Question

Two gases A and B are at absolute temperatures 350 K and 420 K respectively. The ratio of average kinetic energy of the molecules of gas B to that of gas A is

• A. $6 : 5$
• B. $\sqrt{6} : \sqrt{5}$
• C. $36 : 25$
• D. $5 : 6$

Hint:

Kinetic energy depends only on temperature, not on the mass or nature of the gas molecules.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The average kinetic energy of gas molecules is a direct measure of the absolute temperature of the gas, according to kinetic theory.
It is entirely independent of the molar mass or chemical nature of the gas.
Step 2: Key Formula or Approach:
Average translational kinetic energy per molecule $E = \frac{3}{2}k_B T$.
Therefore, $E \propto T$.
The ratio is $\frac{E_B}{E_A} = \frac{T_B}{T_A}$.
Step 3: Detailed Explanation:
Given temperatures: $T_A = 350 \text{ K}$ and $T_B = 420 \text{ K}$.
Using the proportionality relationship: \[ \frac{E_B}{E_A} = \frac{T_B}{T_A} \] Substitute the values: \[ \frac{E_B}{E_A} = \frac{420}{350} \] Simplify the fraction by dividing by 70: \[ \frac{E_B}{E_A} = \frac{420 \div 70}{350 \div 70} = \frac{6}{5} \] The ratio is $6 : 5$.
Step 4: Final Answer:
The ratio is $6 : 5$.