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 \(29.0 \mathrm{~g} / \mathrm{mol}\).

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 \(273.15 \mathrm{K}\).

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: \(210 \mathrm{~mm} \mathrm{Hg} \times 133.322 = \mathrm{28000~Pa}\)

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: \(\rho_2 = \frac{\mathrm{28000~Pa}\cdot 29.0 \mathrm{g/mol}}{(8.314 \mathrm{J/(mol \cdot K)})\cdot 273.15 \mathrm{K}} = 0.362 \mathrm{kg/m^3}\)

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.