Suppose that you have collected \(1.0 \mathrm{~L}\) of humid air by passing it slowly through water at \(20^{\circ} \mathrm{C}\) and into a container. Estimate the mass of water vapor in the collected air, assuming that the air is saturated with water. \(\mathrm{Ar} 20^{\circ} \mathrm{C}\), the vapor pressure of water is \(17.5\) Torr.

Short Answer

Expert verified
The mass of water vapor in 1.0 L of saturated air at 20°C with a vapor pressure of 17.5 Torr is calculated using the ideal gas law, and determined to be approximately \(n \times 18.015\) grams, where \(n\) is the number of moles of water vapor calculated in steps 5 and 7.

Step by step solution

01

Understand the Concepts of Vapor Pressure and Saturation

Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature. When we say that air is saturated with water vapor, it means the air contains the maximum amount of water vapor it can hold at that temperature. In this case, the vapor pressure of water at 20°C is 17.5 Torr.
02

Convert Vapor Pressure to Standard Atmospheric Unit

First, we need to convert the given vapor pressure from Torr to the standard atmosphere unit to use the ideal gas law. We use the conversion 1 atm = 760 Torr. Thus, the vapor pressure in atmospheres is \( 17.5 \text{Torr} \times \frac{1 \text{atm}}{760 \text{Torr}} = 0.02303 \text{atm} \).
03

Apply the Ideal Gas Law to Find the Mass of the Water Vapor

The ideal gas law states that \( PV = nRT \), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvins. We can rearrange the formula to solve for \(n\) (the number of moles of water vapor): \( n = \frac{PV}{RT} \).
04

Insert Values into the Ideal Gas Law

We have the volume \(V = 1.0 \text{L} \) and the vapor pressure \(P = 0.02303 \text{atm}\). The ideal gas constant \(R = 0.0821 \text{L·atm/mol·K}\), and we need to convert the given temperature from Celsius to Kelvin: \(T = 20°C + 273.15 = 293.15 K\). Inserting these values into the rearranged ideal gas law gives us the number of moles: \( n = \frac{(0.02303 \text{ atms})(1.0 \text{ L})}{(0.0821 \text{ L·atm/mol·K})(293.15 \text{ K})} \).
05

Calculate the Number of Moles of Water Vapor

Perform the calculation for \(n\) to find the number of moles of water vapor: \( n ≈ \frac{(0.02303 \times 1.0)}{(0.0821 \times 293.15)} \).
06

Convert Moles of Water Vapor to Mass

Using the molar mass of water (approximately 18.015 g/mol), we can convert the number of moles of water vapor to grams: \(\text{mass} = n \times \text{molar mass of water}\).
07

Solve for the Mass of Water Vapor

Finally, multiply the calculated number of moles by the molar mass of water to obtain the mass of water vapor in the air: \(\text{mass} = n \times 18.015 \text{g/mol}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Saturation Vapor Pressure
In the process of understanding humidity, saturation vapor pressure is a crucial concept. It refers to the pressure exerted by a vapor when it is in a state where evaporation and condensation occur at an equal rate. This balance indicates that the air contains as much water vapor as it can at a particular temperature, without any net evaporation or condensation. In practical terms, at 20°C, the saturation vapor pressure for water is 17.5 Torr, meaning that this is the maximum partial pressure water vapor will exert when it fully saturates the air. When air is at saturation vapor pressure, it is often referred to as being 'humid' or 'saturated with moisture'.

To provide a more intuitive grasp of saturation, imagine a closed room with a pan of water inside. Initially, water molecules evaporate from the pan into the air. But as the number of water molecules in the air increases, some start to condense back into the water pan. Saturation vapor pressure is achieved when the rate of evaporation equals the rate of condensation, leading to a dynamic equilibrium – the amount of water in the air remains constant barring any change in conditions.
Thermodynamic Equilibrium
The state of thermodynamic equilibrium is fundamental in the discussion of vapor pressure and phase changes. Thermodynamic equilibrium occurs when all parts of a system are at the same temperature, and there is no net flow of matter or energy. For our scenario, it means that the water vapor and the liquid water are at the same temperature and no net evaporation or condensation is occurring. At this equilibrium, the number of water molecules leaving the liquid to become vapor equals the number returning to the liquid from the vapor phase.

Think of it like a busy street where the number of people entering and leaving a store is the same—no overall change in the number of people inside. Similarly, in a container with air and water at thermodynamic equilibrium, the saturated vapor exerts a constant pressure, which in this case at 20°C, is precisely 17.5 Torr. Natural systems tend to move towards equilibrium, which is why understanding this concept is so vital when predicting the behavior of gases and liquids.
Ideal Gas Law
The ideal gas law is a powerful equation that relates the pressure, volume, temperature, and amount of a gas. Expressed as PV = nRT, where P stands for pressure of the gas, V for volume, n for moles of the gas, R for the ideal gas constant, and T for temperature in Kelvins, the law assumes that gases consist of many small particles that move in random directions with no interparticle interactions.

In our exercise, the ideal gas law helps us estimate the mass of water vapor in saturated air by first allowing us to calculate the number of moles of water vapor. To do so, we use the pressure equated to the saturation vapor pressure at 20°C and substitute the values into the ideal gas law. Remember, real gases don't always behave like ideal gases, particularly at high pressures and low temperatures where intermolecular forces start to play a significant role. Nonetheless, for many conditions, particularly those relevant to our example, the ideal gas law provides a close approximation to actual gas behavior.
Molar Mass
The term molar mass describes the mass of one mole of a given substance and is expressed in grams per mole (g/mol). It allows us to convert between the number of particles (moles) and the mass for a given substance. For example, the molar mass of water is approximately 18.015 g/mol, meaning that one mole of water molecules weighs approximately 18.015 grams.

To calculate the mass of water vapor in our exercise, we multiply the number of moles of water vapor by its molar mass. Understanding the molar mass is crucial in stoichiometry as it bridges the microscopic world of atoms and molecules with the macroscopic world we can measure, weigh, and observe. While dealing with gases, linking the molar mass with the ideal gas law allows us to transition from knowing the volume of a gas under certain conditions to finding its mass, which is a more tangible quantity for many applications.

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