Question

Consider a \(31.0 \mathrm{L}\) sample of moist air at \(60 .^{\circ} \mathrm{C}\) and one atm in which the partial pressure of water vapor is 0.131 atm. Assume that dry air has the composition 78.0 mole percent \(\mathrm{N}_{2}, 21.0\) mole percent \(\mathrm{O}_{2},\) and 1.00 mole percent Ar. a. What are the mole percentages of each of the gases in the sample? b. The percent relative humidity is defined as \(\% \mathrm{RH}=\) \(P_{H_{2}} o / P_{H_{2} O}^{*}\) where \(P_{H_{2}, O}\) is the partial pressure of water in the sample and \(P_{H, O}^{*}=0.197\) atm is the equilibrium vapor pressure of water at \(60 .^{\circ} \mathrm{C}\). The gas is compressed at \(60 .^{\circ} \mathrm{C}\) until the relative humidity is \(100 . \% .\) What volume does the mixture contain now? c. What fraction of the water will be condensed if the total pressure of the mixture is isothermally increased to 81.0 atm?

Short answer

The mole percentages of each gas in the sample are 67.54% of N2, 18.16% of O2, 0.88% of Ar, and 13.42% of H2O. The final volume of the moist air when relative humidity reaches 100% can be determined using the new total pressure and the ideal gas law. When the total pressure is isothermally increased to 81.0 atm, the fraction of water condensed can be calculated using the equilibrium vapor pressure, the new partial pressure of water vapor, and the total moles of water vapor before compression.

Step-by-step solution

Calculate total moles of gas mixture

The volume of the sample is 31.0L, and we know it's at one atm (1.0 atm) pressure and 60 degrees Celsius. To find the total moles, we use the ideal gas law: \(PV = nRT\) Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant (0.0821 L*atm/mol*K), and T is the temperature in Kelvin. First, convert the given temperature to Kelvin: T = 60 + 273.15 = 333.15 K Next, rearrange the ideal gas law to solve for \(n\): \(n = \frac{PV}{RT}\) Plugging in the values: \(n = \frac{1.0 \space atm * 31.0 \space L}{0.0821 \space L*atm/mol*K * 333.15 \space K}\) Calculating total moles: \(n = \space 1.14\)

Calculate mole percentages of gases

The dry air composition is given as 78.0 mole percent N2, 21.0 mole percent O2, and 1.00 mole percent Ar. To calculate the mole percentages in the moist air we need to consider the total moles including the water vapor. The partial pressure of water vapor (\(P_{H2O}\)) is given as 0.131 atm. Using the ideal gas law formula again, we can find the moles of water vapor: \(n_{H2O}\) = \(\frac{P_{H2O}V}{RT}\) = \(\frac{0.131 \space atm * 31.0 \space L}{0.0821 \space L*atm/mol*K * 333.15 \space K}\) = 0.153 moles Now, we can find the moles of dry air by subtracting the moles of water vapor from the total moles: \(n_{dry}\) = \(n_{total} - n_{H2O}\) = 1.14 - 0.153 = 0.987 moles Now we can calculate the moles of each gas in the dry air: N2, O2, and Ar by multiplying their mole percentages with the total moles of dry air: - Moles of N2: \(0.78 * 0.987\) = 0.770 moles - Moles of O2: \(0.21 * 0.987\) = 0.207 moles - Moles of Ar: \(0.01 * 0.987\) = 0.010 moles Now we can find their mole percentages in the moist air considering the moles of water vapor: - Mole percentage N2: \(\frac{0.770}{1.14}\) × 100 = 67.54% - Mole percentage O2: \(\frac{0.207}{1.14}\) × 100 = 18.16% - Mole percentage Ar: \(\frac{0.010}{1.14}\) × 100 = 0.88% - Mole percentage H2O: \(\frac{0.153}{1.14}\) × 100 = 13.42%

Calculate the new volume at 100% relative humidity

Using the formula for percent relative humidity (\(% \mathrm{RH}\)): \(% \mathrm{RH}\) = \(\frac{P_{H2O}}{P_{H2O}^*}\) We know that the new relative humidity is 100%. So, 1.00 = \(\frac{P_{H2O_{new}}}{0.197 \space atm}\) \(P_{H2O_{new}}\) = 0.197 atm The new total pressure is the summation of partial pressures of each gas. \(P_{total_{new}}\) = \(P_{N2_{new}} + P_{O2_{new}} + P_{Ar_{new}} + P_{H2O_{new}}\) Since the mole percentages remain constant, we can find the new partial pressures using Boyle's law (\(P_{1}V_{1} = P_{2}V_{2}\)): \(P_{N2_{new}}\) = \(\frac{P_{N2_{old}}}{P_{total_{old}}} * P_{total_{new}}\) Calculate \(P_{N2_{new}}\), \(P_{O2_{new}}\), and \(P_{Ar_{new}}\) similarly. Now we can calculate the new volume using the ideal gas law when the total pressure is equal to the new total pressure.

