Multiple Choice
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)
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7. Gases
Root Mean Square Speed
Multiple Choice
What is the root mean square (rms) speed of a nitrogen molecule (N_2, molar mass = 28.01 g/mol) at 298 K? (Use R = 8.314 J/mol·K)
A
Approximately 273 m/s
B
Approximately 492 m/s
C
Approximately 650 m/s
D
Approximately 517 m/s
Verified step by step guidance1
Identify the formula for the root mean square (rms) speed of a gas molecule:
\[v_{rms} = \sqrt{\frac{3RT}{M}}\]
where \(v_{rms}\) is the rms speed, \(R\) is the gas constant, \(T\) is the temperature in kelvin, and \(M\) is the molar mass in kilograms per mole.
Convert the molar mass of nitrogen from grams per mole to kilograms per mole because the gas constant \(R\) is in joules (which involve kilograms):
\[M = 28.01\ \text{g/mol} = 28.01 \times 10^{-3}\ \text{kg/mol}\]
Plug in the known values into the rms speed formula:
- \(R = 8.314\ \text{J/mol·K}\)
- \(T = 298\ \text{K}\)
- \(M = 28.01 \times 10^{-3}\ \text{kg/mol}\)
So,
\[v_{rms} = \sqrt{\frac{3 \times 8.314 \times 298}{28.01 \times 10^{-3}}}\]
Calculate the value inside the square root first, which represents the ratio of thermal energy to molar mass, ensuring units are consistent to get speed in meters per second.
Finally, take the square root of the calculated value to find the rms speed \(v_{rms}\) of the nitrogen molecule at 298 K.
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