To solve the problem, we will work through parts (a) and (b) as follows:
Part (a):
We know the density (\( \rho \)) of dry air is given as:
\[ \rho = 1.12 \times 10^{-3} , \text{g/cm}^3 \]
To find the volume (\( V \)) of air that has a mass of 1.00 kg, we'll use the density formula:
\[ \rho = \frac{m}{V} \]
Rearranging this gives:
\[ V = \frac{m}{\rho} \]
First, we need to convert the mass from kilograms to grams since the density is given in grams per cubic centimeter:
\[ 1.00 , \text{kg} = 1000 , \text{g} \]
Now we can substitute into the volume equation:
\[ V = \frac{1000 , \text{g}}{1.12 \times 10^{-3} , \text{g/cm}^3} \]
Calculating the volume in cubic centimeters:
\[ V = \frac{1000}{1.12 \times 10^{-3}} \approx 892857.14 , \text{cm}^3 \]
Now we convert cubic centimeters to cubic meters. We know that:
\[ 1 , \text{cm}^3 = 1 \times 10^{-6} , \text{m}^3 \]
Thus:
\[ V \approx 892857.14 , \text{cm}^3 \times 1 \times 10^{-6} , \text{m}^3/\text{cm}^3 \approx 0.892857 , \text{m}^3 \]
So the volume of air that has a mass of 1.00 kg is approximately:
\[ \boxed{0.893 , \text{m}^3} \]
Part (b):
To express the density in lb/ft³, we start by converting the density given in g/cm³ to lb/ft³.
First, we convert grams to pounds:
\[ 1 , \text{g} = 0.00220462 , \text{lb} \]
Next, convert cubic centimeters to cubic feet. We know:
\[ 1 , \text{cm}^3 = 3.53147 \times 10^{-5} , \text{ft}^3 \]
Now we can convert the density:
\[ \rho = 1.12 \times 10^{-3} , \text{g/cm}^3 \times \frac{0.00220462 , \text{lb}}{1 , \text{g}} \times \frac{1}{3.53147 \times 10^{-5} , \text{ft}^3/\text{cm}^3} \]
Calculating it step by step:
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Convert g/cm³ to lb/cm³: \[ 1.12 \times 10^{-3} , \text{g/cm}^3 \times 0.00220462 , \text{lb/g} \approx 2.465 \times 10^{-6} , \text{lb/cm}^3 \]
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Convert lb/cm³ to lb/ft³: \[ 2.465 \times 10^{-6} , \text{lb/cm}^3 \times \frac{1, \text{ft}^3}{3.53147 \times 10^{-5}, \text{cm}^3} \approx 69.89 , \text{lb/ft}^3 \]
Thus, the density of dry air expressed as lb/ft³ is approximately:
\[ \boxed{69.89 , \text{lb/ft}^3} \]