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 \(12.5 \times 15.5 \times 80 \mathrm{ft}\) ?

Short answer

The mass of the air in the room is approximately 527.478 kg.

Step-by-step solution

Convert room dimensions to meters

To convert feet to meters, we use the conversion factor \(1 ft = 0.3048 m\). So, \(12.5 ft \times 0.3048 = 3.81 m\) \(15.5 ft \times 0.3048 = 4.724 m\) \(80 ft \times 0.3048 = 24.384 m\)

Calculate the volume of the room

Now that we have the room dimensions in meters, we can calculate the volume of the room by multiplying its length, width, and height: Volume = Length × Width × Height Volume = (3.81 m) × (4.724 m) × (24.384 m) = 443.595 m³

Convert the density of air

The given density of air is 1.19 g/L. We need to convert this to kg per cubic meter. 1 g = 0.001 kg, so 1.19 g = 1.19 × 0.001 kg = 0.00119 kg 1 L = 0.001 m³, so the density of air is: Density = 0.00119 kg / 0.001 m³ = 1.19 kg/m³

Find the mass of the air in the room

To find the mass of the air, we will multiply the volume of the room by the density of the air: Mass = Volume × Density Mass = (443.595 m³) × (1.19 kg/m³) = 527.478 kg The mass of the air in the room is approximately 527.478 kg.

Key concepts

Volume Conversion

When dealing with measurements, it is important to use the same unit system for consistency. In this exercise, the room dimensions are given in feet, and we need to convert them to meters to match the units for density. This is where volume conversion becomes crucial. To convert from feet to meters, use the conversion factor: 1 foot = 0.3048 meters.
Let's take a closer look:

• Length: 12.5 feet = 12.5 × 0.3048 = 3.81 meters
• Width: 15.5 feet = 15.5 × 0.3048 = 4.724 meters
• Height: 80 feet = 80 × 0.3048 = 24.384 meters

Now, you can calculate the room's volume in cubic meters by multiplying these dimensions: Volume = Length × Width × Height = 3.81 m × 4.724 m × 24.384 m = 443.595 m³.
Converting all measurements to the same base units simplifies calculations and reduces errors. Always remember to double-check your conversions to ensure accuracy.

Mass Calculation

With the volume of the room calculated, we can now determine the mass of air contained within it. Mass calculation involves finding the total mass based on the given volume and the density of the substance. In this case, we have already determined the room's volume in cubic meters.
The next step involves using the procedure:

• First, ensure the density is expressed in compatible units. The density of air is given as 1.19 g/L.
• Convert density from grams per liter to kilograms per cubic meter: Knowing that 1 gram = 0.001 kilograms and 1 liter = 0.001 cubic meters,
  the conversion is: 1.19 g/L = 0.00119 kg/m³
• Calculate the mass using the formula: Mass = Volume × Density = 443.595 m³ × 1.19 kg/m³ = 527.478 kg

This calculation tells us the total mass of air in the room, approximately 527.478 kilograms. Always ensure that your final mass unit is kilograms or others as required by your context.

Air Density at Atmospheric Pressure

Air density is a critical factor in many scientific calculations. It describes how much air mass exists per unit volume. Understanding how to work with air density, especially at atmospheric pressure, is important for correctly solving problems involving gases.
At ordinary atmospheric pressure and a temperature of 25°C, the density of air is approximately 1.19 g/L. Since we often work with other unit systems, it is useful to express this value in kg/m³. Here's how to convert it:

• 1 gram per liter is equivalent to 0.001 kilograms per liter.
• 1 liter is equivalent to 0.001 cubic meters, thus the density becomes 1.19 kg/m³.

This density value is an average and depends on conditions like temperature and pressure. Therefore, ensure you use the correct density value reflective of the environment's current conditions. Mastering this will aid in precise and accurate calculations in physics and engineering.