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Administrative data

Endpoint:
vapour pressure
Type of information:
experimental study
Adequacy of study:
key study
Study period:
June 2005
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
other: Study conducted to GLP and in compliance with agreed protocols, with no or minor deviations from standard test guidelines and/or minor methodological deficiencies, which do not effect the quality of the relevant results.

Data source

Reference
Reference Type:
study report
Title:
Unnamed
Year:
2006
Report date:
2006

Materials and methods

Test guideline
Qualifier:
according to guideline
Guideline:
EU Method A.4 (Vapour Pressure)
Deviations:
no
GLP compliance:
yes (incl. QA statement)
Type of method:
effusion method: vapour pressure balance

Test material

Constituent 1
Chemical structure
Reference substance name:
-
EC Number:
481-740-5
EC Name:
-
Cas Number:
848301-67-7
Molecular formula:
main general molecular formula: CnH(2n+2)
IUPAC Name:
2,2,3,4,4,5,6-heptamethylheptane; 2,2,8-trimethyltetradecane; 2,2-dimethyldodecane; 2,2-dimethyltetradecane; 2-methyltetradecane; 3,5,7-triethyldecane; 4-ethyldodecane; 4-propyldodecane; 5,5-dipropylnonane; 5,7-dimethyl-4-propylundecane; 5-butyl-5-propyldecane; 5-butyl-5-propylnonane; 5-ethyl-5-propylnonane; 5-ethyl-7-methyl-4-propylundecane; 5-propyltridecane; 6-butyl-6-methyldodecane; 6-butyldodecane; 6-butylundecane; 7-methyl-4-propylundecane; 8-methyl-5-propyltridecane; 9-butyl-6-propylhexadecane; heptadecane; hexacosane; hexadecane; octane; pentadecane; tetradecane
Details on test material:
- Name of test material (as cited in study report): Distillates (Fischer-Tropsch), C8-26 branched and linear
- Substance type: Organic
- Physical state: Clear colorless liquid
- Analytical purity: 100%
- Impurities (identity and concentrations): Information not available
- Composition of test material, percentage of components: Information not available
- Isomers composition: Information not available
- Purity test date: Information not available
- Lot/batch No.: Information not available
- Expiration date of the lot/batch: Information not available
- Stability under test conditions: Stable under normal temperature and pressure
- Storage condition of test material: Room temperature, in the dark

Results and discussion

Vapour pressure
Temp.:
25 °C
Vapour pressure:
0.54 Pa
Transition / decomposition
Transition / decomposition:
no

Any other information on results incl. tables

Calculation

1) The vapour pressure is related to the observed mass difference by the following equation:

Vp = ρm g / A

where:

Vp = vapour pressure (Pa)

ρm =mass difference (kg)

g = acceleration due to gravity (9.813ms-2)

A = area of the orifice (7.06858x10-6m2)

2) The vapour pressure is related to temperature by the following equation:

Log10[Vp (Pa)] = [slope / temperature (K)]+ intercept

A plot ofLog10[Vp (Pa)] versus reciprocal temperature [1/T(K)] therefore gives a straight line graph.

The vapour pressure of the sample was measured over a range of temperatures to enable extrapolation to 298.15 K.

Results

Run 1:

Temperature (°C) Temperature (K) Reciprocal Temperature (K-1) Mass Difference (µg) Mass Difference (kg) Vapour Pressure (Pa)  Log10Vp
8 281.15 0.003556820 112.82 1,13E-04 0.156623064 -0.805144283
9 282.15 0.003544214 120.63 1,21E-04 0.167465345 -0.776075051
10 283.15 0.003531697 129.40 1,29E-04 0.179640352 -0.745596102
11 284.15 0.003519268 142.34 1,42E-04 0.197604387 -0.704203417
12 285.15 0.003506926 145.44 1,45E-04 0.201907982 -0.694846513
13 286.15 0.003494671 165.01 1,65E-04 0.229076155 -0.640020114
14 287.15 0.003482500 184.26 1,84E-04 0.255800087 -0.592099312
15 288.15 0.003470415 193.88 1,94E-04 0.269155112 -0.569997368

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 -2788.061; Standard deviation in slope 136.534; Intercept 9.104; Standard deviation in intercept 0.480.

The results obtained indicate the following vapour pressure relationship: Log10(Vp (Pa)) = -2788.061/temp(K) + 9.104.

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.247.

Run 2:

Temperature (°C) Temperature (K) Reciprocal Temperature (K-1) Mass Difference (µg) Mass Difference (kg) Vapour Pressure (Pa)  Log10Vp
7 280.15 0.003569516 99.35 9,94E-05 0.137923253 -0.860362507
8 281.15 0.003556820 106.19 1,06E-04 0.147418926 -0.831446758
9 282.15 0.003544214 115.50 1,16E-04 0.160343591 -0.794948395
10 283.15 0.003531697 123.41 1,23E-04 0.171324697 -0.766180026
11 284.15 0.003519268 137.74 1,38E-04 0.191218409 -0.718470300
12 285.15 0.003506926 147.47 1,48E-04 0.204726142 -0.688826698
13 286.15 0.003494671 158.70 1,59E-04 0.220316259 -0.656953452
14 287.15 0.003482500 172.28 1,72E-04 0.239168778 -0.621295516
15 288.15 0.003470415 190.57 1,91E-04 0.264559984 -0.577475845

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 -2845.534; Standard deviation in slope 53.393; Intercept 9.291; Standard deviation in intercept 0.188.

