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Question

"Consider a room with the following dimensions: 10.00ft x 12.00ft x 9.00ft. A “cool-mist” vaporizer, with a capacity 2.50L, is used to add moisture to dry air in a room at 23.4^oC.  The vapor pressure of water at this temperature is 21.587torr and the density is 0.997442g•mL^-1.

(a) If the vaporizer runs until it is empty, what is the vapor pressure of water in the room?

(b) How much water is required to completely saturate the air at 23.4^oC?

(c) A relative humidity of 33% is desirable in heated space on wintry days. What volume of water is left in the vaporizer when the room’s relative humidity reaches that level? (Relative humidity defined as the quotient of the actual vapor pressure, P, with respect to the vapor pressure at saturation, P^o, times 100%:

 "

Answer 1

A

1) calculate the volume of the room (in L):

Vroom​=Length×Width×Height

Vroom=10.00 ft×12.00 ft×9.00 ft=1080.00 ft³

Vroom​=1080.00ft³×28.3168ft³L​= 30522.22L

2) calculate the water vapor (in grams)

The vaporizer has a capacity of 2.50 L of water, the density of water as 0.997442 g/mL;

Density = 0.997442 g/mL = 0.997442 g/cm³

Since 1 liter = 1000 mL, the mass of 2.50 L of water is:

Mass of water=2.50L×1000ml/L​×0.997442g/mL=2493.605g

3) calculate the pressure

using PV=nRT

Where:

1. P = vapor pressure (in atm)
2. V = volume of the room (in liters)
3. n = number of moles of water vapor
4. R = ideal gas constant = 0.0821 L·atm/(mol·K)
5. T = temperature in Kelvin = 23.4∘C + 273.15= 296.55 K

Substituting the known values and solving for P:

P=(138.22 mol)×(0.0821 L\atm/molK)×(296.55 K)/30522.22 L

P=3380.15 Latm/ 30522.22 L = 0.1108 atm

4) convert to TORR

To convert from atm to torr, we use the conversion factor:

1 atm=760 torr

P=0.1108 atm×760 torr/atm = 84.25 torr

B)

1) Find the saturation vapor pressure at 23.4⁰C

the vapor pressure provided is - this is the maximum pressure the water can exert at this temperature when the air is fully saturated.

2) convert to atm

P_{saturation​}= 760torr/atm * 21.587torr ​= 0.02843atm

3) Find the mass of water,

PV=nRT

Where:

1. P = saturation vapor pressure in atm
2. V = volume of the room in liters (we already calculated this as 30,522.22 L)
3. n = number of moles of water vapor required
4. R = ideal gas constant = 0.0821 L·atm/(mol·K)
5. T = temperature in Kelvin = 23.4∘C+273.15 = 296.55 K

solve for n:

n=PV/RT

n=( 0.02843 atm)×(30522.22 L)/(0.0821 Latm/molK)×(296.55 K)

n=35.64mol

Mass of water = 35.64mol×18.015g/mol = 641.55g

4) use the density to find the volume in L

Volume of water= Mass of water/Density = 641.55 g/ 0.997442 g/mL = 643.44 mL = 0.643 L

C)

To determine the volume of water left in the vaporizer when the room's relative humidity reaches 33%, we need to calculate the amount of water vapor required to achieve this specific relative humidity level in the room.

1) We are given the relative humidity as 33%, so we can calculate the actual vapor pressure (PP) using the formula given above, rearranged:

P=(RH×P_sat)÷100

P=33%×21.587 torr÷100

P=0.33 ×21.587 torr =7.13 torr

2) convert to atm

P=760torr/atm x 7.13torr​ = 0.00939atm

3) find the number of moles

PV=nRT

Where:

1. P=0.00939 atm
2. V=30522.22 L (the volume of the room),
3. n is the number of moles of water vapor,
4. R=0.0821 Latm/molK
5. T=296.55 K (temperature in Kelvin).

Rearrange the ideal gas law to solve for n:

n=PV/RT

Substitute the known values:

n=((0.00939 atm)×(30522.22 L))/((0.0821 Latm/molK)×(296.55 K))

n=11.75 mol

5) convert to grams

The molar mass of water is 18.015 g/mol, so the mass of water vapor required is:

Mass of water=11.75 mol×18.015 g/mol=211.43 g

6) convert to L

Using the density of water (0.997442 g/mL), we convert the mass of water to volume:

Volume of water=Mass of water/Density = 211.43 g/ 0.997442 g/mL= 212.83 mL = 0.213 L

7) calculate the remaining water

Initially, the vaporizer had 2.50 L of water. To find how much water is left, we subtract the amount of water vapor already in the room from the total amount of water in the vaporizer:

Volume of water left = 2.50 L − 0.213 L = 2.287 L