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When dealing with gases, the mole fraction is a helpful concept. It indicates the proportion of moles of a specific gas to the total number of moles in a gas mixture. In essence, mole fractions help us to understand the composition of a gaseous mixture.
For example, if a sample of air is composed of nitrogen, oxygen, and argon, their respective mole fractions will tell us how much each gas contributes to the total mixture. In the exercise, these are given as 0.78 for nitrogen (\( \text{N}_2 \)), 0.21 for oxygen (\( \text{O}_2 \)), and 0.01 for argon (Ar).
Mole fraction, often denoted by the symbol \( x \), is unitless and always adds up to 1 for the entire mixture:

• \( x_{N_2} = 0.78 \)
• \( x_{O_2} = 0.21 \)
• \( x_{Ar} = 0.01 \)

The molar mass is the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). It is crucial when calculating properties of gas mixtures, such as air.
Each component of air has its own distinct molar mass:

• Nitrogen (\( \text{N}_2 \)): approximately 28 g/mol
• Oxygen (\( \text{O}_2 \)): approximately 32 g/mol
• Argon (Ar): approximately 40 g/mol

By multiplying the mole fraction of each gas with its molar mass, and summing these products, we can find the average molar mass of the mixture. The formula for the average molar mass (\( M \)) of air presented in the exercise is:\[M = x_{N_2}M_{N_2} + x_{O_2}M_{O_2} + x_{Ar}M_{Ar}\]Substituting the values:\[M = (0.78 \cdot 28) + (0.21 \cdot 32) + (0.01 \cdot 40) \approx 28.96 \, \text{g/mol}\]This weighted average gives us a representative molar mass for the entire air sample.

Density is defined as the mass per unit volume of a substance. To find the density of a gas, we can use the ideal gas law. This law describes the behavior of ideal gases and connects several variables.The ideal gas law is:\[ PV = nRT \]Where:

• \( P \) is the pressure
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant (0.0821 L atm/mol K)
• \( T \) is the temperature in Kelvin

Density (\( \rho \)) is mass (\( m \)) per volume (\( V \)), and can be derived from the ideal gas law as:\[ \rho = \frac{PM}{RT} \]
Here, \( M \) is the molar mass calculated in the previous step. At standard temperature and pressure (i.e., 1 atm, 273.15 K), using this equation allows us to calculate the density of air when given its molar mass:\[\rho = \frac{(1 \, \text{atm})(28.96 \, \text{g/mol})}{(0.0821 \, \text{L atm/mol K})(273.15 \, \text{K})} \approx 1.29 \, \text{g/L}\]

Standard temperature and pressure (STP) is a reference point used in chemistry to provide a common reference for reporting properties of chemicals and gases.
STP is defined as a temperature of 273.15 Kelvin (0 掳C) and a pressure of 1 atmosphere (atm). Using these conditions simplifies comparisons and calculations involving gases.
With these fixed conditions, you can compare how different gases behave or calculate the density and volume of gases under the same conditions consistently. In the exercise, STP conditions were crucial as they provided the necessary constants to calculate the density of air accurately. As a result, we used the well-established values of pressure and temperature to simplify our calculations with the ideal gas law.