Question

At \(10.0^{\circ} \mathrm{C}\) and \(102.5 \mathrm{kPa},\) the density of dry air is \(1.26 \mathrm{~g} / \mathrm{L}\) What is the average "molar mass" of dry air at these conditions?

Short answer

The average molar mass of dry air is approximately 28.8 g/mol.

Step-by-step solution

Write Down the Ideal Gas Law

Recall the ideal gas law equation: \[ PV = nRT \]Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

Convert Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin using the formula: \[ T(K) = T(^{\text{°}}C) + 273.15 \]Plug in the given temperature: \[ T = 10.0 + 273.15 = 283.15 \text{ K} \]

Use the Density Formula

Density \( \rho \) can be related to the molar mass \( M \) by using \[ \rho = \frac{m}{V} \]Here, m is the mass and V is the volume.

Rearrange the Ideal Gas Law

To find the molar mass, rearrange the ideal gas law: \[ n = \frac{PV}{RT} \]Since \( n \) (number of moles) can also be written as \( \frac{m}{M} \) where m is mass and M is molar mass, we get: \[ \rho V = nM \]

Express Molar Mass in Terms of Density

Substitute the relationship between density and mass into the ideal gas law: \[ M = \frac{RT \rho}{P} \]

Substitute the Known Values

Substitute the known values into the formula: \[ R = 8.314 \text{ J/(mol·K)}, \ T = 283.15 \text{ K}, \ \rho = 1.26 \text{ g/L} = 0.00126 \text{ kg/L}, \ P = 102.5 \text{ kPa} = 102500 \text{ Pa} \]Thus: \[ M = \frac{8.314 \times 283.15 \times 0.00126}{102500} \]

Calculate the Molar Mass

Perform the calculation: \[ \ M \approx 0.0288 \text{ kg/mol} = 28.8 \text{ g/mol} \]

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. The equation is given as \[ PV = nRT \]. This equation is incredibly useful because it helps us understand and calculate the behavior of gases under various conditions. Here’s a simple breakdown:

• \(P\): Pressure of the gas.
• \(V\): Volume of the gas.
• \(n\): Number of moles of the gas.
• \(R\): Ideal gas constant, which is always \(8.314 \text{ J/(mol·K)}\).
• \(T\): Temperature of the gas in Kelvin.

We can use this law to find unknown values when other variables are given. In our problem, we need to determine the molar mass. We rearrange the ideal gas law to find the number of moles, \(n\), and simplify that to solve for molar mass.

Density Formula

Density is defined as the mass per unit volume of a substance, and it's given by the formula \( \rho = \frac{m}{V} \). Here's a quick breakdown to make it more digestible:

• \( \rho \): Density, usually in units of \( \text{g/L} \) (grams per liter) for gases.
• \(m\): Mass of the substance.
• \(V\): Volume that the substance occupies.

In the context of our problem, the density of dry air is given as 1.26 g/L. This density can be related to the molar mass through a manipulation of the ideal gas law. By substituting the density into the ideal gas law, we can eventually express the molar mass formula:\[ M = \frac{RT \rho}{P} \].

Temperature Conversion

In many chemistry problems, we need to convert temperatures to Kelvin to use in our calculations, as the ideal gas law requires the temperature to be in Kelvin. The conversion is quite straightforward:\[ T(K) = T(^{\text{°}}C) + 273.15 \].This formula simply adds 273.15 to the Celsius temperature to convert it to Kelvin. For instance, in our specific problem, the given temperature is 10.0°C. Converting it to Kelvin:\[ T = 10.0 + 273.15 = 283.15 \text{ K} \].It's a mandatory step to ensure all values align correctly in the ideal gas law calculations.