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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 \(14.5 \mathrm{ft} \times 16.5 \mathrm{ft} \times 8.0 \mathrm{ft} ?\)

Short Answer

Expert verified
The mass of the air in the room is approximately 64.47 kilograms.

Step by step solution

01

Convert the room dimensions to meters

Before determining the room's volume, we need to convert the dimensions from feet to meters, since the density of air is given in grams per liter, and 1 liter is equivalent to 0.001 cubic meters. Recall that 1 foot is equal to 0.3048 meters. Multiply each dimension by 0.3048: Length: \(14.5 \mathrm{ft} \times 0.3048 = 4.42 \mathrm{m}\) Width: \(16.5 \mathrm{ft} × 0.3048 = 5.03 \mathrm{m}\) Height: \(8.0 \mathrm{ft} × 0.3048 = 2.44 \mathrm{m}\)
02

Calculate the volume of the room in cubic meters

Now that we have the dimensions in meters, multiply them together to find the volume of the room in cubic meters: Volume = Length × Width × Height Volume = \(4.42 \mathrm{m} × 5.03 \mathrm{m} × 2.44 \mathrm{m} = 54.19 \mathrm{m^3}\)
03

Convert the volume to liters

To utilize the given density of air, we need to convert the volume of the room from cubic meters to liters. Since 1 liter is equal to 0.001 cubic meters, we can multiply the volume in cubic meters by 1000 to get the volume in liters: Volume = \(54.19 \mathrm{m^3} × 1000 = 54190 \mathrm{L}\)
04

Calculate the mass of air in grams

Now that we have the volume in liters, we can use the given density of air (1.19 g/L) to determine the mass of the air in the room: Mass = Volume × Density Mass = \(54190 \mathrm{L} × 1.19 \mathrm{g/L} = 64466.1 \mathrm{g}\)
05

Convert the mass to kilograms

Finally, we need to convert the mass of the air from grams to kilograms. To do this, divide the mass in grams by 1000: Mass = \(64466.1 \mathrm{g} ÷ 1000 = 64.47 \mathrm{kg}\) The mass of the air in the room is approximately 64.47 kilograms.

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Key Concepts

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

Unit Conversion
Unit conversion is a crucial process in science and engineering to ensure measurements are in the same system for calculations. This step is fundamental when solving problems that involve different measurement units. For students tackling problems of density, understanding how to go from one unit to another—such as feet to meters or grams to kilograms—is essential for accurate results.

To make this transition smoother, remember the conversion factors. For length, knowing that 1 foot equals 0.3048 meters is key. Multiplying each spatial dimension of an object by this factor changes the unit from feet to meters. Similarly, for mass, the conversion from grams to kilograms is made simple by remembering that 1 kilogram is equal to 1000 grams. Dividing by this factor converts grams into kilograms.

Quick Conversion References

  • For length: 1 ft = 0.3048 m
  • For mass: 1 kg = 1000 g
By mastering these conversions, students can navigate through various physics and chemistry problems with greater ease.
Volume Determination
Determining volume is about finding the amount of three-dimensional space an object occupies. In our context, calculating the volume of a room involves multiplying the room's length, width, and height once you have the measurements in consistent units, typically meters.

The formula for the volume of a rectangular space like a room is straightforward: Volume = Length × Width × Height. Always check that the units are the same for each dimension to avoid errors in your result. Once the volume is found in cubic meters, other conversions may be needed – such as converting to liters for use with density calculations. Since volume is used to determine the mass of air (as well as other substances), having an accurate volume is essential for accurate calculations.

Steps for Volume Calculation

  • Ensure all measurements are in the same unit (meters is standard).
  • Apply the volume formula for the shape you're working with, usually, for rooms, that's a rectangular prism.
  • Remember that for gaseous substances, you may need to convert cubic meters to liters (1 m³ = 1000 L).
By following these steps, you can establish the foundation for understanding more complex concepts in material science and physics.
Mass of Air Calculation
The mass of air within a space can be found by knowing the density of the air and the volume of the space it occupies. To do this, the concept of density—defined as mass per unit volume (usually expressed in g/L or kg/m³)—comes into play. Once the volume is expressed in the proper unit that matches the unit in the density (liters in this case), the mass can be determined by simple multiplication.

In the given problem, the density of air is 1.19 g/L, which is a measure of how much mass of air is present in a given volume (a liter). To find the mass in a larger volume like a room, you multiply the total volume of the room in liters by the density. The resulting value will be in grams, which can then be converted to kilograms if necessary. This process is not only essential in classroom exercises but also in real-life applications such as environmental science and engineering.

Step-by-Step Mass Calculation:

  • Start with the room's volume in a compatible unit (liters).
  • Multiply this volume by the air's density to find the mass in grams.
  • Convert the mass from grams to kilograms if required.
Grasping this calculation ensures that students can approach real-world problems, such as HVAC system design and assessing air quality, with confidence.

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Most popular questions from this chapter

(a) After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A 25.0 -mL portion of the liquid had a mass of \(21.95 \mathrm{~g}\). A chemistry handbook lists the density of benzene at \(15^{\circ} \mathrm{C}\) as \(0.8787 \mathrm{~g} / \mathrm{mL}\). Is the calculated density in agreement with the tabulated value? (b) An experiment requires \(15.0 \mathrm{~g}\) of cyclohexane, whose density at \(25^{\circ} \mathrm{C}\) is \(0.7781 \mathrm{~g} / \mathrm{mL}\). What volume of cyclohexane should be used? (c) A spherical ball of lead has a diameter of \(5.0 \mathrm{~cm}\). What is the mass of the sphere if lead has a density of \(11.34 \mathrm{~g} / \mathrm{cm}^{3}\) ? (The volume of a sphere is \((4 / 3) \pi r^{3}\) where \(r\) is the radius.)

Give the chemical symbol or name for each of the following elements, as appropriate: (a) carbon, (b) nitrogen, (c) titanium, \((\mathbf{d})\) zinc, \((\mathbf{e})\) iron, \((\mathbf{f}) \mathrm{P},(\mathrm{g}) \mathrm{Ca},(\mathbf{h}) \mathrm{He},(\mathbf{i}) \mathrm{Pb},(\mathbf{j}) \mathrm{Ag} .\)

Three beakers contain clear, colorless liquids. One beaker contains pure water, another contains salt water, and another contains sugar water. How can you tell which beaker is which? (No tasting allowed!)

A package of aluminum foil contains \(50 \mathrm{ft}^{2}\) of foil, which weighs approximately 8.0 oz. Aluminum has a density of \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\). What is the approximate thickness of the foil in millimeters?

Classify each of the following as a pure substance or a mixture. If a mixture, indicate whether it is homogeneous or heterogeneous: (a) air, (b) tomato juice, (c) iodine crystals, (d) sand.

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