What is the root mean square speed (in meters per second) of a nitrogen molecule (N_2) at 50.0 °C? (Molar mass of N_2 = 28.0 g/mol)

• A. 402 m/s
• B. 484 m/s
• C. 517 m/s
• D. 563 m/s

Answer: (C)

Verified step by step guidance

1. Convert the temperature from Celsius to Kelvin using the formula: \(T(K) = T(^\circ C) + 273.15\). For 50.0 °C, calculate \(T = 50.0 + 273.15\) K.
2. Convert the molar mass of nitrogen gas from grams per mole to kilograms per mole because SI units require mass in kilograms. Use \(M(kg/mol) = M(g/mol) \times 10^{-3}\). For 28.0 g/mol, calculate \(M = 28.0 \times 10^{-3}\) kg/mol.
3. Recall the formula for the root mean square (rms) speed of a gas molecule: \(v_{rms} = \sqrt{\frac{3RT}{M}}\), where \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass in kilograms per mole.
4. Substitute the values into the formula: use \(R = 8.314\) J/(mol·K), the temperature \(T\) in Kelvin from step 1, and the molar mass \(M\) in kg/mol from step 2.
5. Calculate the value inside the square root first, then take the square root to find the root mean square speed \(v_{rms}\) in meters per second.