CrossxD
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hard problem
Hello guys,
I need to solve how much of butane is liquid and gaseous state in tank at 27°c. I have 10 kg of butane with density 580 kg * m^-3, tank has volume of
20 dm^3 and I know antoine's equation for butane log10(p)=4.356 -1175,580 / (T -2.070)
How should I solve this?
Please dont tell me result.
Than you very much!
CrossxD
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Tsjerk
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I think the volume of the liquid phase lost in order to reach vapor pressure is marginal. So you could start with calculating how much gas space there
is left after having filling the tank with the butane, together with the vapor pressure you know how many grams there are in the gas phase.
Substract the liquid amount with the gas amount and you can see how much was lost in volume of the liquid half, see if this not to far of and you
should be fine.
I know it can be calculated exactly, but I don't know by heart how to do that.
Edit:
I got to a result via the above way and indeed found the loss in volume to be neglect-able. I did assume that butane acts like a ideal even at
elevated pressure. So probably if you would do it properly you will get to even less percentage of butane in the gas phase.
Are you sure your Antoine's equation is correct? I'm of by a factor two when I use yours compared to what I find in vapor pressure tables on butane.
[Edited on 16-10-2016 by Tsjerk]
[Edited on 16-10-2016 by Tsjerk]
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CrossxD
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I dont know but tempeture is in kelvins and pressure in bars
maybe it's becouse of isomerisation of butane
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CrossxD
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Thank you Tsjerk
I've solved it
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Tsjerk
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if it is an ideal gas with 24,5 liter/mol at 1 atm, With your Antoine's equation I get to 0.23% in the gas phase, assuming the loss in volume by the
evaporation is not there.
This volume is about 40 ml out of 17,2431 liter. 20-17,2431 = 2,750 liter, of which 40 ml is about 1.5 % so it would become 0,23 plus 1,5 % which is
0,23345%.
Quite rough but I ques it is close to the truet. Does anyone known the ''official" way to calculate this?
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Magpie
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Assume the butane behaves like an ideal gas and that the ideal gas law applies, ie, PV = nRT; V=nRT/P. The liquid butane evaporates to its vapor
pressure at 27°C, ie, 2.57 bar.
Density of liquid butane = 580kg/m3. weight of butane = 10kg. Volume of tank = 20dm3 = 0.020m3.
T= 27°C = 300°K; From the Antoine equation, P = 2.57bar.
Let x = wt of butane that evaporates in kg
Let V = the volume of the gas in m3
V = (x, kg)(1000g/kg)[8.314 m3-bar x10^-5/(°K-g-mole)][300°K]/[(58.12g/g-mole)2.57 bar] = 0.167x, in m3
tank volume = 0.020 m3 = 0.167x, m3 +
(10kg-x, kg)(m3/580kg)
0.020 = 0.167x +10/580 -x/580
x = 0.0167 kg
% evaporated = (0.0167kg/10kg)100% = 0.167%
The single most important condition for a successful synthesis is good mixing - Nicodem
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