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In humid climates, people constantly dehumidify their cellars to prevent rot and mildew. If the cellar in a house (kept at $20^{\circ} \mathrm{C} )$ has 115 $\mathrm{m}^{2}$ of floor space and a ceiling height of $2.8 \mathrm{m},$ what is the mass of water that must be removed from it in order to drop the humidity from 95$\%$ to a more reasonable 40$\% ?$

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$3.0 \mathrm{kg}$

Physics 101 Mechanics

Chapter 18

Kinetic Theory of Gases

Temperature and Heat

Cornell University

University of Washington

University of Sheffield

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okay. And this problem, we're talking about using a dehumidifier to reduce the vapor pressure from start to reduce the routes of humidity from 95% to 40% which is a lot more reasonable in the house. Uh, so if we assume a temperature of 20 degrees Celsius and ah, room in a house well, rather an entire house of 115 square meters in a height of 2.8 meters, we need to find the total amount of water. Massey needs to be removed from this space to get this community down. But so we're doing a temperature. We basically have a volume. We have a Florida area times height, so the volume is 115 square meters. Times 2.8 meters gives us that value. Um, the rest of this is just using the ideal gas law and a table to get vapor pressure. So it starts out way. Know that at 23 Celsius from table, uh, 18 dash, too. The saturated vapour pressure is 2.33 times 10 to the three. Pascal just consulting the table, which is used a number of times in this chapter of these problems. uh, you can use the ideal gas law two essentially find the number of bowls of water in the air in both these cases of 95% and 40% humidity. Okay. And we just take the difference in the mole amount, mounts and multiply by the molar mass, and that will give us the water mass. So the difference in moles is just going to be assuming the volume and the temperature of the same, which they are is just gonna be the difference in pressures. And this is ideal gas law. Let's just right up there for reference. It's PV equals Little and Artie are is the gas constant? And so this really all follows. The only thing it's really changing the of our tea got a vapor pressure times the difference in humidity just 0.95 minus 2.4. We just plug in our vapor pressure or volume and our temperature and the gas constant, and we can just calculate this change in moles 100 Jen turns out to be 100 69.4 moles. That's the amount of water in mole's that needs to be removed in order to get this change in humidity at constant volume and temperature conditions in the last step here. To get our actual massive water is to just take the changing moles and multiply it by the molar mass for water. So we take our 169.4 moles. We could just look up the molar mass of water, so the number of kilograms per bowl, which is 18 times 10 to the negative three kilograms Permal Simple multiplication gives us kilograms. Gives us exactly three kilograms of water. It has to be removed. It was actually not that much. So just to recap, what we did is we consulted table 18 dash to to get a vapor pressure for the water. And then he used the ideal gas law to determine the change in moles of the gas. If the pressure is changing under constant volume in temperature conditions, natural change in pressure is just our vapor pressure times the changing relative humidity. Once we have the number of bowls, we can multiply by Muller Mass to get the mass of the water, and that's it

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