Antoine constants for Chlorine Dioxide - P in mmHg 1

[nichs2]

I've been searching for a while now (it probably feels like longer than it really is) for antoine constants for chlorine dioxide, to give vapour pressure in mmHg. I've found constants to give vapour pressure in bar, but not mmHg (my consequence analysis software requires them in this form). Or can I convert them somehow? I've tried DIPPR, NIST, various promising spreadsheets on cheresources and the like and google (!). Can anyone help?

 

[orenda1168]

1 Bar = 750.064 mmHg.

Orenda

 

[Montemayor]

I believe you obtained your Antoine relationship from NIST at:

http://www.dfmg.com.tw/member/chemical/cas/10049-04-4.htm

which gives you:

log10(P) = A - (B / (T + C))

where,
P = vapor pressure (bar)
T = temperature (K)

This is derived from data by Stull (in 1947!) and is only valid for the range of 214 to 284 oK (-75 oF to 52 oF). You should abide by the temperature range cited. The Antoine equation with the stated constants is only good for that specific range. This is a characteristic of the Antoine relationship.

If the range is OK, and you desire an Antoine equation, the solution I would give you is to generate a series of vapor pressure values in bars using the above equation. Then, convert the values to mm of mercury in Excel using the conversion factor that orenda cites and let Excel generate the equation by regressing the mm of mercury data. Or, you could use a more sophisticated regression – such as Datafit – and obtain a more accurate relationship.

 

[mbeychok]

nichs2:

Since you had a problem with knowing how to convert bar to mmHg, you might do well to download a copy of the pressure conversion table available at:

http://en.citizendium.org/wiki/Pressure

Milton Beychok
(Visit me at www.air-dispersion.com)
.

 

[nichs2]

Thanks for the quick responses. I should have been clearer, I do actually know how to convert from bar to mmHg, but not how to convert the Antoine constants! Montemayor's idea is a good one, and one I should have thought of. Thanks.

 

[siretb]

It seems to me that you have

log10(Pbar) = A - (B / (T + C))

where,
P = vapor pressure (bar)
T = temperature (K)

Pbar=PmmHg/750.1

so now
log10(PmmHg) = (A+log10(750)) - (B / (T + C))

the B and C are unchanged and the new A constant is the old A(for bar) plus 2.875

 

[nichs2]

That also seems to work siretb, but when I did it the way montemayor suggested, I got values for the constants that were far more in line (i.e. same order of magnitude etc.)with the existing values for other compounds in the database I was adding to. Curious. Amazing how quickly you forget the simple stuff when you don't use it for a while.