Problem 68 - Chemistry The Central Science In Si Units

Question

A sample of \(5.00 \mathrm{~mL}\) of diethylether \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\) density \(=0.7134 \mathrm{~g} / \mathrm{mL}\) ) is introduced into a 6.00 -L vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), whose partial pressures are \(P_{\mathrm{N}_{2}}=21.08 \mathrm{kPa}\) and \(P_{\mathrm{O}_{2}}=76.1 \mathrm{kPa}\). The temperature is held at \(35.0^{\circ} \mathrm{C}\), and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.

Step-by-Step Solution

Answer

(a) The partial pressure of diethylether is 2.06 kPa. (b) The total pressure in the container is 99.24 kPa.

Step 1: Find the Mass of Diethylether

Using the volume and density of diethylether, calculate its mass. The formula to use is \( \text{mass} = \text{volume} \times \text{density} \). For diethylether, this is \( 5.00 \, \text{mL} \times 0.7134 \, \text{g/mL} = 3.567 \, \text{g} \).

Step 2: Determine Moles of Diethylether

Use the molar mass of diethylether, \( \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 \) (molar mass = 74.12 g/mol) to convert the mass to moles: \( \frac{3.567 \, \text{g}}{74.12 \, \text{g/mol}} = 0.0481 \, \text{mol} \).

Step 3: Use Ideal Gas Law for Diethylether Partial Pressure

Apply the ideal gas law \( PV = nRT \) to find the partial pressure of diethylether. We rearrange the formula to solve for \( P \): \( P = \frac{nRT}{V} \). Substituting in the values \( n = 0.0481 \, \text{mol} \), \( R = 8.314 \, \text{J/(mol} \cdot \text{K)} \), \( T = 35.0 + 273.15 = 308.15 \, \text{K} \), and \( V = 6.00 \, \text{L = 0.00600 \, m}^3 \):\( P = \frac{0.0481 \, \text{mol} \times 8.314 \, \text{J/(mol} \cdot \text{K)} \times 308.15 \, \text{K}}{0.00600 \, \text{m}^3} \approx 2058.42 \, \text{Pa} \), which converts to \( 2.06 \, \text{kPa} \) (since \( 1 \, \text{Pa} = 0.001 \, \text{kPa} \)).

Step 4: Calculate Total Pressure in the Container

Sum the partial pressures of all gases to find the total pressure: \( P_{\text{total}} = P_{\text{N}_2} + P_{\text{O}_2} + P_{\text{C}_4\text{H}_{10}\text{O}} \). Substitute in the values: \( P_{\text{total}} = 21.08 \, \text{kPa} + 76.1 \, \text{kPa} + 2.06 \, \text{kPa} = 99.24 \, \text{kPa} \).

Key Concepts

Partial Pressure Calculation

In the realm of gases, understanding how to calculate partial pressures is essential, especially when dealing with gas mixtures. The partial pressure of a gas is the pressure it would exert if it occupied the entire volume on its own. According to Dalton's Law, the total pressure in a mixture of gases is the sum of the partial pressures of the individual gases present. In this scenario, we calculate the partial pressure of diethylether using the ideal gas law.

The ideal gas law is given by the formula: \[ PV = nRT \] Where:

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

To find the partial pressure of diethylether, use the rearranged formula:\[ P = \frac{nRT}{V} \]When the values are substituted, it provides the partial pressure of diethylether in the given conditions.

Molar Mass

Molar mass is a fundamental concept in chemistry, serving as the bridge between grams and moles. It is the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). To calculate the molar mass of a compound like diethylether \(\text{C}_2\text{H}_5\text{OC}_2\text{H}_5\), you need to sum up the atomic masses of each atom present in the molecule.

Here's how you can do it:

• Carbon (C) has an atomic mass of approximately 12.01 g/mol,
• Hydrogen (H) has an atomic mass of approximately 1.01 g/mol,
• Oxygen (O) has an atomic mass of approximately 16.00 g/mol.

For diethylether, this adds up to:\[ 2(12.01) + 5(1.01) + 16.00 + 2(12.01) + 5(1.01) = 74.12 \, \text{g/mol} \]This calculated molar mass allows you to convert between mass and moles easily, which is critical for solving gas law problems.

Density and Volume Relation

Density is another cornerstone in understanding gases and liquids. It's defined as the mass per unit volume of a substance and is usually expressed in g/mL or g/L, depending on the state's density. For diethylether, the density is given as 0.7134 g/mL, which provides a way to convert between volume and mass.

The formula to relate mass, volume, and density is straightforward:\[ \text{mass} = \text{density} \times \text{volume} \]By using this formula, you can determine the mass of diethylether when you know the volume. In our exercise, a 5.00 mL sample of diethylether was used, allowing us to calculate its mass as:\[ 5.00 \, \text{mL} \times 0.7134 \, \text{g/mL} = 3.567 \, \text{g} \]Understanding this relationship between density and volume is crucial for calculations involving conversion to moles, which is a common intermediary step in evaluating gas properties.

Gas Mixtures

Gas mixtures are common in everyday applications and can often be found in enclosed spaces, like the one described in this problem. When dealing with gas mixtures, Dalton’s Law of Partial Pressures becomes instrumental.

Dalton's Law asserts that in a mixture of non-reacting gases, the total pressure is equal to the sum of the partial pressures of the individual gases. Here's how it applies to our exercise situation:

• Partial pressure of \(\text{N}_2\) = 21.08 kPa
• Partial pressure of \(\text{O}_2\) = 76.1 kPa
• Partial pressure of diethylether calculated earlier = 2.06 kPa

Thus, the total pressure is simply the addition of these:\[ P_{\text{total}} = 21.08 + 76.1 + 2.06 = 99.24 \, \text{kPa} \]This calculation showcases the simplicity and power of Dalton’s Law in predicting how gases behave when mixed.

Temperature Conversion

Temperature is a vital parameter when working with gases, because it directly affects their behavior and calculations involving them. In most scientific equations, temperatures need to be converted to Kelvin, because Kelvin is the absolute temperature scale where 0 K is the lowest possible temperature.

To convert from Celsius to Kelvin, one must simply add 273.15 to the Celsius temperature. This is a straightforward and necessary step to ensure accurate application of gas laws, like the ideal gas law.
In our problem, the temperature given is 35.0°C. To convert this to Kelvin:\[ T = 35.0 + 273.15 = 308.15 \, \text{K} \]This conversion allows us to precisely determine the behavior and properties of gases under the given conditions, and it is a step that should never be overlooked when working with gas equations.