Thiophene, tetrahydro-, 1,1-dioxide, 3-(C9-11-isoalkyloxy) derivs., C10-rich

Vapour pressure

Link to relevant study record(s)

Reference

Endpoint:

vapour pressure

Type of information:

experimental study

Adequacy of study:

key study

Study period:

Testing was conducted between 21 September 2011 and 26 September 2011.

Reliability:

1 (reliable without restriction)

Rationale for reliability incl. deficiencies:

other: see 'Remark'

Remarks:

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 the relevant results. The study report was conclusive, done to a valid guideline and the study was conducted under GLP conditions.

Qualifier:

equivalent or similar to guideline

Guideline:

EU Method A.4 (Vapour Pressure)

Deviations:

no

Qualifier:

equivalent or similar 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

Temp.:

25 °C

Vapour pressure:

0.85 Pa

Transition / decomposition:

no

Run 2

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log10Vp
5	278.15	0.003595182	111.53	1.115E-07	0.154832214	-0.810138677
6	279.15	0.003582303	119.31	1.193E-07	0.165632847	-0.780853533
7	280.15	0.003569516	128.50	1.285E-07	0.178390922	-0.748627251
8	281.15	0.003556820	140.08	1.401E-07	0.194466928	-0.711154246
9	282.15	0.003544214	152.16	1.522E-07	0.211237063	-0.675229879
10	283.15	0.003531697	169.23	1.692E-07	0.234934596	-0.629053024
11	284.15	0.003519268	182.51	1.825E-07	0.253370639	-0.596243714
12	285.15	0.003506926	203.98	2.040E-07	0.283176499	-0.547942791
13	286.15	0.003494671	231.04	2.310E-07	0.320742712	-0.493843203
14	287.15	0.003482500	246.41	2.464E-07	0.342080210	-0.465872050
15	288.15	0.003470415	264.39	2.644E-07	0.367041056	-0.435285354

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                                      -3134.990
Standard deviation in slope     71.073

Intercept                                       10.446
Standard deviation in intercept  0.251

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -3134.990/temp(K) + 10.446

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.068.

Run 2 - Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 2 to 5.

Run 3

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log10Vp
5	278.15	0.003595182	105.93	1.059E-07	0.147057979	-0.832511406
6	279.15	0.003582303	115.02	1.150E-07	0.159677228	-0.796757015
7	280.15	0.003569516	123.21	1.232E-07	0.171047046	-0.766884421
8	281.15	0.003556820	139.48	1.395E-07	0.193633975	-0.713018440
9	282.15	0.003544214	154.66	1.547E-07	0.214707704	-0.668152373
10	283.15	0.003531697	164.84	1.648E-07	0.228840152	-0.640467773
11	284.15	0.003519268	180.42	1.804E-07	0.250469183	-0.601245700
12	285.15	0.003506926	198.99	1.990E-07	0.276249101	-0.558699127
13	286.15	0.003494671	219.16	2.192E-07	0.304250228	-0.516769087
14	287.15	0.003482500	241.42	2.414E-07	0.335152811	-0.474757133
15	288.15	0.003470415	265.48	2.655E-07	0.368554256	-0.433498570

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                                      -3212.139
Standard deviation in slope     45.011

Intercept                                       10.709
Standard deviation in intercept  0.159

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -3212.139/temp(K) + 10.709

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.064.

Run 3 - Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 2 to 5.

Run 4

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log10Vp
5	278.15	0.003595182	102.94	1.029E-07	0.142907093	-0.844946215
6	279.15	0.003582303	109.73	1.097E-07	0.152333353	-0.817205000
7	280.15	0.003569516	120.11	1.201E-07	0.166743452	-0.777951212
8	281.15	0.003556820	133.89	1.339E-07	0.185873622	-0.730782237
9	282.15	0.003544214	147.97	1.480E-07	0.205420270	-0.687356705
10	283.15	0.003531697	163.34	1.633E-07	0.226757767	-0.644437827
11	284.15	0.003519268	175.92	1.759E-07	0.244222030	-0.612215162
12	285.15	0.003506926	193.60	1.936E-07	0.268766400	-0.570625026
13	286.15	0.003494671	214.76	2.148E-07	0.298141901	-0.525576983
14	287.15	0.003482500	237.03	2.370E-07	0.329058367	-0.482727062
15	288.15	0.003470415	263.39	2.634E-07	0.365652800	-0.436931096

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                                      -3300.487
Standard deviation in slope     45.763

Intercept                                       11.010
Standard deviation in intercept  0.162

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -3300.487/temp(K) + 11.010

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.060.

Run 4 - Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 2 to 5.

Run 5

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log10Vp
5	278.15	0.003595182	103.74	1.037E-07	0.144017698	-0.841584135
6	279.15	0.003582303	113.32	1.133E-07	0.157317192	-0.803223813
7	280.15	0.003569516	125.70	1.257E-07	0.174503804	-0.758195101
8	281.15	0.003556820	134.39	1.344E-07	0.186567751	-0.729163425
9	282.15	0.003544214	148.57	1.486E-07	0.206253223	-0.685599255
10	283.15	0.003531697	163.84	1.638E-07	0.227451896	-0.643110439
11	284.15	0.003519268	175.13	1.751E-07	0.243125308	-0.614169831
12	285.15	0.003506926	190.20	1.902E-07	0.264046329	-0.578319866
13	286.15	0.003494671	214.36	2.144E-07	0.297586599	-0.526386630
14	287.15	0.003482500	230.94	2.309E-07	0.320603886	-0.494031217
15	288.15	0.003470415	261.69	2.617E-07	0.363292765	-0.439743252

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                                      -3142.998
Standard deviation in slope     53.663

Intercept                                       10.455
Standard deviation in intercept  0.190

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -3142.998/temp(K) + 10.455

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.087.

Run 5 - Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 2 to 5.

Conclusions:

The vapour pressure of the test item has been determined to be 8.5 x 10-1 Pa at 25ºC.

Executive summary:

Introduction

Hazardous physico-chemical properties of the test item have been determined.

Methods employed are designed to be compatible with those specified in Commission Regulation (EC) No 440/2008 of 30 May 2008, Part A: Methods for the determination of physico-chemical properties and the current OECD Guidelines for Testing of Chemicals.

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 a procedure designed to be compatible with Method A4 Vapour Pressure of Commission Regulation (EC) No 440/2008 of 30 May 2008 and Method 104 specified in the OECD Guidelines for Testing of Chemicals, 23 March 2006.

Conclusion

The vapour pressure of the test item has been determined to be 8.5 x 10^-1 Pa at 25ºC.

Description of key information

The vapour pressure of the substance was determined to be 0.85 Pa; study conducted in accordance with OECD 104 and EU Method A.4; Atwal and Tremain (2011)

Key value for chemical safety assessment

Vapour pressure:

0.85 Pa

at the temperature of:

25 °C

Additional information

In a GLP compliant vapour pressure study conducted in accordance with standardised guidelines, the vapour pressure of the test substance was determined using the effusion method: vapour pressure balance. Under the conditions of the test, the vapour pressure of the test substance was determined to be 0.85 Pa which is considered to be a very low volatile based on EASE.