On a humid day in New Orleans, the temperature is 22.0 °C and the partial pressu

Question

On a humid day in New Orleans, the temperature is 22.0 °C and the partial pressure of water vapor is 31.0 torr. The temperature inside the Superdome is also 22.0 °C but the water vapor pressure is only 10.0 torr. The total volume of air in the dome is 2.4x106 m3 and the pressure inside and outside the dome is 1.0 atm.

a) How many metric tons of water are removed each time the inside air is completely replaced with outside air? Hint: How much water in moles is present in the wet air and in the dry air? Use the ideal gas law.
b) Calculate the total amount of heat released when this amount of water is condensed? formula.

Explanation / Answer

The first thing to do is to calculate the amount of water at 22 C and 31 torr, change torr to atm

1 atm = 760 torr

31 torr = 0.0408 atm

temperature = 22 + 273.15 = 295.15 Kelvin

The volume comes from the statement 2.4 x 106 m3, change this to liters by multiplying by 1000 , if you do this you get 2.4 x 109 liters

P * V = n * R * T; where

P is pressure

V is volume

R is gas constant : 0.082 atm L / K mol

T is Temperature (Kelvin or Fahrenheit)

Rearrange the equation to get

n = P * V / (R*T)

n = (0.0408 * 2.4 x 109 liters) / (0.0821 * 295.15) = 4.04 x 106 moles

multiply this by the molar mass of h2o which is 18 g/gmol

mass of h2o = 4.04 x 106 moles * 18 = 7.27x107 grams

1 ton has 1000 000 grams so we have

72.7 tons

Now lets calculate this same thing using 10 torrs

10 torrs are 0.01315 atm

Apply the same equation

n = P * V / (R*T)

n = (0.01315* 2.4 x 109 liters) / (0.0821 * 295.15) = 1.3 x 106 moles

change to grams by multiplying by 18 , the molar mass of h20 then

mass = 18 *1.3 x 106 moles = 2.35 x 107 grams

change this to tons by dividing by 1000 000 to get 23.5 tons

The tons of water removed are

72.7 - 23.5 = 49.26 tons of water removed

B) turn the last value to grams of water and then turn it to moles

49.26 tons = 4.926 x 107 grams

moles =  4.93 x 107 grams / 18 = 2.74 x 106 moles of water

then we use the condensation heat of water is -40.7 KJ / mol

just multiply the moles by this last value to get

2.74 x 106 moles *  -40.7 KJ / mol = -1.114 x 108 KJ, this is the heat released by the condensation of this amount of water

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