Air conditioners not only cool air, but dry it as well. A room in a home measures 6.0 m × 10.0 m × 2.2 m. If the outdoor temperature is 30 °C and the partial pressure of water in the air is 85% of the vapor pressure of water at this temperature, what mass of water must be removed from the air each time the volume of air in the room is cycled through the air conditioner? (Assume that all of the water must be removed from the air.) The vapor pressure for water at 30 °C is 31.8 torr.

Question

Accepted Answer

Hey everyone in this example, we have air conditioners with a dry mode that removes moisture from air like a dehumidifier were given the dimensions of an office building. We're told that the temperature outside is 27°C. And the partial pressure of water in air at this temperature is 73% of the vapor pressure of water. Were given the vapor pressure of water at 27°C being 26. mm of mercury. And we need to determine the mass of water that is removed. So what we want to recall is our formula for pressure times volume equal to the moles of our gas times, the gas constant R times temperature. And because we need to find the massive water removed, we're going to be reorganizing this. Sorry to sulfur. The moles of our water, which would be found by taking our pressure times volume and then dividing this by our gas constant R times temperature. Our first step is to make sure that we have the proper pressure. So according to the prompt, we have the pressure of water Which is equal to a value of 73%. Multiplied by the vapor pressure of water given to us in the prompt as 26.7 mm of Mercury. And so this gives us a pressure of water equal to 19.941 mm of Mercury. Now we should recall that pressure should be in units of a. T. M. So we're going to take this 19 mm of Mercury. And we're going to recall that our conversion factor which tells us we have millimeters of mercury for 1 80 M. And so now we're able to cancel out our units millimeters of mercury were left with a t. M. And this should give us a pressure value equal to 0.2565 A. T. M. We also want our temperature given to be in units of kelvin However, were given temperature in C as 27°C. So we're going to add to 73.15 to get our Kelvin temperature equal to a value of 300.15 Kelvin. We then need our value for volume and according to the prompt, were given the dimensions of the building so we can find volume by taking the length of the building multiplied by the height of the building. Or sorry the width of the building and then multiplied by the height. So according to the prompt we have For length nine m Multiplied by the width of the building. Given us 6.0 m and then multiplied by the height of the building given as 5.5 m. Which gives us a value of 297 m cubed. Now we should recall that volume. We want to be in units of leaders. So we're going to take our two sorry r 2 97 cubed meters. And first convert from meters two centimeters. So recall that our prefix anti tells us that we have 10 to the negative two m because we want to get to volume. We're going to multiply by another conversion factor to go from cubic centimeters, two mL. And then to end up with leaders as our final unit. We're going to convert from milliliters to leaders. So we should recall that for one cc. We have one millimeter and then our prefix milli tells us that we have 10 to the negative third power leaders for one middle leader. So now we're able to cancel out middle leaders centimeters and we're left with leaders as our final unit here. And what we should get is that our volume is equal to a value of 2. times 10 to the fifth power leaders. So now we have volume and all of our other variables. We can go ahead and finally calculate for our molds of water. So this should be again equal to pressure times volume divided by the gas constant R times temperature. So what we should have is our pressure which above we stated at 0.2565 A. T. M's Multiplied by our volume, which we just found as 2.97 times 10 to the 5th power leaders. And then dividing all of this in our numerator by r gas constant. R which we recall is 0. leaders times a t m's divided by moles, times kelvin And then multiplied by our temperature, which we converted to Kelvin as 300 .15 Kelvin. So now we're gonna go ahead and cancel our units of kelvin. We're going to get rid of a. T. M. We're going to get rid of leaders and this leaves us with molds, which is what we want. So what we should get for this quotient is our value for moles of water equal to 309.2957 moles of our water. Now to answer this question, we need to get the mass of our water that is removed. So we're going to take the value that we just found (309. moles of water. And we're going to recall from our periodic tables, the molar mass for one mole of water, Which according to the periodic table is going to be equal to 18. g of water. So this is going to give us a mass equal to 5572 0. g, which we can convert to scientific notation so that we have 5.6 times 10 to the positive third power grams as our mass of water. That is removed. And this would be our final answer here to complete this example for our massive water that is removed according to the calculation. So I hope that everything I explained was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

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Key Concepts

Vapor Pressure

Relative Humidity

Ideal Gas Law