Problem 52 - Chemistry The Central Science In Si Units

Question

(a) Calculate the density of sulfur hexafluoride gas at 94.26 \(\mathrm{kPa}\) and \(21^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a vapor that has a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and \(99.06 \mathrm{kPa}\).

Step-by-Step Solution

Answer

(a) 5.634 g/L; (b) 171.001 g/mol.

Step 1: Convert Temperature and Pressure

First, we need to convert the temperature from Celsius to Kelvin. For part (a), the given temperature is \(21^{\circ}\mathrm{C}\), so the conversion is:\[T_a = 21 + 273.15 = 294.15 \mathrm{K}\]In part (b), the temperature is \(12^{\circ}\mathrm{C}\):\[T_b = 12 + 273.15 = 285.15 \mathrm{K}\]Both pressures are given in kPa, so no conversion is needed.

Step 2: Use the Ideal Gas Law for Part (a)

To calculate the density of sulfur hexafluoride (SF extsubscript{6}) gas, we use the ideal gas law: \( PV = nRT \) and modify it to find density. Density \( \rho \) is defined as mass per unit volume, so:\[\rho = \frac{PM}{RT}\]Where \( M \) is the molar mass of SF extsubscript{6} (146.06 g/mol), \( P = 94.26 \) kPa, \( R = 8.314 \text{ L kPa K}^{-1} \text{mol}^{-1} \), and \( T = 294.15 \) K.

Step 3: Calculate Density for Part (a)

Substitute the values into the density formula:\[\rho_a = \frac{94.26 \times 146.06}{8.314 \times 294.15}\]This simplifies to:\[\rho_a = \frac{13768.596}{2444.661} \approx 5.634 \text{ g/L}\]

Step 4: Rearrange Ideal Gas Law for Molar Mass in Part (b)

Given the density and conditions of the vapor, the formula \( \rho = \frac{PM}{RT} \) can be rearranged to solve for molar mass \( M \):\[M = \frac{\rho RT}{P}\]Here, \( \rho = 7.135 \text{ g/L} \), \( R = 8.314 \text{ L kPa K}^{-1} \text{mol}^{-1} \), \( T = 285.15 \) K, and \( P = 99.06 \) kPa.

Step 5: Calculate Molar Mass for Part (b)

Plug the values into the molar mass equation:\[M = \frac{7.135 \times 8.314 \times 285.15}{99.06}\]Simplifying gives:\[M = \frac{16936.052}{99.06} \approx 171.001 \text{ g/mol}\]

Key Concepts

Density Calculation

Density is a measure of how much mass is contained in a given volume. When dealing with gases, you can calculate the density using the Ideal Gas Law, which is originally given as \(PV = nRT\). But how does this connect to density? To find density, we use the formula:

• \(\rho = \frac{PM}{RT}\)

In this equation:

• \(\rho\) is the density in \(\text{g/L}\),
• \(P\) is the pressure in kPa,
• \(M\) is the molar mass of the gas,
• \(R\) is the universal gas constant (8.314 \(\text{L kPa K}^{-1} \text{mol}^{-1}\)), and
• \(T\) is the temperature in Kelvin.

To demonstrate, consider sulfur hexafluoride (\(SF_6\)) at \(94.26 \text{kPa}\) and \(21^\circ\text{C}\), where \(M\) is \(146.06 \text{g/mol}\) and the temperature conversion gives us \(294.15 \text{K}\). Substituting these into the formula, we calculate the density as:\[\rho = \frac{94.26 \times 146.06}{8.314 \times 294.15} \approx 5.634 \text{ g/L}\]This process shows the power of the Ideal Gas Law in determining the density of gases under various conditions.

Molar Mass Determination

Estimating the molar mass of a gas is crucial for understanding its chemical and physical properties. The given density and conditions, along with the Ideal Gas Law, help in determining this value for a vapor. To calculate the molar mass \(M\) of a vapor, rearrange the density formula from:

• \(\rho = \frac{PM}{RT}\)

to:

• \(M = \frac{\rho RT}{P}\)

Each symbol has its significance:

• \(\rho\) stands for density (\(7.135 \text{ g/L}\)),
• \(R\) is the gas constant,
• \(T\) is the converted temperature in Kelvin, and
• \(P\) is the pressure in kPa.

For example, consider a vapor at \(99.06 \text{kPa}\) and \(12^\circ\text{C}\), resulting in temperature \(285.15\text{ K}\). Substituting into the equation, calculate:\[M = \frac{7.135 \times 8.314 \times 285.15}{99.06} \approx 171.001 \text{ g/mol}\]Determining molar mass through this method helps in identifying unknown gases and understanding their applications.

Temperature Conversion

Temperature conversion is a small yet crucial step when working with gases and the Ideal Gas Law. The Ideal Gas Law requires temperatures to be in Kelvin to ensure the calculations are consistent, as Kelvin offers an absolute measure starting from absolute zero.The conversion from Celsius to Kelvin is simple:

• Add 273.15 to the Celsius temperature, \(T_{K} = T_{C} + 273.15\).

Let’s see this in action with the temperatures provided:

• For \(21^\circ\text{C}\), the conversion is \(21 + 273.15 = 294.15\text{ K}\).
• For \(12^\circ\text{C}\), it becomes \(12 + 273.15 = 285.15\text{ K}\).

Not converting the temperature might lead to incorrect conclusions or errors in gas calculations. Hence, always remember to convert to Kelvin for precision in scientific computations and applications.