Question

A quantity of liquid water comes into equilibrium with the air in a closed container, without completely evaporating, at a temperature of \(25.0^{\circ} \mathrm{C}\). How many grams of water vapor does a liter of the air contain in this situation? The vapor pressure of water at \(25.0^{\circ} \mathrm{C}\) is \(3.1690 \mathrm{kPa}\).

Short answer

The mass of water vapor in a liter of air under these conditions is approximately 0.02286 grams.

Step-by-step solution

Convert the vapor pressure to Pascals

Given that the vapor pressure of water at 25.0°C is 3.1690 kPa, we need to convert it to Pascals (Pa) for use in the ideal gas law formula. To do this, use the conversion factor 1 kPa = 1000 Pa. So, 3.1690 kPa * 1000 Pa/kPa = 3169 Pa.

Set up the ideal gas law formula

The ideal gas law formula is PV = nRT, where P is pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is temperature in Kelvin. We are given the pressure P (3169 Pa) and need to find the number of moles (n) in 1 liter (0.001 m^3) of air. We will use the ideal gas constant R=8.314 J/(mol*K).

Convert the temperature to Kelvin

We are given the temperature in Celsius, so we need to convert it to Kelvin for use in the ideal gas law formula. To do this, add 273.15 to the temperature in Celsius. T = 25.0 + 273.15 = 298.15 K.

Solve for the number of moles (n) using the ideal gas law formula

Now we have all the necessary values to solve for the number of moles (n) using the ideal gas law formula: PV = nRT n = PV/RT n = (3169 Pa * 0.001 m^3)/(8.314 J/(mol*K) * 298.15 K) n ≈ 0.00127 mol

Calculate the mass of water vapor in grams

Now that we have calculated the number of moles of water vapor in a liter of air, we can use the molar mass of water (approximately 18 g/mol) to find the mass of the water vapor in grams. Mass = n * molar mass of water Mass = 0.00127 mol * 18 g/mol Mass ≈ 0.02286 g So, a liter of air contains approximately 0.02286 grams of water vapor at equilibrium with liquid water and 25.0°C.