Question

Calculate the rms speed of an oxygen gas molecule, o2, at 29.0 ∘c . express your answer numerically in meters per second.

Answer 1

The rms speed of a gas can be calculated using the following rule:
Vrms = sqrt ( 3RT / M) where:
R is the gas constant = 8.314
T is the temperature = 29 + 273 = 302 degrees kelvin
M is the molar mass of the gas (oxygen) = 2*16 = 32 grams = 0.032 kg

Substitute in the equation to get Vrms as follows:
Vrms = sqrt [(3*8.314*302) / (0.032)] = 485.05 meters/sec

Answer 2

The rms speed of an oxygen gas molecule : 485.17 m/s

Further explanation

Kinetic Molecular Theory (KMT) states that a gas consists of molecules that move at a constant and random speed. The collisions between molecules are perfectly elastic so that no energy is wasted.

The molecules move in straight lines until they collide

Energy because this motion is expressed as Kinetic energy (KE) which can be formulated as:

 $\displaystyle KE=\frac{1}{2}mv^2$

The average kinetic energy value is only affected by temperature changes. The higher the temperature, the average kinetic energy of the molecule increases

This molecule is very small when compared to the distance between molecules, so the volume of gas contains mostly empty space

Gas particles move randomly (both speed and direction , as vector)

Average velocities of gases can be expressed as root-mean-square averages. (V rms)

$\large{\boxed{\bold{v_{rms}=\sqrt{\frac{3RT}{Mm} } }}$

R = gas constant, T = temperature, Mm = molar mass of the gas particles

From the question

R= 8.314 J/mol K

T = 29 + 273 = 302 K

Mm = molar mass O₂ = 2.16 = 32 g/mol = 32.10-3 kg/mol$\displaystyle v_{rms}=\sqrt{\frac{3.8.314.302}{32.10^{-3}} }\\\\v_{rms}=485.17\:m/s$

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Keywords: Kinetic Molecular Theory, Collisions, The average kinetic energy , the rms speed, gas molecule