Question:

At \( T(K) \), in a closed vessel, a gas obeying the kinetic theory of gases has two molecules. The kinetic energy of these molecules is \( x_1 \) J and \( x_2 \) J respectively. After a few collisions, their kinetic energy is \( y_1 \) J and \( y_2 \) J respectively. Identify the correct relationship/s:

(1) \( x_1 = y_1 \)

(2) \( \frac{x_1 + x_2}{2} = \frac{y_1 + y_2}{2} \)

(3) \( x_2 = y_2 \)

(4) \( (x_1 - x_2)^2 = (y_1 - y_2)^2 \)

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The conservation of kinetic energy during elastic collisions often allows us to use the squares of velocity or energy differences to find relationships between initial and final states.
Updated On: Apr 15, 2025
  • \( I, II, III \) only
  • \( I, II, III \) only
  • \( II, IV \) only
  • \( II \) only
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The Correct Option is D

Approach Solution - 1

The kinetic energy of a gas molecule is directly related to its temperature. In the kinetic theory of gases, collisions between molecules lead to energy redistribution, but the total energy is conserved. Let's evaluate each option: 
Step 1: Analyzing the options 
Option I: \( x_1 = y_1 \) 
This would imply that the kinetic energy of the first molecule does not change after collisions. However, in a typical gas system, collisions result in the redistribution of energy. Therefore, this relationship is not generally true, and hence not valid.
Option II: \( \frac{x_1 + x_2}{2} = \frac{y_1 + y_2}{2} \) 
This suggests that the average kinetic energy of the system remains constant after collisions, which is a valid assumption in an isolated system where the total energy is conserved. Hence, this relationship is correct.
Option III: \( x_2 = y_2 \) 
This would mean that the kinetic energy of the second molecule does not change after collisions, which is also unlikely because energy is typically redistributed between molecules during collisions. So, this relationship is not correct.
Option IV: \( (x_1 - x_2) = (y_1 - y_2) \) 
This would imply that the difference in the kinetic energies between the two molecules remains constant after collisions, which is not a general rule. The energy redistribution typically does not maintain this difference. Hence, this relationship is not valid.
Step 2: Conclusion
The only correct relationship is Option II, where the average kinetic energy of the system remains the same before and after collisions. Therefore, the correct answer is Option (4), \( II \) only.

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Approach Solution -2

To solve the problem, we need to identify the correct relationship between the kinetic energies of two gas molecules at temperature \( T \), following the principles of kinetic theory of gases.

1. Understanding the Problem:
We are given two gas molecules with initial kinetic energies \( x_1 \) and \( x_2 \) respectively at temperature \( T \). After a few collisions, their kinetic energies become \( y_1 \) and \( y_2 \) respectively. We need to determine the correct relationships between these kinetic energies based on the given options.

2. Analyzing the Statements:
Let's evaluate each statement based on the principles of kinetic theory:

  • Statement I: \( x_1 = y_1 \) - This is not necessarily true. Kinetic energy changes based on collisions, so the initial and final kinetic energies may not be the same.
  • Statement II: \( \frac{x_1 + x_2}{2} = \frac{y_1 + y_2}{2} \) - This is true. The average kinetic energy per molecule is related to temperature and remains the same before and after collisions, provided no temperature change occurs in the system.
  • Statement III: \( x_2 = y_2 \) - This is also not necessarily true. Just like \( x_1 \), \( x_2 \) can change depending on the collisions.
  • Statement IV: \( (x_1 - x_2) = (y_1 - y_2) \) - This is not true. There is no reason for the difference in the kinetic energies of the two molecules to remain constant after the collisions.

3. Conclusion:
Only Statement II is correct, as it correctly states that the average kinetic energy remains unchanged, which aligns with the principle of constant temperature in an isolated system.

Final Answer:
The correct answer is (D) II only.

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