![]() ![]() With the water added to the mixture, the portion of heavy molecules (air) has decreased and the portion of light molecules (water vapor) has increased in our cube. Thus, the density of water vapor (18 grams/mole) is less than dry air (28.5 grams/mole). This derivation proves that the density of an air molecule is directly proportional to its molar mass. In the final step, m/V was changed to density. The rest of the variables will remain constant. ![]() The variable M represents the molar mass of the gas molecule. To do this, we will break down the ideal gas equation to solve for density. Now that we know which particles have a greater molar mass, let us prove that density is proportional to molar mass. Water vapor (H2O=18 grams/mole) now takes up some of the space that the nitrogen and oxygen molecules used to occupy. Now, let us add some humidity into the equation. These molecules will move in and out of our cube of air as they please. For details on how these masses were calculated, see our post about the molar masses of gases. It is about 78% nitrogen (N2=28 grams/mole), 21% oxygen (O2=32 grams/mole) and a small amount of other gases. Let’s say we have a cubic foot of dry air. ![]() The law states that the product of the pressure and the volume of a gas is equal to the product of the amount of substance (expressed as moles), the universal gas constant, and the temperature of the gas. So what is an ideal gas anyway? The ideal gas law is used to assume that certain properties about a gas will remain constant. In our previous post, the question was posed: Is it easier to hit a home run on a humid night or a dry night at the same ball park? The following explanation will help us figure this out. ![]() Which is more dense, humid air or dry air? ![]()
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