Question

Case-1: An ideal gas of molecular weight M at temperature T.
Case-2: Another ideal gas of molecular weight 2M at the temperature \(\frac{T}{2}\).
Identify the correct statement in the context of the above two cases.

• A. Average kinetic energy and average speed will be the same in the two cases.
• B. Both the averages are halved
• C. Both the averages are doubled.
• D. Only average speed is halved in the second case.

Answer 1

The Correct Option is B

We are comparing two cases involving ideal gases:

• Case-1: Molecular weight = M, Temperature = T
• Case-2: Molecular weight = 2M, Temperature = T/2

1. Average Kinetic Energy of an ideal gas molecule:

\[ \text{KE}_{\text{avg}} = \frac{3}{2}kT \]

• Depends only on temperature (T), not on molecular mass.
• So, in Case-1: \(\text{KE}_1 = \frac{3}{2}kT\)
• In Case-2: \(\text{KE}_2 = \frac{3}{2}k\cdot\frac{T}{2} = \frac{3}{4}kT\)

⟹ Kinetic energy is halved in Case-2.

2. Average Speed of a gas molecule:

\[ \bar{v} \propto \sqrt{\frac{T}{M}} \]

Let’s calculate the ratio of average speeds in both cases:

\[ \frac{\bar{v}_2}{\bar{v}_1} = \sqrt{\frac{T/2}{2M}} \div \sqrt{\frac{T}{M}} = \sqrt{\frac{T}{4M}} \div \sqrt{\frac{T}{M}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

⟹ Average speed is also halved in Case-2.

Conclusion: In Case-2, both average kinetic energy and average speed are halved compared to Case-1.

Correct Answer: (B) Both the averages are halved.