Question

a clear day where the morning temperature is 26.5 degree celsius and atmospheric pressure is 760 mm Hg, the moisture content of air was measured to be 1.75 percent by mole. Solve for the following:

1. Dew point temperature
2. Molal humidity
3. Absolute humidity
4. Relative humidity

Answer 1

* Temperature = 26.5% Pressure = 760 mmHg Moistuse content of air = 1:75 mol % moles of water moles of humid air. of = 0.0175 moles of welete moles of humid air. 1.75 100 oso, Belice Bendo ao Migration molal humidity moles (1-0.0175 moles of dry 0.0175 of water air 0.0178 moksof water moles of dry ais. :. Molal humidity = of watere 0.0178 moles moles of dry ais. Moleculas weight of watere = 18 g mol. Molecular weight of Air = 29 g mol. .. Absolute Hurudity = 0.0178 moles x 18,2 mol I mole X 29 g mol 0.01105 gram watere gram of dry air. 0.01105_gram watere air Absolute Humidity gram of dry Ang @ Relative humidity = Pw X100 psat Herce, Pw = pastial pressure of watero psat - Vapour pressure of watere

of watere - so, partial presure of watere = ywx PT Yw x PT (Ym = mole fraction 0.0175 X 760 mm Hg Vapour pressure watere at 26.500 = 25.987 mm Hg (From properety table of watere) :. Relative Humidity = 0.0175X760 of X 100 25.987 Relative Humidity 51.18 % From psychometric chart, At Humidity of 0.01105 gram watere gram o of dry air Dew point temperature 1505°C Aus