What would be the absolute humidity of air if 1 cu.m. air contains 4.08gm of vapour at 0°C temperature.
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4.08g water per cubic meter of air is an absolute humidity. It is an amount of water found in an amount of air. This is already in the standard form, but could be converted to others, such as percent mass.
Air density is proportional to pressure and inversely proportional to temperature. Using the specific gas constant R for dry air, which is 287.05 J/kg*K we can calculate the density with ρ = P/R*T. So at 273.15K (0C) and 101325 Pa (1 atm of pressure), we get a density of 1.292 kg/m^3. Now considering the water, 4.08g / 1.292kg + 4.08 g = 0.3% water vapor by mass.
Or possibly you mean relative humidity, which is a measure of the amount of moisture the air holds as a proportion of the maximum amount it could hold at that temperature. This is the number you see as a percentage in weather reports. First, you need to determine the maximum, or saturation, humidity at that temperature. A quick google search yields 5.018 + .32321 T + .0081847 T^2 + .00031243 T^3, in g/m^3 and with T in Celsius, to calculate this for temperatures around 0–40 degrees. Put in 0 for T, and we’re left with 5.018 g/m^3 water vapor in air. So our relative humidity is 4.08(g/m^3)/5.018(g/m^3), or approximately 81% humidity.
Air density is proportional to pressure and inversely proportional to temperature. Using the specific gas constant R for dry air, which is 287.05 J/kg*K we can calculate the density with ρ = P/R*T. So at 273.15K (0C) and 101325 Pa (1 atm of pressure), we get a density of 1.292 kg/m^3. Now considering the water, 4.08g / 1.292kg + 4.08 g = 0.3% water vapor by mass.
Or possibly you mean relative humidity, which is a measure of the amount of moisture the air holds as a proportion of the maximum amount it could hold at that temperature. This is the number you see as a percentage in weather reports. First, you need to determine the maximum, or saturation, humidity at that temperature. A quick google search yields 5.018 + .32321 T + .0081847 T^2 + .00031243 T^3, in g/m^3 and with T in Celsius, to calculate this for temperatures around 0–40 degrees. Put in 0 for T, and we’re left with 5.018 g/m^3 water vapor in air. So our relative humidity is 4.08(g/m^3)/5.018(g/m^3), or approximately 81% humidity.
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Answer:
4.08 gm/m 3
Explanation:
Absolute humidity
= Mass of water vapour
__________________
Volume of air
= 4.08
_____
1
= 4.08 gm/m3
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