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EC number: 482-410-3 | CAS number: -
Vapour pressure
Administrative data
- Endpoint:
- vapour pressure
- Type of information:
- experimental study
- Adequacy of study:
- key study
- Study period:
- Testing was conducted between 20 February 2008 and 03 March 2008.
- 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:
- 2 008
- Report date:
- 2008
Materials and methods
Test guideline
- Qualifier:
- according to guideline
- Guideline:
- EU Method A.4 (Vapour Pressure)
- Deviations:
- yes
- GLP compliance:
- yes (incl. QA statement)
- Remarks:
- Work described was performed in compliance with UK GLP standards (Schedule 1, Good Laboratory Practice Regulations 1999 (SI 1999/3106 as amended by SI 2004/0994)). Date of inspection: 21/08/2007. Date of signature: 29/04/2008
- Type of method:
- effusion method: vapour pressure balance
Test material
- Details on test material:
- Sponsor's identification :Hatcol 1760
Description : clear colourless slightly viscous liquid
Purity : >99%
Date received : 04 February 2008
Storage conditions :room temperature, in the dark
Constituent 1
Results and discussion
Vapour pressure
- Temp.:
- 25 °C
- Vapour pressure:
- 0 Pa
Transition / decomposition
- Transition / decomposition:
- no
Any other information on results incl. tables
Run 1
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 -7026.925
Standard deviation in slope 212.789
Intercept 16.320
Standard deviation in intercept 0.533
The results obtained indicate the following vapour pressure relationship:
Log10(Vp (Pa)) = -7026.925/temp(K) + 16.320.
The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -7.248.
Run 2
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 -8784.915
Standard deviation in slope 278.657
Intercept 20.666
Standard deviation in intercept 0.699
The results obtained indicate the following vapour pressure relationship:
Log10(Vp (Pa)) = -8784.915/temp(K) + 20.666.
The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -8.799.
Run 3
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 -8480.343
Standard deviation in slope 469.643
Intercept 19.878
Standard deviation in intercept 1.177
The results obtained indicate the following vapour pressure relationship:
Log10(Vp (Pa)) = -8480.343/temp(K) + 19.878.
The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -8.565.
Run 4
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 -8080.251
Standard deviation in slope 238.093
Intercept 18.860
Standard deviation in intercept 0.598
The results obtained indicate the following vapour pressure relationship:
Log10(Vp (Pa)) = -8080.251/temp(K) + 18.860.
The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -8.242.
Run 5
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 -7603.848
Standard deviation in slope 399.061
Intercept 17.656
Standard deviation in intercept 1.002
The results obtained indicate the following vapour pressure relationship:
Log10(Vp (Pa)) = -7603.848/temp(K) + 17.656.
The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -7.847.
Run 6
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 -7039.013
Standard deviation in slope 313.177
Intercept 16.190
Standard deviation in intercept 0.786
The results obtained indicate the following vapour pressure relationship:
Log10(Vp (Pa)) = -7039.013/temp(K) + 16.190.
The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -7.419.
Run 7
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 -6714.109
Standard deviation in slope 287.395
Intercept 15.359
Standard deviation in intercept 0.721
The results obtained indicate the following vapour pressure relationship:
Log10(Vp (Pa)) = -6714.109/temp(K) + 15.359.
The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -7.160.
Summary of Results
Run |
Log10[Vp(25ºC)] |
1 |
-7.248 |
2 |
-8.799 |
3 |
-8.565 |
4 |
-8.242 |
5 |
-7.847 |
6 |
-7.419 |
7 |
-7.160 |
Mean |
-7.897 |
Vapour Pressure |
1.268 x 10-8Pa |
The test material did not change in appearance under the conditions used in the determination.
Applicant's summary and conclusion
- Conclusions:
- The vapour pressure of the test material has been determined to be 1.3 x 10-8 Pa at 25ºC.
- Executive summary:
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 Method A4 of Commission Directive 92/69/EEC (which constitutes Annex V of Council Directive 67/548/EEC).
Conclusion
The vapour pressure of the test material has been determined to be 1.3 x 10-8Pa at 25ºC.
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