When water evaporates, it changes from liquid to gas at a temperature below it's boiling point. The volume this amount of water occupies in the gas phase is much greater than in the liquid phase, and depends on its temperature and pressure. Since you didn't specify the temperature and pressure at which the evaporation is taking place, I will use 25 degrees C and 1.00 atm as an example. We also need to know the number of moles of water, which can be calculated from its mass. Water has a density of 1.00g/ml at 25 degrees C, so 18 ml of water = 18 grams. Water has a molar mass of 18 grams/mole, so 18 gram of water = 1.0 mole.
The Ideal Gas Equation PV=nRT, can be used to calculate the volume:
V = nRT/P
n = 1.0 mole
R = ideal gas constant = 0.0821 L-atm/mol-K
T = 25 degrees C = 298 K
P = 1.00 atm
V = (1.0 mol)(0.0821 L-atm/mol-K)(298 K)/(1.00 atm)
V = 24.5 Liters
The molar volume, 22.4 L, is the volume of one mole of any gas at standard temperature and pressure, 273 K and 1.00 atm. If we used 272K (0 degrees C) for temperature the result of this calculation would be 22.4 L. You can substitute a different temperature and/or pressure into the ideal gas equation to find the volume under other conditions.
No comments:
Post a Comment