If the relative humidity of the air is 45% and its temperature is 23 degrees C, how many molecules of water are present in a room measureing 14m by 9m by 8.6m
relative hum = (actual vapor/satd vapor)*100
0.45 = (actual)/20.58 g/m^3
I found the 20.58 g/m^3 at this site.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/relhum.html
Solve for actual vapor and I find something like 9 g/m^3 but you need to do it more accurately.
The volume of the room is 14*9*8.6 = ??m^3 and that times 9 g/m^3 will give you xx grams H2O vapor. Convert that to moles ( moles = g/molar mass) and convert that to molecules H2O using Avogadro's number. Check my thinking.
How many air molecules are in a \rm 14.5 \times 12.0 \times 10.0~ft room? Assume atmospheric pressure of 1.00 \rm atm, a room temperature of 20.0 \rm {}^{\circ}C, and ideal behavior.
Volume conversion:There are 28.2 liters in one cubic foot.
How many air molecules are in a \rm 12.0 \times 12.0 \times 10.0~ft room? Assume atmospheric pressure of 1.00 \rm atm, a room temperature of 20.0 \rm {}^{\circ}C, and ideal behavior.
Volume conversion:There are 28.2 liters in one cubic foot
To calculate the number of water molecules present in a room, we need to understand the concept of relative humidity and use it to determine the amount of water vapor in the air.
Relative humidity is the ratio of the current amount of water vapor present in the air to the maximum amount it could contain at that temperature. It is expressed as a percentage.
First, we need to calculate the maximum amount of water vapor the air can hold at 23 degrees Celsius. This value is known as the saturation vapor pressure.
To find the saturation vapor pressure, we can use a formula called the Magnus formula:
e = 6.112 * exp((17.67 * T) / (T + 243.5))
where:
e = saturation vapor pressure (kPa)
T = temperature in degrees Celsius
Plugging in the temperature (23 degrees Celsius) into the formula:
e = 6.112 * exp((17.67 * 23) / (23 + 243.5))
e ≈ 3.17 kPa
Now, let's calculate the actual vapor pressure using the relative humidity.
Actual vapor pressure = Relative humidity * Saturation vapor pressure
Using the given relative humidity (45%), we can calculate the actual vapor pressure:
Actual vapor pressure = 0.45 * 3.17 kPa
Actual vapor pressure ≈ 1.43 kPa
Next, we can calculate the number of moles of water vapor using the ideal gas law:
PV = nRT
where:
P = pressure (kPa)
V = volume (m³)
n = number of moles
R = ideal gas constant (0.0831 kPa·m³/(K·mol))
T = temperature in Kelvin
Converting the pressure to pascals and the volume to cubic meters:
P = 1.43 kPa * 1000 = 1430 Pa
V = 14m * 9m * 8.6m = 1090.4 m³
Converting the temperature to Kelvin:
T = 23 °C + 273.15 = 296.15 K
Now we can rearrange the ideal gas law and solve for n:
n = PV / (RT)
n = (1430 Pa * 1090.4 m³) / (0.0831 kPa·m³/(K·mol) * 296.15 K)
n ≈ 57.9 mol
Lastly, we can convert moles of water vapor to the number of water molecules by using Avogadro's constant (6.022 × 10^23 molecules/mol):
Number of water molecules = n * Avogadro's constant
Number of water molecules ≈ (57.9 mol) * (6.022 × 10^23 molecules/mol)
Number of water molecules ≈ 3.48 × 10^25 molecules
Therefore, there are approximately 3.48 × 10^25 molecules of water present in the given room.