The relative humidity of air is equal to the ratio of the partial pressure of water in the air to the equilibrium vapor pressure of water at the same temperature. If the relative humidity of the air is 30.% and its temperature is 23°C, how many molecules of water are present in a room measuring 13 m ✕ 6.5 m ✕ 6.0 m? (The vapor pressure of water at 23°C is 21.07 torr.)
Ii believe we have to use the Clausius-
Clapeyron Equation
I don't know how the Clausius-Clapeyron equation fits. I would do this.
0.3 = pH2O/21.07 torr
Solve for pH2O in torr and convert to atm.
Then find the volume of the room, convert to L, then use PV = nRT. Solve for n = number of mols.
Then remember 1 mole of anything contains 6.02E23 molecules.
Yes, you are correct. To determine the number of molecules of water in the room, we can use the Clausius-Clapeyron Equation.
The Clausius-Clapeyron Equation relates the vapor pressure of a substance to its temperature. It is given by:
ln(P₂/P₁) = -(ΔHvap/R)((1/T₂) - (1/T₁))
Where:
P₁ = initial vapor pressure
P₂ = final vapor pressure
ΔHvap = enthalpy of vaporization of water
R = gas constant (8.314 J/mol·K)
T₁ = initial temperature in Kelvin
T₂ = final temperature in Kelvin
First, we need to convert the relative humidity to the actual vapor pressure of water in the air. Given that the vapor pressure of water at 23°C is 21.07 torr, we can calculate the actual vapor pressure:
Vapor pressure = Relative humidity * equilibrium vapor pressure
= 0.30 * 21.07 torr
= 6.321 torr
Next, we need to convert the vapor pressure to P₁ and T₁ in order to use the Clausius-Clapeyron Equation. Let's assume the initial conditions as P₁ = 6.321 torr and T₁ = 23°C = 296K.
Now, we can rearrange the equation to solve for P₂:
ln(P₂/6.321) = -(ΔHvap/R)((1/T₂) - (1/296))
By rearranging, we get:
P₂/6.321 = e^(-(ΔHvap/R)((1/T₂) - (1/296)))
Finally, we can plug in the values and solve for P₂ and T₂.
Using the given volume of the room (13 m * 6.5 m * 6.0 m), we can calculate the total number of moles of water present using the ideal gas law:
PV = nRT,
where:
P = pressure = P₂ (from above)
V = volume of the room
n = number of moles of water
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature = T₂ (from above)
Rearranging the equation, we can solve for n:
n = PV / RT
Finally, we can calculate the number of molecules by multiplying the number of moles by Avogadro's number (6.022 × 10^23 molecules/mol).
To solve this problem, we can use the Clausius-Clapeyron equation to find the partial pressure of water in the air.
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 = initial pressure (equilibrium vapor pressure of water at T1)
P2 = final pressure (partial pressure of water in the air)
ΔHvap = heat of vaporization of water
R = ideal gas constant (0.0821 L*atm/(mol*K))
T1 = initial temperature
T2 = final temperature
Given:
Relative humidity = 30%
Temperature = 23°C = 23 + 273.15 = 296.15 K
Vapor pressure of water at 23°C = 21.07 torr
First, let's calculate the partial pressure of water using the Clausius-Clapeyron equation.
ln(P2 / 21.07) = -ΔHvap / (0.0821) * (1/296.15 - 1/T1)
We know that the relative humidity is defined as the ratio of the partial pressure of water in the air to the equilibrium vapor pressure of water at the same temperature. So, we can write:
Partial pressure of water (P2) = Relative humidity * Vapor pressure of water at 23°C
Partial pressure of water (P2) = 0.30 * 21.07 torr
Now substituting the values back into the equation:
ln((0.30 * 21.07) / 21.07) = -ΔHvap / (0.0821) * (1/296.15 - 1/T1)
ln(0.30) = -ΔHvap / (0.0821) * (1/296.15 - 1/T1)
From here, we can rearrange the equation to solve for ΔHvap:
ΔHvap = -(ln(0.30) * 0.0821 * (1/296.15 - 1/T1))
Now that we have the value of ΔHvap, we can proceed to find the number of water molecules in the room.
The number of water molecules can be calculated using the ideal gas law equation:
PV = nRT
Where:
P = pressure (partial pressure of water in the room)
V = volume of the room (13 m * 6.5 m * 6.0 m = 507 m^3)
n = number of moles of water
R = ideal gas constant (8.3145 J/(mol*K))
T = temperature (in Kelvin)
First, convert the temperature from Celsius to Kelvin:
Temperature (T) = 23 + 273.15 = 296.15 K
Rearranging the equation, we can solve for the number of moles (n) of water:
n = PV / RT
Substituting the values:
n = (Partial pressure of water) * (Volume of the room) / (R) * (Temperature)
n = (P2) * (507 m^3) / (8.3145 J/(mol*K)) * (296.15 K)
Now substitute the value of P2 (calculated earlier) into the equation:
n = (0.30 * 21.07 torr) * (507 m^3) / (8.3145 J/(mol*K)) * (296.15 K)
Finally, calculate the number of moles (n) and convert it to the number of molecules by multiplying by Avogadro's number (6.022 x 10^23 molecules/mol):
Number of molecules of water = (n) * (6.022 x 10^23 molecules/mol)