The results obtained indicate the following vapour pressure relationship: Log10(Vp (Pa)) = -2845.534/temp(K) + 9.291.

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.247.

Run 3:

Temperature (°C) Temperature (K) Reciprocal Temperature (K-1) Mass Difference (µg) Mass Difference (kg) Vapour Pressure (Pa)  Log10Vp
7 280.15 0.003569516 89.30 8,93E-05 0.123971279 -0.906678920
8 281.15 0.003556820 95.82 9,58E-05 0.133022709 -0.876074212
9 282.15 0.003544214 105.02 1,05E-04 0.145794666 -0.836258365
10 283.15 0.003531697 123.52 1,24E-04 0.171477406 -0.765793096
11 284.15 0.003519268 119.67 1,20E-04 0.166132619 -0.779545088
12 285.15 0.003506926 130.79 1,31E-04 0.181570028 -0.740955839
14 287.15 0.003482500 163.19 1,63E-04 0.226549529 -0.644836836
15 288.15 0.003470415 178.06 1,78E-04 0.247192899 -0.606964010

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 -2983.149; Standard deviation in slope 187.129; Intercept 9.739; Standard deviation in intercept 0.659.

The results obtained indicate the following vapour pressure relationship: Log10(Vp (Pa)) = -2983.149/temp(K) + 9.739

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.267.

Run 4:

Temperature (°C) Temperature (K) Reciprocal Temperature (K-1) Mass Difference (µg) Mass Difference (kg) Vapour Pressure (Pa)  Log10Vp
8 281.15 0.003556820 88.55 8,86E-05 0.122930086 -0.910341813
9 282.15 0.003544214 98.07 9,81E-05 0.136146285 -0.865994203
10 283.15 0.003531697 107.48 1,08E-04 0.149209776 -0.826202721
11 284.15 0.003519268 115.71 1,16E-04 0.160635125 -0.794159485
12 285.15 0.003506926 130.36 1,30E-04 0.180973078 -0.742386027
13 286.15 0.003494671 140.84 1,41E-04 0.195522003 -0.708804362
14 287.15 0.003482500 163.62 1,64E-04 0.227146479 -0.643693990
15 288.15 0.003470415 168.86 1,69E-04 0.234420942 -0.630003594

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 -3290.187; Standard deviation in slope 98.126; Intercept 10.795; Standard deviation in intercept 0.345.

The results obtained indicate the following vapour pressure relationship: Log10(Vp (Pa)) = -3290.187/temp(K) + 10.795

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.240.

Run 5:

Temperature (°C) Temperature (K) Reciprocal Temperature (K-1) Mass Difference (µg) Mass Difference (kg) Vapour Pressure (Pa)  Log10Vp
6 279.15 0.003582303 76.78 7,68E-05 0.106590311 -0.972282271
7 280.15 0.003569516 85.87 8,59E-05 0.119209560 -0.923688916
8 281.15 0.003556820 91.54 9,15E-05 0.127080972 -0.895919471
9 282.15 0.003544214 100.31 1,00E-04 0.139255979 -0.856186148
10 283.15 0.003531697 105.23 1,05E-04 0.146086200 -0.835390808
11 284.15 0.003519268 114.53 1,15E-04 0.158996982 -0.798611118
12 285.15 0.003506926 121.27 1,21E-04 0.168353829 -0.773777001
13 286.15 0.003494671 134.64 1,35E-04 0.186914815 -0.728356276
14 287.15 0.003482500 145.65 1,46E-04 0.202199515 -0.694219890
15 288.15 0.003470415 155.60 1,56E-04 0.216012664 -0.665520786

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 -2675.919; Standard deviation in slope 55.686; Intercept 8.620; Standard deviation in intercept 0.196.

The results obtained indicate the following vapour pressure relationship: Log10(Vp (Pa)) = -2675.919/temp(K) + 8.620

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.355.

Summary of results

Run Log10[Vp(25°C)]
1 -0.247
2 -0.253
3 -0.267
4 -0.240
5 -0.355
Mean                -0.272
Vapour Pressure  5.4 x 10-1 Pa

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

Applicant's summary and conclusion

Conclusions:
vapour pressure = 0,54 Pa (at 25°C)
Executive summary:

According to EU Method A.4 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.

The vapour pressure of the test material has been determined to be 5.4 x 10-1 Pa at 25 ºC (arithmetic mean of five determinations) - the test material did not change in appearance under the conditions used in the determination.