Question

The ratio of the average translational kinetic energies of hydrogen and oxygen at the same temperature is

• A. 1:8
• B. 1:4
• C. 1:1
• D. 1:6

Hint:

A key result from the kinetic theory of gases is that the average kinetic energy of gas molecules depends only on the temperature. While heavier molecules (like oxygen) move slower on average than lighter molecules (like hydrogen) at the same temperature (since $v_{rms} \propto 1/\sqrt{M}$), their average kinetic energies ($1/2 m v^2$) are identical.

Answer 1

The Correct Option is C

Step 1: Recall the equipartition of energy theorem.
The equipartition of energy theorem states that for a system in thermal equilibrium, the total energy is shared equally among all its degrees of freedom. Specifically, the average kinetic energy associated with each translational degree of freedom is $\frac{1}{2}k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.

Step 2: Determine the average translational kinetic energy of a molecule.
A gas molecule in three-dimensional space has three translational degrees of freedom (corresponding to motion along the x, y, and z axes). According to the equipartition theorem, the total average translational kinetic energy of a single molecule is: \[ KE_{avg, trans} = 3 \times \left(\frac{1}{2}k_B T\right) = \frac{3}{2}k_B T. \]

Step 3: Analyze the dependence of translational kinetic energy.
The formula $KE_{avg, trans} = \frac{3}{2}k_B T$ shows that the average translational kinetic energy depends only on the absolute temperature $T$. It does not depend on the mass of the molecule, the type of gas (monatomic, diatomic, etc.), or any other property of the gas.

Step 4: Calculate the ratio for hydrogen and oxygen.
We are given that the hydrogen ($H_2$) and oxygen ($O_2$) gases are at the same temperature. Since the average translational kinetic energy depends only on temperature, the molecules of both gases will have the same average translational kinetic energy. \[ (KE_{avg, trans})_{H_2} = \frac{3}{2}k_B T. \] \[ (KE_{avg, trans})_{O_2} = \frac{3}{2}k_B T. \] The ratio is therefore: \[ \frac{(KE_{avg, trans})_{H_2}}{(KE_{avg, trans})_{O_2}} = \frac{\frac{3}{2}k_B T}{\frac{3}{2}k_B T} = 1. \] The ratio is 1:1.