Question

The helium and argon are put in the flask at the same room temperature (300 K). The ratio of average kinetic energies (per molecule) of helium and argon is : (Give : Molar mass of helium = 4 g/mol, Molar mass of argon = 40 g/ mol )

• A. 1 : 10
• B. 10 : 1
• C. \( 1 : \sqrt{10} \)
• D. 1 : 1

Hint:

The average kinetic energy per molecule of an ideal gas depends only on the temperature, not on the mass or type of the gas, as long as they are at the same temperature. This is a direct consequence of the equipartition theorem.

Answer 1

The Correct Option is D

Step 1: Recall the formula for the average kinetic energy per molecule.
The average kinetic energy of a molecule in an ideal gas is directly proportional to the absolute temperature and is given by the equipartition theorem: \[ K_{avg} = \frac{f}{2} k_B T \] where \( f \) is the number of degrees of freedom of the molecule, \( k_B \) is the Boltzmann constant, and \( T \) is the absolute temperature.
Step 2: Determine the degrees of freedom for helium and argon.
Helium (He) and argon (Ar) are both noble gases and are monatomic. Monatomic gases have only three translational degrees of freedom (motion along the x, y, and z axes).
Therefore, for both helium and argon, \( f = 3 \).
Step 3: Calculate the average kinetic energy for helium and argon.
For helium at temperature \( T = 300 \) K: \[ K_{avg, He} = \frac{3}{2} k_B (300) \] For argon at temperature \( T = 300 \) K: \[ K_{avg, Ar} = \frac{3}{2} k_B (300) \]
Step 4: Find the ratio of the average kinetic energies.
The ratio of the average kinetic energies per molecule of helium and argon is: \[ \frac{K_{avg, He}}{K_{avg, Ar}} = \frac{\frac{3}{2} k_B (300)}{\frac{3}{2} k_B (300)} = 1 \]
Thus, the ratio is 1 : 1. The molar masses of helium and argon are irrelevant for determining the average kinetic energy per molecule at the same temperature.