Question

Given that a sample of air is made up of nitrogen, oxygen, and argon in the mole fractions \(78 \% \mathrm{~N}_{2}, 21 \% \mathrm{O}_{2}\), and \(1.0 \% \mathrm{Ar}\), what is the density of air at standard temperature and pressure?

Short answer

The density of air at standard temperature and pressure is found by calculating the molar mass of air and using the ideal gas law. The molar mass of air is obtained as 28.96 g/mol by using the given mole fractions of nitrogen, oxygen, and argon. Applying the ideal gas law and substituting values, the density of air at STP is calculated to be 1.29 g/L.

Step-by-step solution

Calculate the molar mass of air

First, we need to calculate the molar mass of air using the given mole fractions of nitrogen (N₂), oxygen (O₂), and argon (Ar). The molar mass of air (M) can be calculated by multiplying the molar mass of each component by its mole fraction and then summing up the results. \(M = \%(N_2) * M(N_2) + \%(O_2) * M(O_2) + \%(Ar) * M(Ar)\) where \(\%(N_2)\), \(\%(O_2)\) and \(\%(Ar)\) represent the mole fractions of nitrogen, oxygen, and argon, respectively, and \(M(N_2)\), \(M(O_2)\) and \(M(Ar)\) represent the molar mass of nitrogen (28 g/mol), oxygen (32 g/mol), and argon (40 g/mol), respectively.

Calculate the density of air at STP

To find the density of air at STP, we'll use the ideal gas law equation: \(PV = nRT\) At standard temperature and pressure (STP), the temperature (T) is 273.15 K and the pressure (P) is 1 atm (101325 Pa). The universal gas constant (R) is 8.314 J/(mol·K). Rearrange the equation to solve for the density of air (ρ): \(\rho = \frac{nM}{V} = \frac{PM}{RT}\) Using the molar mass of air (M) obtained in Step 1, calculate the density of air at STP.

Substitute values and solve for the density of air at STP

Substitute the values for P, M, R, and T into the density equation: \(\rho = \frac{(101325 \, \text{Pa})(M)}{(8.314 \, \frac{\text{J}}{\text{mol·K}})(273.15 \, \text{K})}\) First, calculate the molar mass of air (M) from Step 1: \begin{align*} M &= 0.78 * 28 + 0.21 * 32 + 0.01 * 40 \\ &= 21.84 + 6.72 + 0.4 \\ &= 28.96 \, \frac{\text{g}}{\text{mol}} \end{align*} Now, plug in the value of M into the density equation and solve for ρ: \begin{align*} \rho &= \frac{(101325 \, \text{Pa})(28.96 \, \frac{\text{g}}{\text{mol}})}{(8.314 \, \frac{\text{J}}{\text{mol·K}})(273.15 \, \text{K})} \\ &= \frac{2,934,675.2 \, \frac{\text{g·Pa}}{\text{mol·K}}}{2,270.9 \, \frac{\text{J}}{\text{mol·K}}} \\ &= 1.29 \, \frac{\text{g}}{\text{L}} \end{align*} Therefore, the density of air at standard temperature and pressure is 1.29 g/L.

Key concepts

Molar Mass Calculation

Understanding how to calculate molar mass is crucial when determining the properties of gases such as air. Molar mass, measured in grams per mole (g/mol), tells you the mass of one mole of a substance. For air, which is a mixture of gases, this involves a weighted average of the molar masses of each component according to their mole fractions. In our case, air consists principally of nitrogen (\( ext{N}_2 \)), oxygen (\( ext{O}_2 \)), and argon (\( ext{Ar} \)).
To find the molar mass of this air mixture, use the following formula:

• Multiply the molar mass of each gas by its respective mole fraction.
• Sum up all these values to get the total molar mass of the mixture.

For instance, with nitrogen's molar mass at 28 g/mol and its mole fraction at 78%, oxygen's molar mass at 32 g/mol with 21%, and argon's at 40 g/mol with 1%, the calculation turns into:\[M = 0.78 imes 28 + 0.21 imes 32 + 0.01 imes 40 = 28.96 \text{ g/mol}\] This value enables us to analyze further properties of air.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that relates the properties of gases. It's expressed as:\[ PV = nRT \]where:

• \( P \) is pressure,
• \( V \) is volume,
• \( n \) is the number of moles,
• \( R \) is the universal gas constant (8.314 J/(mol·K)),
• \( T \) is temperature in Kelvin.

This equation is idealized as it assumes gases behave perfectly under a large range of conditions. However, it provides excellent approximations under typical conditions, like at standard temperature and pressure. One can derive various other properties of gases, such as density, using this equation by rearranging for different variables based on what is known. For instance, for density (\( \rho \)), we can rearrange to:\[ \rho = \frac{PM}{RT} \]This equation becomes highly useful to calculate things like the density of air.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure (STP) is an essential reference set of conditions in the study of gases. At STP, the temperature is set at 273.15 K (0°C) and the pressure at 1 atmosphere (101325 Pa). These conditions provide a common ground for comparing the behaviors and properties of gases, avoiding discrepancies that arise at different temperatures and pressures.
When working with the ideal gas law, STP allows for simplified calculations, as constants like temperature and pressure are known and can directly be substituted into the equation. This results in clear, comparable measurements such as density or volume of a gas at these standard conditions. Understanding STP is crucial for anyone dealing with gas calculations.

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture. It's the ratio of the number of moles of a particular substance to the total number of moles of all substances present in the mixture. It is generally expressed as a percentage.
For example, in our air mixture, the mole fraction helps us understand the composition:

• 78% nitrogen means that out of every 100 moles of air, 78 are nitrogen.
• 21% oxygen corresponds to 21 out of every 100 moles being oxygen.
• 1% argon means 1 mole out of every 100 moles is argon.

Mole fraction is not only essential for calculating molar mass but also vital in reactions involving gas mixtures, as it directly influences how each component behaves under given conditions. Knowing the mole fraction helps in accurate modeling of air's properties like density under STP conditions.