m-phenylenebis(methylamine)

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 2009-07-30 and 2009-08-20

Reliability:

1 (reliable without restriction)

Rationale for reliability incl. deficiencies:

other: GLP guideline study

Qualifier:

according to guideline

Guideline:

OECD Guideline 104 (Vapour Pressure Curve)

Deviations:

no

GLP compliance:

yes (incl. QA statement)

Remarks:

Date of Inspection: 2008-08-19; Date of Signature: 2009-03-04

Type of method:

effusion method: vapour pressure balance

Temp.:

25 °C

Vapour pressure:

0.69 Pa

Remarks on result:

other: Initially the test material was a clear colourless liquid. During the period the sample was under vacuum, the presence of an increasing amount of white solid was observed. See below for further details on results.

Run 1

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log10Vp
21	294.15	0.003399626	380.50	3.805E-07	0.528231484	-0.277175718
22	295.15	0.003388108	400.57	4.006E-07	0.556093785	-0.254851958
23	296.15	0.003376667	427.03	4.270E-07	0.592827044	-0.227071992
24	297.15	0.003365304	456.18	4.562E-07	0.633294713	-0.198394138
25	298.15	0.003354016	494.72	4.947E-07	0.686798107	-0.163170911
26	299.15	0.003342805	532.27	5.323E-07	0.738927127	-0.131398390
27	300.15	0.003331667	572.80	5.728E-07	0.795193151	-0.099527369
28	301.15	0.003320604	613.24	6.132E-07	0.851334231	-0.069899904

 

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                                       -2688.184
Standard deviation in slope       61.971

Intercept                                          8.854
Standard deviation in intercept    0.208

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -2688.184/temp(K) + 8.854

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.162.

Run 2

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log10Vp
21	294.15	0.003399626	387.29	3.873E-07	0.537657743	-0.269494095
22	295.15	0.003388108	405.86	4.059E-07	0.563437661	-0.249154128
23	296.15	0.003376667	429.33	4.293E-07	0.596020034	-0.224739142
24	297.15	0.003365304	458.08	4.581E-07	0.635932399	-0.196589048
25	298.15	0.003354016	491.33	4.913E-07	0.682091918	-0.166157096
26	299.15	0.003342805	528.17	5.282E-07	0.733235276	-0.134756649
27	300.15	0.003331667	560.32	5.603E-07	0.777867713	-0.109094254
28	301.15	0.003320604	604.15	6.042E-07	0.838714982	-0.076385599

 

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                                       -2479.964
Standard deviation in slope       63.555

Intercept                                          8.154
Standard deviation in intercept    0.214

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) =-2479.964/temp(K) + 8.154

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.164.

Run 3

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log10Vp
22	295.15	0.003388108	404.27	4.043E-07	0.561230333	-0.250858864
23	296.15	0.003376667	431.62	4.316E-07	0.599199140	-0.222428818
24	297.15	0.003365304	463.47	4.635E-07	0.643415100	-0.191508751
25	298.15	0.003354016	492.73	4.927E-07	0.684035477	-0.164921374
26	299.15	0.003342805	527.97	5.280E-07	0.732957625	-0.134921133
27	300.15	0.003331667	561.52	5.615E-07	0.779533621	-0.108165149
28	301.15	0.003320604	614.84	6.148E-07	0.853555441	-0.068768265

 

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                                       -2638.728
Standard deviation in slope       60.718

Intercept                                          8.688
Standard deviation in intercept    0.204

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -2638.728/temp(K) + 8.688

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.163.

Conclusions:

The vapour pressure of the test material has been determined to be 0.69 Pa at 25 degrees Centigrade

Description of key information

GLP guideline study reporting the vapour pressure of the test material as 0.69 Pa at 25 °C (OECD 104)

Key value for chemical safety assessment

Vapour pressure:

0.69 Pa

at the temperature of:

25 °C