Cesium tungsten oxide

Vapour pressure

Administrative data

Endpoint:

vapour pressure

Type of information:

experimental study

Adequacy of study:

key study

Study period:

07 Feb 2017 - 18 Feb 2017

Reliability:

1 (reliable without restriction)

Rationale for reliability incl. deficiencies:

guideline study

Remarks:

GLP guideline study

Data source

Reference

Reference Type:

study report

Title:

Unnamed

Year:

2017

Report date:

2017

Materials and methods

Test guideline

Test guideline 1

Qualifier:

according to guideline

Guideline:

OECD Guideline 104 (Vapour Pressure Curve)

Test guideline 2

Qualifier:

according to guideline

Guideline:

EU Method A.4 (Vapour Pressure)

GLP compliance:

yes (incl. QA statement)

Remarks:

Department of Health, London, UK

Type of method:

effusion method: vapour pressure balance

Test material

Test material information

Constituent 1

Test material form:

solid: nanoform, no surface treatment

Results and discussion

Vapour pressure

Temp.:

25 °C

Vapour pressure:

ca. 0.001 Pa

Remarks on result:

other: Result is average of extrapolated measured values from 5 runs.

Any other information on results incl. tables

Results
Recorded temperatures, mass differences and the resulting calculated values of vapour pressure are shown in the following tables:

Table 1 – Run 1 results

Temperature [°C]	Temperature [K]	Reciprocal temperature [K^-1]	Mass difference [µg]	Mass difference [kg]	Vapour pressure [Pa]	Log Vp
45	318.15	0.003143171	4.14	4.140E-09	0.005747381	-2.240530038
46	319.15	0.003133323	4.37	4.370E-09	0.006066680	-2.217048942
47	320.15	0.003123536	4.33	4.330E-09	0.006011149	-2.221042482
48	321.15	0.003113810	5.86	5.860E-09	0.008135181	-2.089632763
49	322.15	0.003104144	5.00	5.000E-09	0.006941281	-2.158560374
50	323.15	0.003094538	5.98	5.980E-09	0.008301772	-2.080829195
51	324.15	0.003084992	4.08	4.080E-09	0.005664085	-2.246870216
52	325.15	0.003075504	7.43	7.430E-09	0.010314744	-1.986541565
53	326.15	0.003066074	7.27	7.270E-09	0.010092623	-1.995995968
54	327.15	0.003056702	8.85	8.850E-09	0.012286067	-1.910587108
55	328.15	0.003047387	9.97	9.970E-09	0.013840914	-1.858835220

 

A plot of Log₁₀(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 1 gives the following statistical data using an unweighted least squares treatment.
Slope: -3.72E03
Standard error in slope: 749
Intercept: 9.42
Standard error in intercept: 2.32
The results obtained indicate the following vapour pressure relationship: log (Vp(Pa)) = -3.72E03 / temp (K) + 9.42

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -3.06.

Table 2 – Run 2 results

Temperature [°C]	Temperature [K]	Reciprocal temperature [K^-1]	Mass difference [µg]	Mass difference [kg]	Vapour pressure [Pa]	Log Vp
45	318.15	0.003143171	2.27	2.270E-09	0.003151342	-2.501504522
46	319.15	0.003133323	4.21	4.210E-09	0.005844559	-2.233248283
47	320.15	0.003123536	4.08	4.080E-09	0.005664085	-2.246870216
48	321.15	0.003113810	4.25	4.250E-09	0.005900089	-2.229141449
49	322.15	0.003104144	4.87	4.870E-09	0.006760808	-2.170001418
50	323.15	0.003094538	5.42	5.420E-09	0.007524349	-2.123531092
51	324.15	0.003084992	5.88	5.880E-09	0.008162946	-2.088153053
52	325.15	0.003075504	6.31	6.310E-09	0.008759897	-2.057501019
53	326.15	0.003066074	7.44	7.440E-09	0.010328626	-1.985957443
54	327.15	0.003056702	8.11	8.110E-09	0.011258758	-1.948509525
55	328.15	0.003047387	8.39	8.390E-09	0.011647470	-1.933768418

 

A plot of Log₁₀(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 2 gives the following statistical data using an unweighted least squares treatment.
Slope: -4.93E03
Standard error in slope: 548
Intercept: 13.1
Standard error in intercept: 1.70
The results obtained indicate the following vapour pressure relationship: log (Vp(Pa)) = -4.93E03 / temp (K) + 13.1

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -3.42.

