- #1
Hendrick
- 43
- 0
Density of water vapour in the air involving compression
A volume Vi = 23.2 L of air at temperature T = 22 °C is compressed at constant temperature to a volume Vf = 11.0 L. The relative humidity of the air before compression is H = 59.4 %. The saturated vapour pressure of water at this temperature is 2.67 kPa, corresponding to a vapor density ρw = 19.6 g m^–3.
A)What would be the density of water vapour in the air following compression if it behaved like an ideal gas instead of experiencing partial condensation?
B)What is the density of water vapour in the air following compression?
C)What mass of water condenses out because of the compression?
PV = nRT
P1 * V1 = P2 * V2
P = rho * R/M * T
n = m/M
n = number of moles of substance
M = molar mass
m = mass
P = pressure
V = volume
rho = density
R = universal gas constant
T = absolute temperature
A)P1 * V1 / V2 = P2
2670 * (23.2/11) = P2
rho = P2 * M/(R * T)
= P2 * 0.018/(8.314 * (273 + 22))
= 0.04132825
Using the molar mass of water (0.018 kg mol^-1) and P1 as the saturated vapour pressure of water at 22 °C (2.67 kPa)
The actual answer is 24.6 g m^-3
----------------------------------------------------------------
B)rho = P * M/(R * T)
= 2670 * 0.018/(8.314 * (273 + 22))
= 0.019595291
Using the molar mass of water (0.018 kg mol^-1) and P as the saturated vapour pressure of water at 22 °C (2.67 kPa)
Actual answer is 19.6 g m^-3
-- is this just the value of vapor density ρw = 19.6 g m^–3. from the line at the beginning of the problem "corresponding to a vapor density ρw = 19.6 g m^–3."
----------------------------------------------------------------
C) PV = nRT
n = PV/RT
n = m/M
P = rho * R/M * T
=> PV/RT = m/M
=> [rho * R/M * T ] *V/RT = m/M
rho * V = m
19.6 * (11/1000) = 0.2156
Using ρw = 19.6 g m^–3, Vf = 11.0 L
The actual answer is 0.0545 g
-- Help would be much appreciated
Homework Statement
A volume Vi = 23.2 L of air at temperature T = 22 °C is compressed at constant temperature to a volume Vf = 11.0 L. The relative humidity of the air before compression is H = 59.4 %. The saturated vapour pressure of water at this temperature is 2.67 kPa, corresponding to a vapor density ρw = 19.6 g m^–3.
A)What would be the density of water vapour in the air following compression if it behaved like an ideal gas instead of experiencing partial condensation?
B)What is the density of water vapour in the air following compression?
C)What mass of water condenses out because of the compression?
Homework Equations
PV = nRT
P1 * V1 = P2 * V2
P = rho * R/M * T
n = m/M
n = number of moles of substance
M = molar mass
m = mass
P = pressure
V = volume
rho = density
R = universal gas constant
T = absolute temperature
The Attempt at a Solution
A)P1 * V1 / V2 = P2
2670 * (23.2/11) = P2
rho = P2 * M/(R * T)
= P2 * 0.018/(8.314 * (273 + 22))
= 0.04132825
Using the molar mass of water (0.018 kg mol^-1) and P1 as the saturated vapour pressure of water at 22 °C (2.67 kPa)
The actual answer is 24.6 g m^-3
----------------------------------------------------------------
B)rho = P * M/(R * T)
= 2670 * 0.018/(8.314 * (273 + 22))
= 0.019595291
Using the molar mass of water (0.018 kg mol^-1) and P as the saturated vapour pressure of water at 22 °C (2.67 kPa)
Actual answer is 19.6 g m^-3
-- is this just the value of vapor density ρw = 19.6 g m^–3. from the line at the beginning of the problem "corresponding to a vapor density ρw = 19.6 g m^–3."
----------------------------------------------------------------
C) PV = nRT
n = PV/RT
n = m/M
P = rho * R/M * T
=> PV/RT = m/M
=> [rho * R/M * T ] *V/RT = m/M
rho * V = m
19.6 * (11/1000) = 0.2156
Using ρw = 19.6 g m^–3, Vf = 11.0 L
The actual answer is 0.0545 g
-- Help would be much appreciated
Last edited: