To calculate the mass of 1 liter of moist air at a temperature of 27 degrees Celsius, with a barometric pressure of 753.6 mm of Hg and a dew point of 16.1 degrees Celsius, we need to follow several steps. This involves determining the partial pressures of the dry air and water vapor, as well as using the ideal gas law. Let’s break this down step by step.

Step 1: Determine the Saturated Vapor Pressure

We know that the saturated vapor pressure of water at the dew point (16.1 degrees Celsius) is given as 13.6 mm of Hg. This means that at this temperature, water vapor exerts a pressure of 13.6 mm of Hg when it is saturated.

Step 2: Calculate the Partial Pressure of Dry Air

The total atmospheric pressure (barometric pressure) is the sum of the partial pressures of dry air and water vapor. We can express this as:

P_total = P_dry_air + P_water_vapor

Rearranging this gives us:

P_dry_air = P_total - P_water_vapor

Substituting the known values:

P_dry_air = 753.6 mm of Hg - 13.6 mm of Hg = 740 mm of Hg

Step 3: Calculate the Density of Dry Air

To find the density of dry air at the given conditions, we can use the ideal gas law, which states:

P = ρRT

Where:

• P = pressure in atm
• ρ = density in g/L
• R = ideal gas constant (0.0821 L·atm/(K·mol))
• T = temperature in Kelvin

First, convert the pressure from mm of Hg to atm:

P_dry_air = 740 mm of Hg × (1 atm / 760 mm of Hg) = 0.974 atm

Next, convert the temperature from Celsius to Kelvin:

T = 27 + 273.15 = 300.15 K

Now, we can rearrange the ideal gas law to solve for density:

ρ = P / (RT)

Substituting the values:

ρ_dry_air = 0.974 atm / (0.0821 L·atm/(K·mol) × 300.15 K) = 0.0394 mol/L

To convert moles to grams, we multiply by the molar mass of dry air (approximately 29 g/mol):

Density of dry air = 0.0394 mol/L × 29 g/mol = 1.1446 g/L

Step 4: Calculate the Density of Water Vapor

Using the same ideal gas law approach for water vapor:

P_water_vapor = 13.6 mm of Hg = 0.01789 atm

Using the same temperature:

ρ_water_vapor = P_water_vapor / (RT)

Substituting the values:

ρ_water_vapor = 0.01789 atm / (0.0821 L·atm/(K·mol) × 300.15 K) = 0.000724 mol/L

Now, converting moles to grams:

Density of water vapor = 0.000724 mol/L × 18 g/mol = 0.013032 g/L

Step 5: Calculate the Total Density of Moist Air

The total density of moist air can be calculated by adding the densities of dry air and water vapor:

Density_moist_air = ρ_dry_air + ρ_water_vapor

Density_moist_air = 1.1446 g/L + 0.013032 g/L = 1.157632 g/L

Step 6: Calculate the Mass of 1 Liter of Moist Air

Since we have the density of moist air, we can find the mass of 1 liter:

Mass = Density × Volume

Mass = 1.157632 g/L × 1 L = 1.157632 g

Thus, the mass of 1 liter of moist air at the given conditions is approximately 1.16 grams.