Problem 88 - University Physics with Modern Physics

Question

Meteorology. The vapor pressure is the pressure of the vapor phase of a substance when it is in equilibrium with the solid or liquid phase of the substance. The relative humidity is the partial pressure of water vapor in the air divided by the vapor pressure of water at that same temperature, expressed as a percentage. The air is saturated when the humidity is 100\(\%\) (a) The vapor pressure of water at \(20.0^{\circ} \mathrm{C}\) is \(2.34 \times 10^{3}\) Pa. If the air temperature is \(20.0^{\circ} \mathrm{C}\) and the relative humidity is \(60 \%,\) what is the partial pressure of water vapor in the atmosphere (that is, the pressure due to water vapor alone)? (b) Under the conditions of part (a), what is the mass of water in 1.00 \(\mathrm{m}^{3}\) of air? (The molar mass of water is 18.0 \(\mathrm{g} / \mathrm{mol}\) . Assume that water vapor can be treated as an ideal gas.)

Step-by-Step Solution

Answer

(a) 1404 Pa; (b) 10.42 g/m³.

Step 1: Determine the Partial Pressure of Water Vapor

The relative humidity is given by \( R = \frac{P}{P_0} \times 100\% \), where \( P \) is the partial pressure of the water vapor, and \( P_0 \) is the vapor pressure of water at the temperature. Rearrange the formula to find \( P \): \( P = \frac{R \times P_0}{100\%} \). Substituting the given values: \( R = 60\% \) and \( P_0 = 2.34 \times 10^3 \; \text{Pa} \), we get \( P = \frac{60\% \times 2.34 \times 10^3}{100\%} = 1.404 \times 10^3 \; \text{Pa} \).

Step 2: Calculate the Mass of Water Using the Ideal Gas Law

First, use the ideal gas law, \( PV = nRT \), to find the number of moles of water vapor, \( n \). Rearrange to find \( n = \frac{PV}{RT} \). Here, \( P = 1.404 \times 10^3 \; \text{Pa} \), \( V = 1.00 \; \text{m}^3 \), \( R = 8.31 \; \text{J/(mol K)} \), and \( T = 293.15 \; \text{K} \). Thus, \( n = \frac{1.404 \times 10^3 \times 1.00}{8.31 \times 293.15} = 0.579 \; \text{moles} \).

Step 3: Convert Moles of Water to Mass

The mass of water vapor is calculated using \( n \times M \), where \( M \) is the molar mass of water (18.0 g/mol). Therefore, the mass \( m = 0.579 \times 18.0 = 10.42 \; \text{g} \).

Key Concepts

Relative Humidity

Relative humidity is an important concept in understanding how much moisture is present in the air. It is defined as the ratio of the partial pressure of water vapor currently in the air to the equilibrium vapor pressure of water at the same temperature, expressed as a percentage. It gives us a sense of how "full" the air is of water vapor.

This measurement is crucial because it tells us about air saturation. For instance, at 100% relative humidity, the air is saturated with water and can hold no more, leading to precipitation like rain. To calculate relative humidity, you would typically use the formula:

• \( R = \frac{P}{P_0} \times 100\% \)

where \( P \) is the partial pressure of water vapor and \( P_0 \) is the vapor pressure. If you know the vapor pressure of water and the current relative humidity, you can find the partial pressure of water vapor.

Knowing the relative humidity helps in predicting weather patterns, understanding comfort levels, and even in designing HVAC systems for better climate control.

Ideal Gas Law

The ideal gas law is a fundamental principle used to describe the behavior of gases. It relates four key variables: pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the amount of gas in moles (\(n\)) through the equation:

• \( PV = nRT \)

The constant \(R\) is the ideal gas constant, and its value depends on the units used for pressure, volume, and temperature. This equation assumes that gases behave ideally, meaning they have perfectly elastic collisions and no intermolecular forces.

In practical applications like this exercise, the ideal gas law can be used to calculate the number of moles of water vapor in a given volume of air when the partial pressure, volume, and temperature are known. By rearranging the formula to solve for \(n\), you can find the amount of substance:

• \( n = \frac{PV}{RT} \)

This versatility makes the ideal gas law a powerful tool in chemistry and physics for analyzing gaseous systems under various conditions.

Molar Mass of Water

The molar mass of water is a critical value used in many calculations involving water. It represents the mass of one mole of water molecules, which is approximately 18.0 g/mol. A mole is a unit in chemistry that represents a large number of atoms or molecules—about \(6.022 \times 10^{23}\) entities, also known as Avogadro's number.

Water's molar mass is derived from combining the atomic masses of its constituent elements, hydrogen and oxygen. Each hydrogen atom has a mass of about 1 g/mol, and there are two hydrogen atoms in water. Oxygen has a mass of approximately 16 g/mol. Adding these together gives:

• 2 g/mol (from hydrogen) + 16 g/mol (from oxygen) = 18 g/mol

In practical terms, knowing the molar mass allows you to convert between grams of water and moles, which is particularly useful when using the ideal gas law to find the mass from the amount in moles. This understanding is essential for tasks like determining the mass of water vapor in a volume of air.