4-chlorophenol

Vapour pressure

Link to relevant study record(s)

Reference

Endpoint:

vapour pressure

Type of information:

experimental study

Adequacy of study:

key study

Study period:

16-09-2019 to 16-10-2019

Reliability:

1 (reliable without restriction)

Rationale for reliability incl. deficiencies:

other: Guideline study performed under GLP. All relevant validity criteria were met.

Qualifier:

according to guideline

Guideline:

EU Method A.4 (Vapour Pressure)

Deviations:

no

Qualifier:

according to guideline

Guideline:

OECD Guideline 104 (Vapour Pressure Curve)

Deviations:

no

GLP compliance:

yes

Type of method:

dynamic method

Temp.:

20 °C

Vapour pressure:

3.9 Pa

Remarks on result:

other: 2.9 x 10^{-2} mm Hg

Temp.:

25 °C

Vapour pressure:

7.6 Pa

Remarks on result:

other: 5.7 x 10^{-2} mm Hg

Table 1.0 – Results of the Vapour Pressure Isothermal TGA Analysis

Temperature [°C]	Weight loss [µg/min]	νT [g/cm^2/h]	logνT	logPT	PT [Pa]
30	7.54399	9.00x10^-4	-3.05	1..20	16
	6.77287	8.08x10^-4	-3.09	1.15	14
35	12.6071	1.50x10^-3	-2.82	1.45	28
	12.1827	1.45x10^-3	-2.84	1.43	27
40	21.3986	2.55x10^-3	-2.59	1.71	51
	22.3650	2.67x10^-3	-2.57	1.73	54

 

Table 2.0 - Vapour Pressure of the Test Item

Temperature [°C]	logPT	PT [Pa]	PT [mm Hg]
20	0.59	3.9	2.9x10^-2
25	0.88	7.6	5.7x10^-2

Discussion

Fitting these data using the least squares method gives a value of 3.9 Pa for the vapour pressure of the test substance at 20°C and 7.6 Pa for the vapour pressure of the test substance at 25°C. The vapour pressure regression curve of the log P_T of the test item as function of the reciprocal temperatures had an equation given by:-

log P_T = -5181 x 1/T + 18.26 (r = 0.996, n = 6).

where:-

Evaporation rate (ν_T) = Δm / Ft
and:
Δm: weight loss of the test item [g]
F: surface of the sample plate [cm²]
t: elapsed time for the weight loss [h]

Vapour pressure equation log P_T = c log ν_T + d

and:
c: constant specific for the experimental arrangement
d: constant specific for the experimental arrangement

Vapour pressure regression curve log P_T = a 1/T + b
and:
T: temperature [K]
a: slope [K]
b: intercept

Conclusions:

The vapour pressure of the substance has been determined to be 3.9 Pa at 20°C and 7.6 Pa at 25°C.

Executive summary:

The vapour pressure was determined using EU Method A.4 and OECD TG 109 using isothermal TGA analysis under GLP. The measured results (In p) are plotted versus the reciprocal temperature (in Kelvin) and yield a rectilinear curve for the limited temperature range. The vapour pressure at 20°C and 25°C was calculated from this curve. Fitting the data using the least squares method gives a value of 3.9 Pa for the vapour pressure of the test substance at 20°C and a value of 7.6 Pa for the vapour pressure of the test substance at 25°C.

Description of key information

Vapour Pressure: 3.9 Pa at 20C and 7.6 Pa at 25C, EU Method A.4 and OECD 104, 2019.

Key value for chemical safety assessment

Vapour pressure:

7.6 Pa

at the temperature of:

25 °C

Additional information

EU Method A.4 and OECD 104, 2019: The vapour pressure at 20 degrees C is 3.9 Pa and the vapour pressure at 25 degrees C is 7.6 Pa.