Calculate the fraction of water condensed

We know the equilibrium vapor pressure of water at 60 degrees Celsius, \(P_{H, O}^{*}=0.197\) atm. This is the maximum pressure water vapor can have at 60 degrees Celsius before condensation. When the total pressure is isothermally increased to 81.0 atm, the new partial pressure of water vapor (\(P_{H_{2}, O_{new}}\)) can be found using the mole percentages. From the new partial pressure of water vapor and the equilibrium vapor pressure, we can calculate how many moles of water vapor will condense. Finally, the fraction of condensed water can be found by dividing the moles of condensed water vapor by the total moles of water vapor before compression.

Key concepts

Understanding the Ideal Gas Law

The ideal gas law is a fundamental concept in chemistry that relates the pressure, volume, temperature, and the number of moles of a gas. It is commonly written as the formula:
\[ PV = nRT \]
where P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles of the gas, R is the ideal gas constant (typically 0.0821 L*atm/mol*K), and T is the temperature in Kelvin (K). This equation can be used to solve a variety of problems involving gases.
To address the textbook exercise, one applies the ideal gas law to calculate the total number of moles of a gas mixture. By incorporating the given temperature, volume, and atmospheric pressure, one can directly calculate the number of moles, which sets the stage for further calculations, such as determining the mole percentages of individual gases within the mixture.

• To find the moles: rearrange the ideal gas law to solve for n, and plug in the known values of P, V, R, and T.
• Account for the presence of water vapor by adjusting the moles of dry air accordingly.

Relative Humidity and its Significance

Relative humidity is a measure of the water vapor content in the air relative to the maximum amount the air can hold at that temperature. It's expressed as a percentage and is calculated using the formula:
\[ \% RH = \frac{P_{H2O}}{P_{H2O}^*} \times 100 \]
where \(P_{H2O}\) is the partial pressure of water vapor currently in the air, and \(P_{H2O}^*\) is the equilibrium vapor pressure of water at the same temperature. High relative humidity indicates a large amount of water vapor, which often correlates with a higher likelihood of precipitation and the sensation of 'mugginess'.
In the textbook problem, relative humidity is used to determine at what volume water vapor starts to condense when air is compressed at a constant temperature. This requires understanding the relationship between relative humidity, vapor pressure, and changes in the system's volume.

Calculating Volume at 100% Relative Humidity

At 100% relative humidity, the air is 'saturated' with water vapor, meaning any increase in moisture or decrease in temperature would lead to condensation. By equating the desired relative humidity to the equilibrium vapor pressure, one can find the new conditions under which the air becomes saturated.

Mole Percentage in Gas Mixtures

Mole percentage is an expression of the concentration of a particular component within a mixture, defined as the number of moles of that component divided by the total number of moles of all components, multiplied by 100%. It's a way to understand the composition of a mixture in terms of its constituents.
In the context of the problem, we're asked to find the mole percentages of gases in moist air.

• First, calculate the total moles of the gas mixture using the ideal gas law.
• Subtract the moles of water to find the moles of dry air.
• Apply the given mole percentages of the dry components (N2, O2, and Ar) to determine their respective mole amounts.
• Account for the water vapor when determining the new mole percentages of all gases present.

The mole percentages provide insight into the behavior and properties of the gas mixture, especially under changing conditions such as compression or temperature variation.

The Role of Vapor Pressure in Gas Mixtures

Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. Every liquid has a characteristic equilibrium vapor pressure that depends on temperature; when the vapor pressure equals the external pressure, the liquid will boil.
For the case at hand, we use vapor pressure to determine when water vapor in the air will condense. The equilibrium vapor pressure of water at a certain temperature is the maximum amount of pressure that the water vapor can exert without condensing.

Isothermal Compression and Condensation

When we increase the total pressure of the mixture isothermally to 81.0 atm, we compare this new pressure to the vapor pressure to understand how much water will condense. Any excess pressure beyond vapor pressure will push the water vapor to transition into a liquid, thereby changing the composition and characteristics of the gas mixture and allowing us to calculate the fraction of water condensed.