Table 3 – Run 3 results

Temperature [°C]	Temperature [K]	Reciprocal temperature [K^-1]	Mass difference [µg]	Mass difference [kg]	Vapour pressure [Pa]	Log Vp
45	318.15	0.003143171	3.51	3.510E-09	0.004872779	-2.312223262
46	319.15	0.003133323	3.97	3.970E-09	0.005511377	-2.258739872
47	320.15	0.003123536	4.80	4.800E-09	0.006663630	-2.176289141
48	321.15	0.003113810	4.09	4.090E-09	0.005677968	-2.245807071
49	322.15	0.003104144	4.56	4.560E-09	0.006330448	-2.198565536
50	323.15	0.003094538	5.50	5.500E-09	0.007635409	-2.117167689
51	324.15	0.003084992	5.75	5.750E-09	0.007982473	-2.097862534
52	325.15	0.003075504	6.25	6.250E-09	0.008676601	-2.061650361
53	326.15	0.003066074	6.79	6.790E-09	0.009426260	-2.025660604
54	327.15	0.003056702	7.58	7.580E-09	0.010522982	-1.977861173
55	328.15	0.003047387	8.78	8.780E-09	0.012188889	-1.914035863

 

A plot of Log₁₀(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 3 gives the following statistical data using an unweighted least squares treatment.
Slope: -3.83E03
Standard error in slope: 308
Intercept: 9.72
Standard error in intercept: 0.953
The results obtained indicate the following vapour pressure relationship: log (Vp(Pa)) = -3.83E03 / temp (K) + 9.72

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -3.12.

Table 4 – Run 4 results

Temperature [°C]	Temperature [K]	Reciprocal temperature [K^-1]	Mass difference [µg]	Mass difference [kg]	Vapour pressure [Pa]	Log Vp
45	318.15	0.003143171	2.47	2.470E-09	0.003428993	-2.464833425
46	319.15	0.003133323	2.81	2.810E-09	0.003901000	-2.408824059
47	320.15	0.003123536	4.03	4.030E-09	0.005594672	-2.252225333
48	321.15	0.003113810	4.29	4.290E-09	0.005955619	-2.225073087
49	322.15	0.003104144	4.75	4.750E-09	0.006594217	-2.180836769
50	323.15	0.003094538	4.87	4.870E-09	0.006760808	-2.170001418
51	324.15	0.003084992	4.50	4.500E-09	0.006247153	-2.204317865
52	325.15	0.003075504	5.85	5.850E-09	0.008121299	-2.090374513
53	326.15	0.003066074	5.93	5.930E-09	0.008232359	-2.084475685
54	327.15	0.003056702	7.48	7.480E-09	0.010384156	-1.983628781
55	328.15	0.003047387	7.83	7.830E-09	0.010870046	-1.963768617

 

A plot of Log₁₀(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 4 gives the following statistical data using an unweighted least squares treatment.
Slope: -4.71E03
Standard error in slope: 470
Intercept: 12.4
Standard error in intercept: 1.46
The results obtained indicate the following vapour pressure relationship: log (Vp(Pa)) = -3.8E03 / temp (K) + 12.4

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -3.40.

Table 5 – Run 5 results

Temperature [°C]	Temperature [K]	Reciprocal temperature [K^-1]	Mass difference [µg]	Mass difference [kg]	Vapour pressure [Pa]	Log Vp
45	318.15	0.003143171	3.90	3.900E-09	0.005414199	-2.266465772
46	319.15	0.003133323	3.51	3.510E-09	0.004872779	-2.312223262
47	320.15	0.003123536	3.16	3.160E-09	0.004386890	-2.357843296
48	321.15	0.003113810	4.46	4.460E-09	0.006191623	-2.208195520
49	322.15	0.003104144	4.38	4.380E-09	0.006080562	-2.216056268
50	323.15	0.003094538	4.31	4.310E-09	0.005983384	-2.223053109
51	324.15	0.003084992	4.75	4.750E-09	0.006594217	-2.180836769
52	325.15	0.003075504	5.34	5.340E-09	0.007413288	-2.129989122
53	326.15	0.003066074	5.86	5.860E-09	0.008135181	-2.089632763
54	327.15	0.003056702	6.25	6.250E-09	0.008676601	-2.061650361
55	328.15	0.003047387	7.16	7.160E-09	0.009939914	-2.002617356

 

A plot of Log₁₀(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 5 gives the following statistical data using an unweighted least squares treatment.
Slope: -3.14E03
Standard error in slope: 427
Intercept: 7.54
Standard error in intercept: 1.32
The results obtained indicate the following vapour pressure relationship: log (Vp(Pa)) = -3.14E03 / temp (K) + 7.54

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -3.00.

The vapour pressure values at 25 °C extrapolated from each graph are shown below.

Table 6 – Summary of vapour pressure data

Run	Log10 [Vp [25 °C]]
1	-3.06
2	-3.42
3	-3.12
4	-3.40
5	-3.00
Mean	-3.20
Vapour pressure	6.32E-04 Pa

 

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

A total of 5 runs were completed for the main sequence. Equilibrium with regard to vapor pressure was assessed to have been reached over the 5 runs. Thus all 5 runs have been used to calculate the definitive vapor pressure value for the test item.

The results may represent rounded values obtained by calculations based on the exact raw data.

Conclusion
The vapor pressure of the test item has been determined to be 6.3 x 10-4 Pa at 25 ºC.