A volume Vi = 23.1 L of air at temperature T = 21 ¢XC is compressed at constant temperature to a volume Vf = 7.7 L. The relative humidity of the air before compression is R = 56.4 %. Saturated water vapour in air at this temperature has a vapour pressure vp = 2.5 kPa and a density of £l = 18.4 g.m−3
a)What is the density the water vapour would have had following compression if it behaved like an ideal gas instead of experiencing partial condensation?
b) What mass of water condenses out because of the compression?
To answer these questions, we can use the ideal gas law and the concept of partial pressure.
a) To find the density of the water vapor following compression if it behaved like an ideal gas, we need to calculate the partial pressure of the water vapor.
1. Convert the relative humidity (R) to a decimal:
R = 56.4% = 0.564
2. Calculate the partial pressure of the water vapor using the saturation vapor pressure (vp):
Partial pressure of water vapor (Pv) = R * vp
Pv = 0.564 * 2.5 kPa
Pv = 1.41 kPa
3. Use the ideal gas law to calculate the density of the water vapor:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since we are considering the water vapor as an ideal gas, we can use the ideal gas law to calculate the density.
To find the number of moles, we can use the equation n = PV / RT:
n = (Pv * Vf) / (R * (T + 273.15))
The water vapor density (ρv) can then be calculated as the mass of the water vapor (m) divided by the volume (Vf):
ρv = m / Vf
b) To find the mass of water that condenses out because of the compression, we can compare the initial and final densities of the water vapor.
1. Calculate the initial mass of the water:
m_initial = ρ_initial * Vi
2. Calculate the final mass of the water:
m_final = ρ_final * Vf
3. The mass of water that condenses out is the difference between the initial and final masses:
m_condensed = m_initial - m_final
Let's calculate these values:
Given:
Vi = 23.1 L (initial volume)
Vf = 7.7 L (final volume)
T = 21 °C = 21 + 273.15 K (temperature)
R = 56.4% (relative humidity)
vp = 2.5 kPa (vapor pressure)
ρ_initial = 18.4 g/m³ (initial water vapor density)
First, let's calculate the density of the water vapor following compression:
a) Density of water vapor following compression:
Pv = R * vp = 0.564 * 2.5 kPa = 1.41 kPa
n = (Pv * Vf) / (R * (T + 273.15))
n = (1.41 kPa * 7.7 L) / (56.4% * (21 + 273.15) K)
n = 0.01248 mol
ρv = m / Vf
m = ρv * Vf
m = (n * Molar mass of water) / Vf
(Using the molar mass of water, Molar mass of water = 18.015 g/mol)
m = (0.01248 mol * 18.015 g/mol) / 7.7 L
m ≈ 0.029 g/L
Now, let's calculate the mass of water that condenses out because of the compression:
b) Mass of water that condenses out:
m_initial = ρ_initial * Vi
m_initial ≈ 18.4 g/m³ * 23.1 L ≈ 425.04 g
m_final = ρ_final * Vf
m_final ≈ 0.029 g/L * 7.7 L ≈ 0.223 g
m_condensed = m_initial - m_final
m_condensed ≈ 425.04 g - 0.223 g ≈ 424.82 g
Therefore, the answers are:
a) The density the water vapor would have had following compression if it behaved like an ideal gas instead of experiencing partial condensation is approximately 0.029 g/L.
b) The mass of water that condenses out because of the compression is approximately 424.82 g.
To find the answers to these questions, we need to apply the ideal gas law and the concept of partial pressure.
a) To determine the density of water vapor after compression if it behaved like an ideal gas, we can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Since the volume and temperature are constant during compression, the pressure will change. We can calculate the final pressure using the partial pressure of water vapor.
The partial pressure of water vapor (Pvapor) can be calculated using the relative humidity (R) and the vapor pressure (vp):
Pvapor = R * vp
Now, using the ideal gas law equation, we can determine the density (ρ) by rearranging the equation:
ρ = PN / RT
where PN is the partial pressure of the water vapor.
b) To find the mass of water that condenses out due to compression, we need to calculate the initial and final number of moles. The difference between these two values will give us the number of moles that condensed.
The number of moles (n) can be calculated using the ideal gas law:
n = P * V / (R * T)
First, calculate the initial number of moles using the initial volume, temperature, and pressure. Then, calculate the final number of moles using the final volume, temperature, and pressure.
The difference in the number of moles will give the number of moles of water that condensed. Multiply this by the molar mass of water (18 g/mol) to find the mass of water condensate.
Note: It's important to work with consistent units throughout the calculations. In this case, the given units are L, kPa, ¢XC, and g.m−3.
Hope this helps! Let me know if you have any further questions.