Dry air is bubbled through 25.0 liters of water at a rate of 15 L (STP)...

Dry air is bubbled through 25.0 liters of water at a rate of 15 L (STP) / Min. The air leaving the liquid is saturated with water at 25 C and 1.5 atm. How long will it take for all the water to vaporize ? answer: 67.8 days

Solutions

Expert Solution

Solution:

Psat H2O @25°C = 31.6874 hPa x (100Pa / 1hPa) x (1atm / 101325Pa) = 0.03127 atm

the vapor leaving has a total pressure of 1.5atm.. and by "daltons law of partial pressures"
Ptotal = Pdry air + PH2O

so that..
Pdry air leaving = 1.5 atm - 0.03127atm = 1.4687 atm

now since the volume of "dry air" = the volume of "water", because both "gases" fill their container... think about it... if we determine the volume of "dry air" leaving, we'll know, PH2O, VH2O, TH2O and all we need is nH2O...

so let's find the volume of "dry air" being removed like this

start with...
P1V1 / (n1T1) = P2V2 / (n2T2)

rearrange...
V2 = V1 x (P1/P2) x (T2/T1) x (n2/n1)

assuming all the "dry air in" = "dry air out".. ie.. the air isn't dissolving in the water... n2 = n1 (for the "dry air"!).. so that...
V2 = V1 x (P1/P2) x (T2/T1)

so volume of "dry air" being removed is
V2 = (15L/min) x (1atm / 1.4687 atm) x (298.15K / 273.15K) = 11.1 L/min

and now we know the volume of the H2O vapor being removed..11.1 L/min... because the volume H2O vapor = volume of "dry air"

we can convert that, along with Psat H2O, and T=25°C to moles / min of H2O vapor..
PV = nRT
n = PV/(RT) = (0.03127atm) x (11.1L/min) / ((0.08206 Latm/molK) x (298.15K))
n = 0.01419 moles H2O vapor / min

and then.. moles water initially present...
25L H2O x (1000mL / 1L) x (1g / 1mL) x (1 mol / 18.01g) = 1388 mol

and finally..
1388 mol H2O x (1min / 0.01419 mol H2O) = 9.78x10^4 min

which is nothing but 67.91 days.

queen_honey_blossom answered 1 year ago

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