Question

The air is said to be "thinner" in higher altitudes than at sea level. Compare the density of air at sea level where the barometric pressure is \(755 \mathrm{~mm} \mathrm{Hg}\) and the temperature is \(0^{\circ} \mathrm{C}\) with the density of air on top of \(\mathrm{Mt}\). Everest at the same temperature. The barometric pressure at that altitude is \(210 \mathrm{~mm} \mathrm{Hg}\) (3 significant figures). Take the molar mass of air to be \(29.0 \mathrm{~g} / \mathrm{mol}\).

Short answer

Answer: The air density at sea level is 1.251 kg/m³, while the air density at the top of Mt. Everest is 0.362 kg/m³. Therefore, the air is significantly thinner, or less dense, at the top of Mt. Everest compared to sea level.

Step-by-step solution

Convert temperature to Kelvin

Since we will use the Ideal Gas Law, we need to convert the temperature from Celsius to Kelvin. To do this, we add \(273.15\) to the temperature in Celsius: \(0^{\circ} \mathrm{C} + 273.15 = 273.15 \mathrm{K}\) Temperature at both locations is equal to \(273.15 \mathrm{K}\).

Convert pressure to Pascals

We need to convert the given pressures from mm Hg to Pascals (Pa) to use in the Ideal Gas Law. To convert from mm Hg to Pascals, we multiply the pressure value by 133.322. Sea level pressure: \(755 \mathrm{~mm} \mathrm{Hg} \times 133.322 = \mathrm{100667~Pa}\) Mt. Everest pressure: \(210 \mathrm{~mm} \mathrm{Hg} \times 133.322 = \mathrm{28000~Pa}\)

Calculate the density of air at both locations

We will use the Ideal Gas Law, \(PV=nRT\), to find the density of air at both locations. First, we will rearrange the equation to solve for the number of moles (n) and then convert the moles to mass using the molar mass of air. Density = \(\frac{m}{V} = \frac{n \cdot M}{V}\), where \(m\) is the mass, \(n\) is the number of moles, \(V\) is the volume, and \(M\) is the molar mass of air. We can rearrange the Ideal Gas Law formula to solve for \(n\): \(n = \frac{PV}{RT}\) Then, we can substitute this expression for \(n\) in the density formula: Density = \(\frac{(PV \cdot M)}{(RT)(V)} = \frac{PM}{RT}\) Now we can calculate the density of air at both locations using their respective pressure values and the molar mass \(29.0 \mathrm{g/mol}\): Sea level density: \(\rho_1= \frac{\mathrm{100667~Pa}\cdot 29.0 \mathrm{g/mol}}{(8.314 \mathrm{J/(mol \cdot K)})\cdot 273.15 \mathrm{K}} = 1.251 \mathrm{kg/m^3}\) Mt. Everest density: \(\rho_2 = \frac{\mathrm{28000~Pa}\cdot 29.0 \mathrm{g/mol}}{(8.314 \mathrm{J/(mol \cdot K)})\cdot 273.15 \mathrm{K}} = 0.362 \mathrm{kg/m^3}\)

Compare the densities

We found that the density of air at sea level is \(1.251 \mathrm{kg/m^3}\) and the density of air on top of Mt. Everest is \(0.362 \mathrm{kg/m^3}\). Therefore, the air is, indeed, "thinner" at higher altitude (on top of Mt. Everest) compared to sea level because of the lower density.

Key concepts

Air Density

Air density refers to the mass of air molecules present in a given volume. The density of air is influenced by several factors:

• Pressure: Higher pressure means more air molecules are packed into a space, increasing the density.
• Temperature: Warmer air is less dense because its molecules move faster and spread apart.
• Altitude: As altitude increases, pressure decreases, leading to lower air density.

In the problem, we see air density calculated at two different altitudes using the Ideal Gas Law. At sea level with higher pressure, the air density is about 1.251 kg/m³, while on Mt. Everest with lower pressure, it is 0.362 kg/m³. This illustrates why people often describe mountaintop air as "thin."

Pressure Conversion

Pressure conversion is essential when dealing with gas laws, as pressure must be in a consistent unit for formulas. To work within the Ideal Gas Law, pressure often needs to be converted to Pascals (Pa). One common conversion factor is:

• 1 mm Hg = 133.322 Pa

In the exercise solution, pressures were given in mm Hg. For sea level, 755 mm Hg was successfully converted to 100667 Pa. On Mt. Everest, the pressure was 210 mm Hg, converted to 28000 Pa. These converted values ensure that pressure is correctly integrated into the Ideal Gas Law calculations, aiding accurate determination of air density.

Molar Mass of Air

Molar mass is the mass of one mole of a substance, allowing us to relate volume of gas to its quantity. Air's molar mass is a measure of the combined masses of the various gases that compose it, most notably nitrogen and oxygen.

• In this problem, the molar mass of air was given as 29.0 g/mol.
• This value is crucial for converting moles to mass in gas equations like the Ideal Gas Law.

Using the formula for density involving molar mass, the air density can be calculated by combining pressure, volume, and molar mass of air at a given temperature. The consistency of the molar mass across conditions helps compare densities at different pressures and altitudes, such as at sea level versus atop Mt. Everest. This understanding assists in explaining why air density decreases with increasing altitude.