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Chapter 15: Problem 17

Compressed air with mass \(8.8 \mathrm{kg}\) is stored in a \(0.050-\mathrm{m}^{3}\) cylinder. (a) What's the density of the compressed air? (b) What volume would the same gas occupy at a typical atmospheric density of \(1.2 \mathrm{kg} / \mathrm{m}^{2} ?\)

Short Answer

Expert verified
The density of the compressed air is \(176\, \mathrm{kg/m}^{3}\) and the volume it would occupy at atmospheric density is \(7.33\, \mathrm{m}^{3}\).

Step by step solution

01

Calculate the Density

The density of the compressed air can be found using the formula: \(\rho = \frac{m}{V} \) where \(\rho\) is the density, \(m\) is the mass and \(V\) is the volume. Substituting the provided values we get: \(\rho = \frac{8.8\, \mathrm{kg}}{0.050\, \mathrm{m}^{3}}\)
02

Solve for the Density

Performing the division results in \(\rho = 176\, \mathrm{kg/m}^{3}\). Thus, the density of the compressed air is \(176\, \mathrm{kg/m}^{3}\).
03

Calculate the Volume

The volume of gas at atmospheric density is calculated using the formula: \(V = \frac{m}{\rho_{atm}} \) where \(V\) is the volume, \(m\) is the mass and \(\rho_{atm}\) is the atmospheric density. Substituting the provided values we get: \(V = \frac{8.8\, \mathrm{kg}}{1.2\, \mathrm{kg/m}^{3}}\)
04

Solve for the Volume

Performing the division results in \(V = 7.33\, \mathrm{m}^{3}\). Thus, the volume of the gas at atmospheric density is \(7.33\, \mathrm{m}^{3}\).

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