Question

2. Calcule la masa de vapor de agua presente en un salón de 520 m de volumen que contiene aire a 27 °C en un día donde la humedad relativa era de 55 %. (Presión de vapor a 27 °C = 26.74 Torr) 3. Dado que la densidad de aire a 0.987 bar y 27 °C es de 1.146 kg/m", calcule los moles totales de aire y la masa total del aire que se encuentra en 400 m. Determine además la fracción molar y la presión parcial de Oxigeno y Nitrógeno asumiendo que el aire se compone solamente de estos dos gases.

2. Calculate the mass of water vapor present in a 520 m3 volume room containing air at 27 ºC on a day where the relative humidity was 55%. (Vapor pressure at 27 ºC = 26.74 Torr)

3.Given that the air density at 0.987 bar and 27 ºC is 1,146 kg / m3, calculate the total moles of air and the total mass of the air found in 400 m 3. Also determine the molar fraction and the partial pressure of Oxygen and Nitrogen assuming that air is made up of only these two gases.

Answer 1

2.Solution

From the question at 27 oC, the water has a vapor pressure of 26.74 mmHg

Hence for 26.74 mmHg *(101325 Pa/760 mmHg)

= 3565.04 Pa

As we know that the partial pressure of water = vapor pressure of water *relative humidity

= 3565.04 Pa *0.55

= 1960.772 Pa

Now n = PV/(RT)

= (1960.772 Pa* 520 m3)/ (8.314 Pa*m3/mol/K *300.15 K)

= 408.58 moles of water

=408.58 mol water *(18 g/mol)
= 7354.44 grams