Density of water vapour in the air following compression if it were an Ideal Gasby Hendrick Tags: compression, density, ideal, vapour, water 

#1
Apr1707, 06:05 AM

P: 43

1. The problem statement, all variables and given/known data
A volume Vi = 23.2 L of air at temperature T = 22 °C is compressed at constant temperature to a volume Vf = 11.0 L. The relative humidity of the air before compression is H = 59.4 %. The saturated vapour pressure of water at this temperature is 2.67 kPa, corresponding to a vapor density ρw = 19.6 g m^–3. A)What would be the density of water vapour in the air following compression if it behaved like an ideal gas instead of experiencing partial condensation? B)What is the density of water vapour in the air following compression? C)What mass of water condenses out because of the compression? 2. Relevant equations PV = nRT P1 * V1 = P2 * V2 P = rho * R/M * T n = m/M n = number of moles of substance M = molar mass m = mass P = pressure V = volume rho = density R = universal gas constant T = absolute temperature 3. The attempt at a solution A)P1 * V1 / V2 = P2 2670 * (23.2/11) = P2 rho = P2 * M/(R * T) = P2 * 0.018/(8.314 * (273 + 22)) = 0.04132825 Using the molar mass of water (0.018 kg mol^1) and P1 as the saturated vapour pressure of water at 22 °C (2.67 kPa) The actual answer is 24.6 g m^3  B)rho = P * M/(R * T) = 2670 * 0.018/(8.314 * (273 + 22)) = 0.019595291 Using the molar mass of water (0.018 kg mol^1) and P as the saturated vapour pressure of water at 22 °C (2.67 kPa) Actual answer is 19.6 g m^3  is this just the value of vapor density ρw = 19.6 g m^–3. from the line at the beginning of the problem "corresponding to a vapor density ρw = 19.6 g m^–3."  C) PV = nRT n = PV/RT n = m/M P = rho * R/M * T => PV/RT = m/M => [rho * R/M * T ] *V/RT = m/M rho * V = m 19.6 * (11/1000) = 0.2156 Using ρw = 19.6 g m^–3, Vf = 11.0 L The actual answer is 0.0545 g  Help would be much appreciated 



#2
Apr1707, 11:21 AM

Sci Advisor
HW Helper
P: 6,588

The vapour pressure of fully saturated air is 2670 Pa of which this water vapour contributes 59.4% or 1586 Pa. Using n = PV/RT, the amount of water vapour is 1586*.0232/8.314*295 = .015 moles which has a mass of .27 g. So the density of water vapour is .27g/23.2 L = 11.6 x 10^3 g/L = 11.6 x g/m^3 . If you reduce the volume to 11 litres, without changing the amount of water vapour, what is the density? AM 



#3
Apr1707, 09:35 PM

P: 43

I think I understand how to do part A) now: n = PV/RT m = nM rho = m/v n = (0.594*2670)*(23.2/1000)/8.314*(273+22) m = n*(0.018*1000) [for molar mass in g mol^1] rho = m/(11/1000) = 24.5489813 = 24.5 g m^3 (3sf)  Is part B) just the vapor density of water stated in the beginning of the equation? If so, why?  I don't know where to start on C), I tried taking the change in volume and using a 'm = rho*v' equation with rho being the answer in B but it's incorrect thanks for the help on A), I didn't even think of finding n and substituting it on find the density... 



#4
Apr1807, 03:30 AM

Sci Advisor
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P: 6,588

Density of water vapour in the air following compression if it were an Ideal GasUsing n = PV/RT will give you the no. of moles from which you can work out the density. Part C) is just the difference between A and B. AM 



#5
Apr1807, 07:16 AM

P: 43

Why is the air pressure 213704 Pa? I don't know why I must use the saturated vapour pressure. Part C): Do you mean the difference in masses between A and B? I didn't think it would involve part A) as I thought it was a hypothetical question: 



#6
Apr1807, 10:44 AM

Sci Advisor
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P: 6,588

What you have to do to get it exact is to work out the pressure of the air only at 11 L. It is 101325  2670 at 23.2 L. You have to use the saturation vapour pressure because if it is 59.4% saturated at 1 atm, it will be completely saturated when compressed to less than half the volume. You know this because when you compress water vapour from 1580 Pa at 22 C in 23.2 L to 11 L at 22C, the vapour pressure increases to ________ Pa. But we know that the maximum vapour pressure is 2670 Pa. You should be able to determine from that how much precipitates out. AM 



#7
Apr1907, 06:30 AM

P: 43

Ok, I'm pretty confused about the meaning of part A)
 I still don't quite get why you have to do all of the calculations as I thought the answer was just from  Thanks for your help 



#8
Apr2007, 09:33 AM

Sci Advisor
HW Helper
P: 6,588

In part A, you just assume that there is no maximum vapour pressure.  Then find the amount of water vapour and the density of this vapour at 11 L. If none precipitated out, what would the vapour pressure be? Can it actually be that high? Why not (what is the maximum?)? So if that is the maximum, how much water is in the vapour state? How does that compare to the amount that was present in the unsaturated air (all in vapour state)? What is the difference? Where is that difference (ie. the amount not in the vapour state?)? AM 



#9
Apr2007, 09:56 AM

P: 43

Thanks very much for your help Andrew.
I have everything calculationwise but your explanation aided my comprehension of the problem...  Thank you 


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