.050E-07

0.423404258

-0.373244779

-5

268.15

0.003729256

353.82

3.538E-07

0.491192808

-0.308748001

-4

269.15

0.003715400

410.44

4.104E-07

0.569795874

-0.244280700

-3

270.15

0.003701647

468.82

4.688E-07

0.650842271

-0.186524248

-2

271.15

0.003687996

525.72

5.257E-07

0.729834049

-0.136775880

-1

272.15

0.003674444

596.89

5.969E-07

0.828636242

-0.081636076

0

273.15

0.003660992

673.41

6.734E-07

0.934865607

-0.029250817

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.

 

Slope:

-4.01 x 103

Standard error in slope:

78.6

Intercept:

14.7

Standard error in intercept:

0.293

 

The results obtained indicate the following vapor pressure relationship:

 

Log10(Vp (Pa)) = -4.01 x 103/temp(K) + 14.7

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of 1.20.

Summary of Results

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

Run

Log10[Vp(25 ºC)]

1

1.17

2

1.16

3

1.13

4

1.18

5

1.19

6

1.20

Mean

1.17

Vapor Pressure

14.9 Pa

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

Conclusions:
The vapor pressure of the test item has been determined to be 15 Pa at 25 ºC.
Executive summary:

The vapor pressure of the test substance, IFF 215 (Floriane) was determined to be 15 Pa at 25 ºC according to OECD Test Guideline 104 using a vapor pressure balance method.

Description of key information

The vapor pressure of the test substance, IFF 215 (Floriane) was determined to be 15 Pa at 25 ºC according to OECD Test Guideline 104 using a vapor pressure balance method.

Key value for chemical safety assessment

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

Additional information