Calculate the root mean square velocity of carbon monoxide (CO) at 316 K. (Use R = 8.314 J/mol·K and the molar mass of CO = 28.01 g/mol)

• A. 450 m/s
• B. 500 m/s
• C. 515 m/s
• D. 483 m/s

Answer: (D)

Verified step by step guidance

1. First, understand the formula for root mean square velocity (v_rms) which is given by: $\sqrt{\frac{3RT}{M}}$, where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass in kg/mol.
2. Convert the molar mass of carbon monoxide (CO) from grams per mole to kilograms per mole. Since 1 g = 0.001 kg, the molar mass of CO is $28.01\times 0.001$ kg/mol.
3. Substitute the values into the formula: R = 8.314 J/mol·K, T = 316 K, and M = 0.02801 kg/mol. The equation becomes: $\sqrt{\frac{3\times 8.314\times 316}{0.02801}}$.
4. Calculate the numerator of the fraction: $3\times 8.314\times 316$.
5. Finally, divide the result from the previous step by the molar mass in kg/mol, and take the square root of the entire expression to find the root mean square velocity.