A rigid 41.3-Liter container contains dry air at 32°C and 0.9 atm. Calculate the

Question

A rigid 41.3-Liter container contains dry air at 32°C and 0.9 atm. Calculate the required mass of water, in grams, that is required in order to saturate the air in water. Assume that pressure and temperature stay constant during the water's addition. Use three decimal figures in your answer.

Question 6 0 1 point A rigid 41.3-Liter container contains dry air at 32°C and 0.9 atm. Calculate the required mass of water, in grams, that is required in order to saturate the air in water. Assume that pressure and temperature stay constant during the water's addition. Use three decimal figures in your answer. Answer: 26.742 x (1.471)

Explanation / Answer

Saturated conditions refer to partial pressure of water = vapor pressure of liquid

Vapor pressure of water at 32 deg.c can be calculated using Antoine equation

Antone equation for water is

Log Psat(mm HG)= 8.07131- 1730.63/(t+233.426)

Where t is in deg.c

At t= 32 deg.c

Log Psat(mm HG)= 8.07131- 1730.63/(32+233.426)

Psat= 35.57 mm Hg

Hence partial pressure of water vapor=35.57 mm Hg

Converting this into atm, partial pressure of water vapor = 35.57/760 atm=0.047 atm

Moles of water/moles of dry air = partial pressure of water vapor/partial pressure of dry air (1)

Moles of dry air can be calculated from gas law equation

PV= nRT

Where R =0.0821 L.atm/mole.K, n= PV/RT

P= pressure in atm=0.9 atm, V= 41.3L, T= 32 deg.c =32+273= 305K

Hence moles of dry air, n= 0.9*41.3/(0.0821*305)=1.48 moles

Hence from Eq.1

Moles of water/1.48= 0.047/0.9

Mole of water =1.48*0.047/0.9 =0.0773 moles of water vapor

Mass of water to be added = moles* molar mass =0.0773*18 gm =1.3914 gm

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