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The kinetic interpretation of temperature is a fundamental concept in statistical mechanics. Consider two different gases, Gas A and Gas B, at the same temperature. Gas A consists of molecules with twice the mass of those in Gas B. Compare the average kinetic energies of molecules in Gas A and Gas B.

Option: 1

1


Option: 2

70


Option: 3

96


Option: 4

88


Answers (1)

best_answer

In the kinetic theory of gases, the average kinetic energy of a gas molecule is given by:

\text { Average Kinetic Energy }=\frac{3}{2} k_B T

Where: - k_B is the Boltzmann constant (1.38 × 10^{-23} J/K) - T is the temperature in Kelvin

Since both gases are at the same temperature (T_A = T_B), we can compare the average kinetic energies of Gas A and Gas B by considering the masses of their molecules.

Given that the molecules in Gas A have twice the mass of those in Gas B (m_A = 2m_B), we can calculate the ratio of their average kinetic energies:

\frac{\text { Average Kinetic Energy }_{\mathrm{A}}}{\text { Average Kinetic Energy }_{\mathrm{B}}}=\frac{\frac{3}{2} k_B T}{\frac{3}{2} k_B T}=1

This implies that the ratio of the average kinetic energies is 1, which means that the molecules in Gas A and Gas B have the same average kinetic energy despite the difference in their masses.

Therefore, the average kinetic energies of molecules in Gas A and Gas B are equal.

This result highlights the principle that temperature is a measure of the average kinetic energy of particles in a substance, regardless of their masses.
Therefore, the correct option is 1.

Posted by

Deependra Verma

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