Question

An ideal gas has a density of \(0.0899 \mathrm{~g} / \mathrm{L}\) at \(20.00^{\circ} \mathrm{C}\) and \(101.325 \mathrm{kPa}\). Identify the gas.

Short answer

Based on the given density, temperature, and pressure values, the ideal gas with a molar mass of approximately 28.96 g/mol is identified as nitrogen (N₂).

Step-by-step solution

Convert the given values to appropriate units

Given density: \(0.0899 \mathrm{~g} / \mathrm{L}\) Temperature: \(20.00^{\circ} \mathrm{C}\) Pressure: \(101.325 \mathrm{kPa}\) First, let's convert the given values to the units compatible with the ideal gas law: 1. Convert density to \(\mathrm{kg/m^3}\): \(0.0899 \mathrm{~g} / \mathrm{L} \times 1000 \mathrm{~mg} / \mathrm{g} \times 1 \mathrm{~L} / 1000 \mathrm{~mL} = 89.9 \mathrm{~kg/m^3}\) 2. Convert temperature to kelvins (K): \(T_{K} = 20.00 + 273.15 = 293.15 \mathrm{K}\) 3. Convert pressure to pascals (Pa): \(101.325 \mathrm{kPa} \times 1000 \mathrm{~Pa} / \mathrm{kPa} = 101325 \mathrm{~Pa}\) Now we have the following values: Density: \(89.9 \mathrm{~kg/m^3}\) Temperature: \(293.15 \mathrm{K}\) Pressure: \(101325 \mathrm{~Pa}\)

Rewrite the ideal gas law using density and molar mass

The ideal gas law is written as \(PV = nRT\). We can rewrite the equation in terms of density and molar mass: \(n = m/M\), where m is the mass of the gas and M is the molar mass. \(m/V = \rho\), where \(\rho\) is the density of the gas. Therefore, \(\rho = nM/V\), and we can substitute this into the ideal gas law: \(P = \rho RT/M\)

Solve for the molar mass of the gas

Now, we will solve for the molar mass (M) using the given values for pressure, density, and temperature, as well as the ideal gas constant (R): \(M = \frac{\rho RT}{P} = \frac{89.9 \mathrm{~kg/m^3} \times 8.314 \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 293.15 \mathrm{K}}{101325 \mathrm{~Pa}}\) \(M \approx 28.96 \mathrm{~g/mol}\)

Identify the gas

Finally, we will identify the gas based on its molar mass. A gas with a molar mass of approximately \(28.96 \mathrm{~g/mol}\) is nitrogen (N₂), which has a molar mass of \(28.02 \mathrm{~g/mol}\). Our result is very close to the actual value, so it is reasonable to identify the gas as nitrogen (N₂).

Key concepts

Gas Density Conversion

Understanding the conversion of gas density is critical, and often first step in applying the Ideal Gas Law. Density is a measure of how much mass is contained within a certain volume. Typically, you might encounter gas density given in units like grams per liter (g/L). This is adequate for simple calculations but to utilize the Ideal Gas Law effectively, we often need density in kilograms per cubic meter (kg/m³). To convert, remember:

• 1 gram (g) = 1000 milligrams (mg)
• 1 liter (L) = 1000 milliliters (mL) = 0.001 cubic meters (m³)

For example, if a gas has a density of 0.0899 g/L, you convert to kg/m³ by calculating:\[ 0.0899 \times 1000 \text{ mg/g} \times 1 \text{ L/mL} = 89.9 \text{ kg/m}³ \]This result provides the density in units compatible with the Ideal Gas Law.

Molar Mass Calculation

The Ideal Gas Law is expressed as \( PV = nRT \), where:

• \( P \) is the pressure
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant
• \( T \) is the temperature in Kelvin

Moles \( n \) can be expressed as mass \( m \) over molar mass \( M \), i.e., \( n = \frac{m}{M} \). So, rearranging the Ideal Gas Law, we integrate this relation:\[ P = \frac{\rho RT}{M} \]This indicates that the pressure \( P \) of a gas is directly related to its density \( \rho \), temperature \( T \), and inversely to its molar mass \( M \).To find the molar mass, solve:\[ M = \frac{\rho RT}{P} \]For example, substitute the values: density \( \rho = 89.9 \text{ kg/m}³ \), temperature \( T = 293.15 \text{ K} \), pressure \( P = 101325 \text{ Pa} \), and the ideal gas constant \( R = 8.314 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \). Performing the calculation gives \( M \approx 28.96 \text{ g/mol} \). This molar mass helps in identifying the gas.

Nitrogen Identification

Identifying a gas involves comparing its calculated properties to known data. In this exercise, we estimate a molar mass close to 28.96 g/mol for the gas. Referencing periodic tables or gas charts can assist in determining the gas identity. Nitrogen, a common atmospheric component, has a molar mass of approximately 28.02 g/mol. This small difference between the calculated and known values is typically due to rounding or minor experimental errors and is quite acceptable in practical chemistry applications. Factors influencing gas identification include:

• The accuracy of measurement and unit conversions
• Standard environmental conditions held during calculation

Given the close match in molar mass, it is reasonable and scientifically sound to identify the unknown gas as nitrogen (N₂) in this scenario.