Question

In: Chemistry

“Dry air” is defined as air with no water vapor, and the molecular weight of air,...

“Dry air” is defined as air with no water vapor, and the molecular weight of air, Mair = 28.97 kg/kmol, is for dry air (zero humidity). “Wet air” is typically defined as air with 100% humidity.

(a) Calculate the mol fraction of water vapor in wet air at STP conditions. Give your answer in units of PPM to three significant digits.

(b) Compare the molecular weight of dry air and wet air at STP conditions. Which air is heavier? Explain. Comment about how humid air feels compared to dry air. In particular, in the summer time most people say that high-humidity air feels “thicker” or “heavier” than low-humidity air at the same temperature. Does this feeling agree with your calculations?

Solutions

Expert Solution

(a) Dry air has a molar mass of 0.029 kg. It is denser than water vapour, which has a molar mass of 0.018 kg. Therefore, humid air is lighter than dry air.

If the total atmospheric pressure is P and the water vapour pressure is p, the partial pressure of the dry air component is P - p .

The weight ratio of the two components, water vapour and dry air is:

kg water vapour / kg dry air = 0.018 ×p / ( 0.029 ×(P - p ) ) = 0.62 ×p / (P - p )

At room temperature P - p is nearly equal to P, which at ground level is close to 100,000 Pa,

and at 100% humidity p = 1

so, approximately: kg water vapour / kg dry air = 0.62 ×10-5

(b) The addition of water vapor to air (making the air humid) reduces the density of the air, which may at first appear counter-intuitive. This occurs because the molar mass of water (18 g/mol) is less than the molar mass of dry air(around 29 g/mol). For any gas, at a given temperature and pressure, the number of molecules present is constant for a particular volume (Avogadro's Law). So when water molecules (water vapor) are added to a given volume of air, the dry air molecules must decrease by the same number, to keep the pressure or temperature from increasing. Hence the mass per unit volume of the gas (its density) decreases.

in the summer time most people say that high-humidity air feels “thicker” or “heavier” than low-humidity air at the same temperature but this feeling does not agree with our calculations


Related Solutions

A tank contains 21 kg of dry air and 0.3 kg of water vapor at 30°C...
A tank contains 21 kg of dry air and 0.3 kg of water vapor at 30°C and 100 kPa total pressure. Determine (a) The specific humidity (b) The relative humidity (c) The specific enthalpy of the mixture in the tank as (kJ/kg-mixture) (calculate by hand) (d) The specific enthalpy of the mixture in the tank as (kJ/kg-dry air) (calculate by hand) (e) Explain (in 1-2 sentences) the difference between the values in c and d – which value agrees with...
A mixture of dry air and water vapour is at a temperature of 25C. It is...
A mixture of dry air and water vapour is at a temperature of 25C. It is under a pressure of 100kPa. The dew point temperature is 15C. What is: a) Partial pressure of the water vapour? b) Relative humidity? c) Specific humidity? d) Specific volume of the dry air?
The dry weight of a natural stone sample is 336 g, the water-saturated weight is 340...
The dry weight of a natural stone sample is 336 g, the water-saturated weight is 340 g, and the water-saturated sample is 196 g in water. The volume of the 50 gr powder sample obtained by grinding the same stone is 18.7 cm3 measured at the picnometer. Calculate the unit weight, specific weight, and weight and volume water absorption rates of this stone.
1- what is the effect of water vapor on the air entering the alveoli?
1- what is the effect of water vapor on the air entering the alveoli?
A person inhales air richer in O2 and exhales air richer in CO2 and water vapor....
A person inhales air richer in O2 and exhales air richer in CO2 and water vapor. During each hour of sleep, a person exhales a total of 300. L of this CO2-enriched and H2O-enriched air. (a) If the partial pressures of CO2 and H2O in exhaled air are each 20.0 torr at 26.5°C, calculate the masses of CO2 and of H2O exhaled in 1.00 hour of sleep. CO2 ___ ?g H2O_____ ?g (b) How many grams of body mass does...
Q1-Explain on a molecular basis the difference between the vapor pressure of water and an aqueous...
Q1-Explain on a molecular basis the difference between the vapor pressure of water and an aqueous solution of sodium chloride at the same temperature Q2-what is the atomic weight of an element that crystallizes in a face-centered cubic unit cell with an atomic radius of 1.37 A and a density of 12.023 g/cm^3
A gas mixture containing water vapor has a dry basis molar composition of 8.5% CO, 10.5%...
A gas mixture containing water vapor has a dry basis molar composition of 8.5% CO, 10.5% CO2,0.5% O2 and 80.5% N2. The gas s at 620c and 1 atm with a dewpoint of 57c at a flow rate of 1270mol/h.The gas is cooled to 425C by spraying water at 20C into the gas.Assume that the spray unit is adiabatic .calculat the required water feed rate in kg /h.
Dry air is bubbled through 25.0 liters of water at a rate of 15 L (STP)...
Dry air is bubbled through 25.0 liters of water at a rate of 15 L (STP) / Min. The air leaving the liquid is saturated with water at 25 C and 1.5 atm. How long will it take for all the water to vaporize ? answer: 67.8 days
Compound A is a water-soluble solid at room temperature with a molecular weight of 138.12 amu....
Compound A is a water-soluble solid at room temperature with a molecular weight of 138.12 amu. Describe one (or more) experiment(s) that you would perform in lab to determine if Compound A is a strong acid, weak acid, strong base, weak base, or none of these in water. State what you would do, what you would measure, and how you would decide the nature of Compound A in water?
Given the mole of air as 9.2x10-5 mol, the volume of water vapor, after correction, is...
Given the mole of air as 9.2x10-5 mol, the volume of water vapor, after correction, is 5.0 mL while the temperature is 75 oC, what is the vapor pressure of water in mmHg if the atmospheric pressure (Patm ) is 0.989 atm.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT