Question

Calculate the rms speed of an oxygen gas molecule, o2, at 21.0 ?c .

Answer 1

The root mean square (rms) speed of an oxygen gas molecule (O2) at 21.0 °C is approximately 484 meters per second.

Step-by-step explanation:

The root mean square speed of gas molecules can be calculated using the formula
`\( \text{rms speed} = \sqrt{(3kT)/(m)} \)`, where
`\( k \)` is the Boltzmann constant,
`\( T \)` is the temperature in Kelvin, and
`\( m \)` is the mass of the molecule. For an oxygen molecule
`(\( O_2 \))`, the molar mass
`(\( m \))` is approximately 32 grams/mol. Converting this to kilograms
`(\( 32 * 10^(-3) \) kg/mol)` and plugging in the values into the formula, we find the rms speed.

At 21.0 °C, the temperature in Kelvin
`(\( T \)) is \( 21.0 + 273.15 = 294.15 \) K`. Substituting the values into the formula:
`\( \text{rms speed} = \sqrt{(3 * (1.38 * 10^(-23)) * 294.15)/(32 * 10^(-3))} \), the`calculation yields an rms speed of approximately 484 m/s.

In summary, the root mean square speed of an oxygen gas molecule at 21.0 °C is approximately 484 meters per second.