A greenhouse contains 216 m3 of air at a temperature of 22°C, and a humidifier in it vaporizes 3.57 L of wate?

Question by Santiago: A greenhouse contains 216 m3 of air at a temperature of 22°C, and a humidifier in it vaporizes 3.57 L of wate?
A greenhouse contains 216 m3 of air at a temperature of 22°C, and a humidifier in it vaporizes 3.57 L of water.
(a) What is the pressure of water vapor in the greenhouse, assuming that none escapes and that the air was originally completely dry? (d of H2O = 1.00 g/mL.)
0.022 atm

(b) What total volume of liquid water would have to be vaporized to saturate the air (i. e., achieve 100% relative humidity)? (See this table.)

table:

http://www.webassign.net/silber/5-3tab.gif

Best answer:

Answer by cattbarf
Gee, this looks like yesterdays puzzle. Why can’t you use yesterday’s answers.

First, find the moles of “air” initially in the greenhouse. Use the ideal gas law for this. You will get about 9000 g-moles.

Next, find the moles of water that are evaporated. You have 3570 grams. You will get about 198 moles.

Now you have to assume that the initial pressure in the greenhouse was something. 1 atm is as good an assumption as any. The greenhouse’s gas content has gone from 9000 to 9198 moles. So the pressure will change in proportion to this. So if the final pressure is due to air+water,and Pw is the water portion,
1+Pw /1 = 9198/9000.

Go to your vapor pressure table to look up the saturated vapor pressure at 22 deg C.(about 20 mm Hg). In the greenhouse, the 0.022 atm represents water vapor. To express that in mm Hg, multiply it by 760. I think you will get about 17 mm. So to saturate the air, you need (20/17) x 3.57 L

What do you think? Answer below!

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