Problem 143 - General Chemistry

Question

The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through the liquid at a given temperature and pressure. In an experiment, a 5.40-L sample of nitrogen gas, \(\mathrm{N}_{2}\), at \(20.0^{\circ} \mathrm{C}\) and \(745 \mathrm{mmHg}\) is bubbled through liquid isopropyl alcohol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\), at \(20.0^{\circ} \mathrm{C}\). Nitrogen containing the vapor of \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) at its vapor pressure leaves the vessel at \(20.0^{\circ} \mathrm{C}\) and \(745 \mathrm{mmHg} .\) It is found that \(0.6149 \mathrm{~g} \mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) has evaporated. How many moles of \(\mathrm{N}_{2}\) are in the gas leaving the liquid? How many moles of alcohol are in this gaseous mixture? What is the mole fraction of alcohol vapor in the gaseous mixture? What is the partial pressure of the alcohol in the gaseous mixture? What is the vapor pressure of \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) at \(20.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

0.217 moles of N2 and 0.0102 moles of alcohol are present. The mole fraction of alcohol is 0.045, with a partial pressure of 33.5 mmHg. The vapor pressure of alcohol is 33.5 mmHg at 20°C.

Step 1: Calculate Moles of Nitrogen

First, let's determine how many moles of nitrogen gas (_2) are present. We use the ideal gas law equation:\[PV = nRT\]Rearrange this to find moles:\[n = \frac{PV}{RT}\]Given values are:- Volume, \(V = 5.40 \text{ L}\)- Pressure, \(P = 745 \text{ mmHg} = \frac{745}{760} \text{ atm}\)- Temperature, \(T = 20.0^{\circ} \text{C} = 293.15 \text{ K}\)- Gas constant, \(R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\)Substitute these into the equation:\[n = \frac{\left(\frac{745}{760}\right) \times 5.40}{0.0821 \times 293.15} = 0.217 \text{ moles of } \text{N}_2\]

Step 2: Calculate Moles of Alcohol Evaporated

We know the weight of evaporated isopropyl alcohol (_{3} _{8} O) is 0.6149 g. To find moles, use the molecular weight of isopropyl alcohol (~60.1 g/mol).\[n = \frac{\text{mass}}{\text{molecular weight}} = \frac{0.6149}{60.1} \approx 0.0102 \text{ moles of } \text{C}_3 _8 O\]

Step 3: Calculate Mole Fraction of Alcohol

To find the mole fraction of alcohol in the exiting gaseous mixture, we use:\[\text{Mole fraction, } x = \frac{\text{moles of alcohol}}{\text{moles of alcohol} + \text{moles of } \text{N}_2}\]Substitute the values:\[x = \frac{0.0102}{0.0102 + 0.217} \approx 0.045\]

Step 4: Calculate Partial Pressure of Alcohol

The partial pressure can be determined by multiplying the mole fraction by the total pressure:\[P_{C_3 _8 O} = x \times P_{\text{total}} = 0.045 \times 745 \text{ mmHg} \approx 33.5 \text{ mmHg}\]

Step 5: Determine Vapor Pressure of Alcohol at 20°C

The partial pressure of the alcohol vapor at the given conditions is equal to its vapor pressure at 20°C because it's in equilibrium with its liquid. Therefore:\[\\text{Vapor pressure of } \text{C}_3 _8 O \text{ at 20°C is } \approx 33.5 \text{ mmHg}\]

Key Concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. In our exercise, the vapor pressure of isopropyl alcohol at 20°C is determined through an experiment. This is achieved by allowing nitrogen gas to bubble through the liquid alcohol and capturing the vapor in the exiting gas. The vapor pressure is important because it represents the tendency of molecules to escape the liquid or solid phase, which in turn affects how the alcohol evaporates in different conditions. At 20°C, we calculated the vapor pressure of isopropyl alcohol to be approximately 33.5 mmHg. This means that at this temperature, the alcohol's vapor can exert a pressure equivalent to 33.5 mmHg, demonstrating its volatility.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry used to relate the pressure, volume, temperature, and number of moles of a gas. This equation is expressed as:\[PV = nRT\]Where:

• \(P\) is the pressure of the gas
• \(V\) is the volume of the gas
• \(n\) is the number of moles of gas
• \(R\) is the ideal gas constant (0.0821 L atm K⁻¹ mol⁻¹)
• \(T\) is the temperature in Kelvin

In the given exercise, the Ideal Gas Law was used to calculate the amount of nitrogen gas by rearranging the formula to solve for \(n\) (moles). This equation played a central role in finding out that the 5.40 L of nitrogen at 20°C and 745 mmHg contains approximately 0.217 moles of nitrogen. Understanding how the ideal gas law functions helps us predict how a gas will behave under various conditions.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is calculated by dividing the moles of the component by the total moles of all components in the mixture. The formula is:\[x = \frac{n_i}{n_{\text{total}}}\]Where \(x\) is the mole fraction, \(n_i\) is the number of moles of the component of interest, and \(n_{\text{total}}\) is the total number of moles in the mixture. In the context of the exercise, the mole fraction of alcohol vapor in the exiting gaseous mixture was calculated by dividing the moles of alcohol (about 0.0102 moles) by the total moles in the mixture (the sum of nitrogen and alcohol moles). As a result, the mole fraction of alcohol was found to be approximately 0.045. This calculation is crucial as it helps in understanding the composition and behavior of gases in the mixture.

Partial Pressure Calculation

Partial pressure is the pressure exerted by a single gas component in a mixture of gases. It is directly proportional to its mole fraction in the mixture and can be calculated using Dalton's Law of Partial Pressures:\[P_i = x_i \times P_{\text{total}}\]Where:

• \(P_i\) is the partial pressure of the gas
• \(x_i\) is the mole fraction of the gas
• \(P_{\text{total}}\) is the total pressure of the gas mixture

In our example, using the calculated mole fraction of alcohol (0.045) and the total pressure (745 mmHg), the partial pressure of the alcohol was determined to be approximately 33.5 mmHg. This matches the vapor pressure, confirming the alcohol's presence in equilibrium with its liquid. Understanding partial pressures is essential for predicting how gases interact in a mixture, which is particularly useful in real-world applications like chemical engineering and environmental science.