Question

The ratio of kinetic energy of a molecule of neon to that of the oxygen gas at 27°C is

• A. \( \frac{3}{2} \)
• B. \( \frac{3}{5} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{5}{3} \)

Hint:

The kinetic energy of gases is directly related to their temperature, and the ratio depends on the molar masses of the gases.

Answer 1

The Correct Option is B

We are tasked with finding the ratio of the kinetic energy of a molecule of neon (Ne) to that of oxygen (Oâ‚‚) at \(27^\circ {C}\). 
Step 1: Kinetic energy of a gas molecule The average kinetic energy of a gas molecule is given by: \[ E_k = \frac{f}{2} k_B T \] where \(f\) is the degrees of freedom, \(k_B\) is the Boltzmann constant, and \(T\) is the temperature in Kelvin. 
Step 2: Degrees of freedom
For neon (monatomic gas), \(f = 3\).
For oxygen (diatomic gas), \(f = 5\).
Step 3: Ratio of kinetic energies
The ratio of the kinetic energy of neon to that of oxygen is: \[ \frac{E_{k, {Ne}}}{E_{k, {O}_2}} = \frac{\frac{3}{2} k_B T}{\frac{5}{2} k_B T} = \frac{3}{5} \] Step 4: Match with the options The ratio \(\frac{3}{5}\) matches option (2). 
Final Answer: \(\boxed{2}\)