What is the root mean square (rms) speed of O_2 molecules at 375 K? (Use R = 8.314 J mol^{-1} K^{-1}, molar mass of O_2 = 32.0 g/mol, and express your answer in m/s.)

• A. 325 m/s
• B. 273 m/s
• C. 615 m/s
• D. 484 m/s

Answer: (D)

Verified step by step guidance

1. Identify the formula for the root mean square (rms) speed of a gas molecule, which is given by: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where \(R\) is the gas constant, \(T\) is the temperature in kelvin, and \(M\) is the molar mass in kilograms per mole.
2. Convert the molar mass of oxygen from grams per mole to kilograms per mole because the units of \(R\) are in joules (which involve kilograms): \[ M = 32.0\ \text{g/mol} = 32.0 \times 10^{-3}\ \text{kg/mol} \]
3. Substitute the given values into the formula: \[ R = 8.314\ \text{J mol}^{-1} \text{K}^{-1}, \quad T = 375\ \text{K}, \quad M = 32.0 \times 10^{-3}\ \text{kg/mol} \]
4. Calculate the expression inside the square root first: \[ \frac{3RT}{M} = \frac{3 \times 8.314 \times 375}{32.0 \times 10^{-3}} \]
5. Finally, take the square root of the result from the previous step to find the rms speed: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] This will give the rms speed in meters per second (m/s).