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

Short Answer

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

Step by step solution

01

Convert room dimensions to meters

To convert feet to meters, we use the conversion factor \(1 ft = 0.3048 m\). So, \(12.5 ft \times 0.3048 = 3.81 m\) \(15.5 ft \times 0.3048 = 4.724 m\) \(80 ft \times 0.3048 = 24.384 m\)
02

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
03

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
04

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

The concepts of accuracy and precision are not always easy to grasp. Here are two sets of studies: (a) The mass of a secondary weight standard is determined by weighing it on a very precise balance under carefully controlled laboratory conditions. The average of 18 different weight measurements is taken as the weight of the standard. (b) A group of 10,000 males between the ages of 50 and 55 is surveyed to ascertain a relationship between calorie intake and blood cholesterol level. The survey questionnaire is quite detailed, asking the respondents about what they eat, smoking and drinking habits, and so on. The results are reported as showing that for men of comparable lifestyles, there is a \(40 \%\) chance of the blood cholesterol level being above 230 for those who consume more than 40 calories per gram of body weight per day, as compared with those who consume fewer than 30 calories per gram of body weight per day. Discuss and compare these two studies in terms of the precision and accuracy of the result in each case. How do the two studies differ in nature in ways that affect the accuracy and precision of the results? What makes for high precision and accuracy in any given study? In each of these studies, what factors might not be controlled that could affect the accuracy and precision? What steps can be taken generally to attain higher precision and accuracy?

(a) To identify a liquid substance, a student determined its density. Using a graduated cylinder, she measured out a 45-mLsample of the substance. She then measured the mass of the sample, finding that it weighed \(38.5 \mathrm{~g}\) She knew that the substance had to be either isopropyl alcohol (density \(0.785 \mathrm{~g} / \mathrm{mL}\) ) or toluene (density \(0.866 / \mathrm{mL})\). What are the calculated density and the probable identity of the substance? (b) An experiment requires \(45.0 \mathrm{~g}\) of ethylene glycol, a liquid whose density is \(1.114 \mathrm{~g} / \mathrm{mL}\). Rather than weigh the sample on a balance, a chemist chooses to dispense the liquid using a graduated cylinder. What volume of the liquid should he use? (c) A cubic piece of metal measures \(5.00 \mathrm{~cm}\) on each edge. If the metal is nickel, whose density is \(8.90 \mathrm{~g} / \mathrm{cm}^{3}\), what is the mass of the cube?

Carry out the following operations, and express the answers with the appropriate number of significant figures. (a) \(12.0550+9.05\) (b) \(257.2-19.789\) (c) \(\left(6.21 \times 10^{3}\right)(0.1050)\) (d) \(0.0577 / 0.753\)

Automobile batteries contain sulfuric acid, which is commonly referred to as "battery acid." Calculate the number of grams of sulfuric acid in \(0.500 \mathrm{~L}\) of battery acid if the solution has a density of \(1.28 \mathrm{~g} / \mathrm{mL}\) and is \(38.1 \%\) sulfuric acid by mass.

(a) What is the difference between a hypothesis and a theory? (b) Explain the difference between a theory and a scientific law. Which addresses how matter behaves, and which addresses why it behaves that way?

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