Question

Assuming ideal gas behavior, the density of \( \text{O}_2 \) gas at 300 K and 1.0 atm is ......... g L\(^{-1}\) (rounded up to two decimal places).

Hint:

To find the density of a gas, use the equation \( \rho = \frac{PM}{RT} \), where \( P \) is pressure, \( M \) is molar mass, and \( R \) is the ideal gas constant.

Answer 1

Correct Answer: 1.29 - 1.31

Step 1: Using the ideal gas equation. 
The ideal gas equation is: \[ PV = nRT \] where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature. To find the density, we rearrange the equation to express \( n/V \) as the molar concentration: \[ \frac{n}{V} = \frac{P}{RT} \] The molar mass of \( \text{O}_2 \) is 32 g/mol, so the density (\( \rho \)) is given by: \[ \rho = \frac{P}{RT} \times M \] Substituting the given values: \[ P = 1.0 \, \text{atm}, \, T = 300 \, \text{K}, \, R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1}, \, M = 32 \, \text{g/mol} \] \[ \rho = \frac{1.0 \times 32}{0.0821 \times 300} = 1.299 \, \text{g/L} \]

Step 2: Conclusion. 
The density of \( \text{O}_2 \) at 300 K and 1.0 atm is 1.299 g L\(^{-1}\).