# Calculating the mass of water removed by air conditioning condensation

I cannot figure out how to calculate the mass.

The question is:

"Suppose the percent relative humidity is $$\mathrm{80.0\%}$$ at $$\mathrm{91.4\ ^\circ F}$$ ($$\mathrm{33.0\ ^\circ C}$$) in a house with a volume $$\mathrm{245\ m^3}$$. Then the air conditioner is turned on. Due to the condensation of of water vapor on the cold coils of the air conditioner, water vapor is also removed from the air as it cools. After the temperature has reached $$\mathrm{77\ ^\circ F}$$ ($$\mathrm{25\ ^\circ C}$$), the percent relative humidity is measured to be $$\mathrm{15.0\ \%}$$. What mass of water vapor has been removed from the house? (Reminder: take into account the difference in saturated water vapor pressure at two temperatures.)"

The answer is $$6.00\ \mathrm{kg}$$.

Relative humidity= (actual pressure of $$\ce{H2O}$$ pressure/partial pressure of $$\ce{H2O}$$ if saturated) * 100

So, my first step was to find the actual pressure of the initial and final conditions.

$$X/355.1 = 0.80$$

$$X = 284.08~\mathrm{Torr}$$

Then $$X/23.76 = 0.15$$

$$X= 3.564$$

$$\mathrm{284.08~Torr- 3.564~Torr = 280.516~Torr}$$ (water removed, I think)

I attempted to use the ideal gas formula ($$pV= nRT$$) but kept getting the wrong answer. How do I approach the rest of the problem?

• Are you using the right gas constant since you're in torr? Did you convert the moles to grams by multiplying the mols by 18.02 g/mol? Which temprature did you use? When did you include the relative humidity? Apr 20, 2016 at 3:18

Step 0) Realize that sometimes the answer given for a homework problem is wrong.

In this case it appears that the correct answer is $\pu{6.11 kg}$ of water, not $\pu{6.00 kg}$ of water.

The approach to the problem given below is to simply calculate the total water mass in the house before and after cooling, then subtract the latter from the former to get the mass of water removed by condensation.

Step I) Assemble parameters applicable before and after cooling:

$R =$ gas constant $= \pu{8.31m^3~bar~mol^{-1}~K^{-1}}$

$V = \pu{245m^3}$

$MW_{water}=\pu{18.0g~mol^{-1}}$

${n=\frac{PV}{RT}}$

Step II) Calculate water mass before cooling:

$T = \pu{33^oC = 306~K}$

$P_{saturation} = \mathrm{50.3~mbar^{(1)}}$

$P = 80\%$ of $P_{saturation} = \mathrm{40.2~mbar}$

${n=\mathrm{\frac{0.0402~bar\dot\ 245~m^3}{\mathrm{8.31~m^3~bar~mol^{-1}~K^{-1}}\dot\ 306~K}=387~mol~\ce{H2O}}}$

$\mathrm{387~mol~\ce{H2O}\dot\ 18.0~g~mol^{-1}=6.97~kg~\ce{H2O}}$

Step III) Calculate water mass after cooling:

$T = \mathrm{25^oC=298~K}$

$P_{saturation} = \mathrm{31.7~mbar^{(1)}}$

$P = 15\%$ of $P_{saturation} = \mathrm{4.76~mbar}$

${n=\mathrm{\frac{0.00476~bar\dot\ 245~m^3}{\mathrm{8.31~m^3~bar~mol^{-1}~K^{-1}}\dot\ 298~K}=47.9~mol~\ce{H2O}}}$

$\mathrm{47.9~mol~\ce{H2O}\dot\ 18.0~g~mol^{-1}=0.862~kg~\ce{H2O}}$

Step IV) Calculate total water removed from the house:

$$\mathrm{6.97~kg~\ce{H2O} - 0.862~kg~\ce{H2O} = 6.11~kg~\ce{H2O}}$$

Reference:
1) Lide, David R., ed. (2006). CRC Handbook of Chemistry and Physics (87th ed.). Boca Raton, FL: CRC Press. ISBN 0-8493-0487-3. ....