Rank: Guest 
|
Posted By prads
How do we calculate the airborne concentration of a substance if we know the room dimension, the molecular weight of the substance and the specefic gravity?
For ex: 1 gallon of acetone breaks and evaporates in a 30'x30'x10' room. Mol wt is 58 and specefic gravity is 0.6.
Any help!
Rgds,
P
|
|
|
|
|
|
Rank: Guest
|
Posted By Paul Leadbetter
Prads
You need to multiply the volume of acetone by the density to derive the weight of the solvent; don't forget to allow for any difference in units (you have quoted the volume in gallons [US or imperial?] but the specific gravity is in grammes per cc). By the way, my reference book gives the SG as 0.8.
58 g of acetone will produce 22.4 litres of vapour at STP so from the weight of acetone calculated as above, you can determine the volume of vapour and then adjust for the actual temperature of the room using Boyles' Law. From the dimensions of the room, you can calculate the volume and you can then determine the acetone concentration by dividing the volume of acetone vapour by the volume of the room (again, allow for any difference in units).
Paul
|
|
|
|
|
|
Rank: Guest
|
Posted By prads
Hi Paul,
How did u derive that 58 g of acetone will produce 22.4 litres of vapour? Can you please solve this question step-by-step for better understanding. Assume that the volume is 1 US gallons.
Thanks,
P
|
|
|
|
|
|
Rank: Guest
|
Posted By Paul Leadbetter
Prads
22.4 l at STP is the molar volume and 58 g of acetone is 1 mole. Give me a while to do the calculation; I will assume an ambient temperature of 20ºC unless you tell me different.
Paul
|
|
|
|
|
|
Rank: Guest
|
Posted By Haggis JM
1 mole of a substance occupies 22.4L in the vapour phase at standard T & P.
Molecular weight of Acetone = 58.08 g
|
|
|
|
|
|
Rank: Guest
|
Posted By Paul Leadbetter
Prads
My calculation will probably be easier to understand if I can use the flexibility of MS Word as this forum is not ideal for complex layouts; drop me an e-mail at paul@paul-leadbetter.co.uk and I will reply direct.
Paul
|
|
|
|
|
|
Rank: Guest
|
Posted By Flic
Don't forget that this will only be the first approximation. The vapour concentration in the room is unlikely to be uniform - it will tend to be higher just above the puddle from which the acetone was evaporating.
Flic
|
|
|
|
|
|
You cannot post new topics in this forum.
You cannot reply to topics in this forum.
You cannot delete your posts in this forum.
You cannot edit your posts in this forum.
You cannot create polls in this forum.
You cannot vote in polls in this forum.