Problem 132 - Chemistry

Question

A \(1.20-\mathrm{g}\) sample of water is injected into an evacuated \(5.00-\mathrm{L}\) flask at \(65^{\circ} \mathrm{C} .\) What percentage of the water will be vapor when the system reaches equilibrium? Assume ideal behavior of water vapor and that the volume of liquid water is negligible. The vapor pressure of water at \(65^{\circ} \mathrm{C}\) is \(187.5 \mathrm{mmHg}\)

Step-by-Step Solution

Answer

67.6% of the water will be vapor when the system reaches equilibrium at \(65^{\circ} \mathrm{C}\)

Step 1: Convert mass of water to mol

The molar mass of water is 18.015g/mol, thus the number of moles in a 1.20g sample of water can be calculated as follows: \(1.20g \times (1 mol/18.015g) = 0.0666 mol\)

Step 2: Apply the Ideal Gas Law

The Ideal Gas Law states: \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. We will use the volume of the flask and the number of moles calculated earlier, along with the gas constant (0.0821 L-atm/mol-K) and the temperature, which must be converted to Kelvin (338K) to use in this formula. However, before that, the pressure given in mmHg must be converted to atm. The conversion factor is 760 mmHg = 1 atm. Thus, the pressure of the water vapor is \(187.5 mmHg \times (1 atm / 760 mmHg) = 0.2467 atm\). The total pressure inside the flask when the system is at equilibrium will be equal to this pressure of the water vapor, because the flask was initially evacuated.

Step 3: Calculate moles of water that will be vapor

We can rearrange the Ideal Gas Law to solve for the number of moles of water that will be at equilibrium in the vapor state by isolating n, giving us \(n = PV/RT\). Substituting the given/converted values we have, \(n = (0.2467 atm) * (5.00 L) / [(0.0821 L-atm/mol-K) * (338 K)] = 0.0450 mol\). This is the number of moles of water that will be vapor when the system reaches equilibrium.

Step 4: Calculate the percentage of the water that will be vapor

The percentage of the water that will be vapor is simply the ratio of moles of water that are vapor to the total moles of water, which you then multiply by 100 to get a percentage. So, the percentage of water that will be vapor is \(0.0450 mol / 0.0666 mol * 100% = 67.6%\)

Key Concepts

Ideal Gas Law

The Ideal Gas Law is fundamental to understanding how gases behave under various conditions. It is expressed as \( PV = nRT \), where:

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

In the exercise, we need to find out how much water turns into vapor when the container reaches equilibrium. We first need to convert all relevant measurements to compatible units for the formula. The universal gas constant \( R \) often has a value of 0.0821 L-atm/mol-K when dealing with atmosphere (atm) as a pressure unit. Using the Ideal Gas Law allows us to calculate the number of moles of water vapor formed at equilibrium by rearranging the equation to solve for \( n \) given the vapor pressure, volume, and temperature of the system.

Molar Mass

Molar mass helps us convert between the mass of a substance and the number of moles. The molar mass of a molecule is the combined mass of all the atoms in a molecule, measured in grams per mole (g/mol). For water, which has the chemical formula \( H_2O \), the molar mass is typically around 18.015 g/mol.

In the exercise, to determine the number of moles in the given mass of water, we use the relation:\[\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.20 \, g}{18.015 \, g/mol} = 0.0666 \, mol\]This conversion is crucial because in many cases, calculations in chemistry require the amount of substance to be in moles. Moles allow for easier mapping to chemical equations and theoretical models like the Ideal Gas Law.

Equilibrium

Equilibrium in a closed container system, like the one described in the exercise, is a point where the opposing processes of liquid evaporation and vapor condensation occur at the same rate. Thus, the state of the system does not change over time.

For water in a flask, equilibrium is reached when the vapor pressure of the water complements the conditions set forth by the Ideal Gas Law and temperature.

When we reach this point of equilibrium, the pressure is constant and equal to the vapor pressure. Any amount of water vapor formed more than equilibrium will condense back into liquid, stabilizing the amount in each phase.Moreover, in terms of the Ideal Gas Law, this is when the real pressure modeling the vapor conforms to \( P = \text{vapor pressure} \). The system's state remains constant unless the external variables such as temperature or volume are altered.

Conversion Factors

Conversion factors are truly the unsung heroes in calculations involving different units. They allow us to switch units while maintaining the quantitative relationships between values without changing the underlying meaning.

• For pressures, we use 1 atm = 760 mmHg to convert vapor pressures between atmospheres and millimeters of mercury.
• To switch temperature units from Celsius to Kelvin, we use \( T(K) = T(°C) + 273.15 \).

For the exercise in question, these conversions included turning the given vapor pressure from mmHg to atm:\[P = 187.5 \, \text{mmHg} \times \left( \frac{1 \, \text{atm}}{760 \, \text{mmHg}} \right) = 0.2467 \, \text{atm}\]Additionally, converting temperature from Celsius to Kelvin ensures proper use with the Ideal Gas Law, which requires temperature in Kelvin to maintain proportional relationships between pressure and temperature (absolute scale). These conversions set the stage for accurate calculations and conclusions.