Question

The density of air at ordinary atmospheric pressure and \(25^{\circ} \mathrm{C}\) is \(1.19 \mathrm{~g} / \mathrm{L}\). What is the mass, in kilograms, of the air in a room that measures \(12.5 \times 15.5 \times 8.0 \mathrm{ft}\) ?

Short answer

The mass of the air in the room with dimensions \(12.5 \times 15.5 \times 8.0 ft\) and a density of \(1.19 g/L\) at \(25^{\circ}C\) is approximately 52.58 kg.

Step-by-step solution

Convert dimensions to meters

First, we need to convert the room's dimensions from feet to meters. We can use the following conversion factor: 1 foot = 0.3048 meters. Thus: \(12.5 ft = 12.5 * 0.3048 m = 3.81 m\) \(15.5 ft = 15.5 * 0.3048 m = 4.724 m\) \(8.0 ft = 8.0 * 0.3048 m = 2.4384 m\)

Find the volume of the room

Now we will find the volume of the room in cubic meters, by multiplying the dimensions: Volume = length × width × height Volume = 3.81 m × 4.724 m × 2.4384 m = 44.186 m³

Convert density of air to kg/m³

The given density of air is 1.19 g/L. To convert it to kg/m³, we need to first convert grams to kilograms and liters to cubic meters: 1 g = 0.001 kg 1 L = 0.001 m³ Density of air = 1.19 g/L × (0.001 kg/g) × (1 m³ / 0.001 L) = 1.19 kg/m³

Calculate the mass of the air

Now we have the density of air in kg/m³ and the volume of the room in m³. We can use the formula for density to find the mass of air in the room: Density = mass/volume mass = Density × volume mass = 1.19 kg/m³ × 44.186 m³ = 52.581 kg Therefore, the mass of the air in the room is approximately 52.58 kilograms.

Key concepts

Unit Conversion

Understanding unit conversion is essential in many scientific calculations, including density problems. Converting from one unit to another correctly is key to achieving the right answer. In our example, the conversion from feet to meters was necessary because the standard scientific units of measurement are in the metric system, specifically meters for length.

Here's a tip: to avoid errors in unit conversion, use conversion factors. A conversion factor is a ratio that equates two different types of units, such as the one used in our problem, where 1 foot is equivalent to 0.3048 meters. By multiplying each dimension of the room in feet by this factor, we can easily convert our measurements to meters, which can then be used in subsequent volume and density calculations.

Volume Calculation

Volume calculation is another fundamental aspect of solving many physics and chemistry problems. It determines the amount of space that a three-dimensional shape occupies. For rectangular rooms or objects, the volume is found by multiplying the length by the width by the height. It is crucial to use consistent units when calculating volume.

Consider this: when you multiply the converted dimensions of the room together, as shown in our solution, you derive the volume in cubic meters (m³), which is the standard scientific unit for volume. Understanding how to calculate volume allows you to relate it to other physical properties, such as density or mass.

Density of Air

The density of air is often a required value in problems dealing with gases and fluid dynamics. It can vary with temperature and pressure, but at standard conditions (ordinary atmospheric pressure and 25 degrees Celsius), the density of air is approximately 1.19 g/L.

Why the Density Matters

Knowing the density of air is necessary when you're trying to calculate the mass of air in a given space. In our example, we converted the density of air from grams per liter to kilograms per cubic meter to correspond with the volume unit we calculated. This conversion ensures that all units will cancel appropriately in the mass calculation for a coherent and correct solution.

Mass of Air Calculation

Determining the mass of air within a space involves using the density of air and the volume of the space. The formula for mass based on density is Mass = Density × Volume, where the mass is what you want to find, the density should be in kg/m³, and the volume should be in m³.

Apply it like this: By substituting the density of air we converted into the formula along with the calculated volume of the room, the product is the mass of the air in the room. This calculation provides insight into properties of air in our environment and can be applied to broader contexts, including HVAC system designs, understanding weather patterns, and various engineering applications.