The total volume of a new house is 800 m3. Before the heat is turned on, the air temperature inside is 10°C, and the relative humidity is 50%. After the air is warmed to 20°C, how much water vapor must be added to the air to make the relative humidity 50%?
m =
kg
To determine the amount of water vapor that needs to be added to the air, we can use the concept of specific humidity.
Specific humidity is the mass of water vapor per unit mass of air. It is expressed in kilograms of water vapor per kilogram of dry air.
To find the specific humidity, we need the initial specific humidity and the final specific humidity.
Given information:
Initial air temperature (T1) = 10°C
Final air temperature (T2) = 20°C
Initial relative humidity (RH1) = 50%
Volume of the house = 800 m^3
First, we need to calculate the initial and final values of specific humidity:
1. Calculate the saturation vapor pressure at both temperatures:
The saturation vapor pressure can be calculated using the Clausius-Clapeyron equation:
P = P0 * exp((L / Rv) * ((1 / T0) - (1 / T)))
Where:
P0 = saturation vapor pressure at T0
L = latent heat of vaporization (2.257 × 10^6 J/kg)
Rv = gas constant for water vapor (461.5 J/kg K)
For T = 10°C:
P0 = 611.2 Pa (saturation vapor pressure at 10°C)
For T = 20°C:
P0 = 2339.1 Pa (saturation vapor pressure at 20°C)
2. Calculate the actual vapor pressure at the initial temperature (RH1):
P_actual_1 = (RH1 / 100) * P0
3. Calculate the specific humidity at the initial temperature (SH1):
SH1 = (0.622 * P_actual_1) / (P1 - P_actual_1)
Where:
P1 = total atmospheric pressure (assume 101325 Pa)
Similarly, calculate the specific humidity (SH2) at the final temperature using the saturation vapor pressure at 20°C and the actual vapor pressure corresponding to 50% relative humidity.
4. Calculate the mass of dry air in the house:
The mass of dry air (m_dry) can be calculated using the ideal gas law:
PV = nRT
Assuming the air behaves ideally, we can rearrange the equation to solve for the mass of dry air:
m_dry = (PV) / (R * T)
Where:
P = atmospheric pressure (assume 101325 Pa)
V = volume of the house
5. Calculate the mass of water vapor to be added (m):
m = (m_dry * SH2) - (m_dry * SH1)
Now, plug in the values and calculate:
P0(10°C) = 611.2 Pa
P0(20°C) = 2339.1 Pa
P_actual_1 = (50 / 100) * 611.2 = 305.6 Pa
SH1 = (0.622 * 305.6) / (101325 - 305.6) = 0.001875 kg/kg
SH2 = (0.622 * P_actual_2) / (101325 - P_actual_2)
m_dry = (101325 * 800) / (287.058 * (10 + 273.15)) ≈ 12212.82 kg
m = (12212.82 * SH2) - (12212.82 * 0.001875)
Solving this equation will give you the mass of water vapor (m) needed to reach 50% relative humidity at 20°C in the house.