Problem 172 - Chemistry

Question

Which of the following has a greater mass: a sample of air of volume \(V\) at a certain temperature \(T\) and pressure \(P\) or a sample of air plus water vapor having the same volume and at the same temperature and pressure?

Step-by-Step Solution

Answer

A sample of air of volume V at a certain temperature T and pressure P has a greater mass than the same volume of air plus water vapor at the same temperature and pressure.

Step 1: Understand Avogadro’s law

The first step in tackling this problem is understanding Avogadro’s law. It states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. This means that whether the volume 'V' is filled with air or a mixture of air and water vapor, the total number of molecules will be same.

Step 2: Understand the nature of gases

While the number of molecules are equal, we must also evaluate the nature of the molecules making up the gases. Air is primarily composed of Nitrogen and Oxygen, whereas water vapor is composed of Hydrogen and Oxygen. Oxygen has a molecular weight of 16, Nitrogen has 14, and Hydrogen has 1. A single molecule of water vapor (H2O) would hence be lighter than a molecule of Nitrogen (N2) or Oxygen (O2) which are the primary constituents of air.

Step 3: Final Evaluation

Since the total number of molecules are same and a molecule of water vapor weighs less than Nitrogen or Oxygen, replacing some of the air with water vapor reduces the overall mass. Hence, a sample of air of volume 'V' at a certain temperature and pressure would have a greater mass than a sample of air plus water vapor having the same volume and at the same temperature and pressure.

Key Concepts

Understanding the Mass of Gases

When comparing the mass of different gases, it's crucial to consider not just the volume they occupy, but also the identity of the gas molecules. Avogadro's Law simplifies part of this comparison by ensuring that equal volumes of gas, at the same temperature and pressure, contain the same number of molecules. Therefore, if we have one volume of pure air and the same volume of air mixed with water vapor, the number of molecules present in both samples will be identical.
However, the mass of these molecules can differ significantly due to their molecular weights. Air is primarily composed of nitrogen and oxygen, which have relatively higher molecular weights. This weight translates to a greater mass for the same volume.
On the other hand, when water vapor, which is lighter compared to the nitrogen and oxygen in air, is introduced into the mixture, it takes the place of some heavier air molecules. This substitution means that the overall mass of the air and water vapor mixture will be less than that of pure air under identical conditions.

Comparing Molecular Weights

Molecular weight, which is sometimes referred to as molar mass, is an essential factor when comparing the mass of gases. It is the sum of the atomic weights of all atoms in a molecule. For instance, oxygen molecules (O2) are heavier because each oxygen atom has a molecular weight of 16 AMU (Atomic Mass Units), and therefore a molecule of oxygen weighs 32 AMU (16 x 2).
Nitrogen molecules (N2), similarly possess a molecular weight of 28 AMU (14 x 2).

• Oxygen (O2) – 32 AMU
• Nitrogen (N2) – 28 AMU
• Water vapor (H2O) – 18 AMU

Water vapor (H2O), which is part of the gas mixture in this problem, has only 18 AMU due to its composition of hydrogen and oxygen (1 + 1 + 16). Therefore, when lighter molecules like water vapor replace heavier molecules like nitrogen or oxygen, the total molecular weight of the gas mixture decreases, resulting in a lighter sample mass for the same volume.

Properties of Gas Mixtures

Gases exhibit interesting properties when mixed, particularly when comparing their collective physical characteristics to those of their individual components. Avogadro's Law guides us that a mixture will have the same number of molecules per unit volume, provided the temperature and pressure are consistent. This regulatory nature ensures that while the molecular count remains unchanged, the mass and density of the mixture can alter based on the individual gases involved.
Here's how this works in our problem:

• The introduction of water vapor, a lighter gas, does not alter the number of molecules but changes the mixture's mass due to its lower molecular weight.
• Despite being a lighter gas, water vapor retains the gas law principles that dictate behavior concerning volume, temperature, and pressure.
• This reduction in mass affects the density and thereby ensures the mass of air alone is greater than the mass of the same air interspersed with water vapor.

Thus, understanding these properties helps in accurately assessing and predicting the behavior of gas mixtures in various conditions, making such knowledge invaluable in fields such as meteorology, engineering, and environmental science.