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Chapter 1: Problem 60

Carbon monoxide (CO) detectors sound an alarm when peak levels of carbon monoxide reach 100 parts per million (ppm). This level roughly corresponds to a composition of air that contains \(400,000\) \mug carbon monoxide per cubic meter of air \(\left(400,000 \mu \mathrm{g} / \mathrm{m}^{3}\right) .\) Assuming the dimensions of a room are 18 $\mathrm{ft} \times 12 \mathrm{ft} \times 8 \mathrm{ft}$ , estimate the mass of carbon monoxide in the room that would register 100 \(\mathrm{ppm}\) on a carbon monoxide detector.

Short Answer

Expert verified
The mass of carbon monoxide in the room that would register 100 ppm on a carbon monoxide detector is approximately 19.54 grams.

Step by step solution

01

Convert room dimensions to meters

To convert the room dimensions from feet to meters, we will use the conversion factor: 1 ft ≈ 0.3048 m. Therefore, the room dimensions are as follows: Length: 18 ft × 0.3048 ≈ 5.4864 m Width: 12 ft × 0.3048 ≈ 3.6576 m Height: 8 ft × 0.3048 ≈ 2.4384 m
02

Calculate the room volume

Now that we have the room dimensions in meters, we can calculate the volume of the room by multiplying the length, width, and height: Volume = Length × Width × Height Volume = 5.4864 m × 3.6576 m × 2.4384 m ≈ 48.8562 m³
03

Calculate the mass of CO at 100 ppm in the room

We are given that 100 ppm corresponds to 400,000 µg/m³ of CO. To estimate the mass of CO in the room that would register 100 ppm, we must multiply this concentration by the volume of the room: Mass of CO = Concentration × Room Volume Mass of CO = 400,000 µg/m³ × 48.8562 m³ ≈ 19,542,480 µg Finally, convert the mass of CO into grams (1 g = 1,000,000 µg): Mass of CO ≈ 19,542,480 µg × (1 g/1,000,000 µg) ≈ 19.54 g So, the mass of carbon monoxide in the room that would register 100 ppm on a carbon monoxide detector is approximately 19.54 grams.

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