Consider monoatomic ideal gas molecules of equal mass \( m \) in thermal equilibrium, at temperature \( T \). Which one of the following equations is correct? (the angular brackets denote average, \( k_B \) is Boltzmann constant, \( v \) is velocity, and \( v_x \) is the x-component of velocity)
Show Hint
- \( \frac{3}{2} mv^2 = \frac{3}{2} k_B T \)
- \( \frac{1}{2} mv^2 = \frac{3}{2} k_B T \)
- \( \frac{1}{2} mv^2 = \frac{3}{2} k_B T \)
- \( \frac{1}{2} mv_x^2 = \frac{3}{2} k_B T \)
The Correct Option is B
Solution and Explanation
Step 1: Understanding the relationship between kinetic energy and temperature.
In a monoatomic ideal gas, the average kinetic energy of a molecule is related to the temperature by the equation: \[ \langle \frac{1}{2} mv^2 \rangle = \frac{3}{2} k_B T. \]
Step 2: Analyzing the options.
(A) \( \frac{3}{2} mv^2 = \frac{3}{2} k_B T \): Incorrect — This is not correct, the factor should be \( \frac{1}{2} \).
(B) \( \frac{1}{2} mv^2 = \frac{3}{2} k_B T \): Correct — This is the correct equation for the total energy of the gas molecules in thermal equilibrium.
(C) \( \frac{1}{2} mv^2 = \frac{3}{2} k_B T \): Incorrect — This is a repetition of option B.
(D) \( \frac{1}{2} mv_x^2 = \frac{3}{2} k_B T \): Incorrect — This refers to the x-component of the velocity, which is incorrect for the total kinetic energy.
Step 3: Conclusion.
The correct answer is (B) \( \frac{1}{2} mv^2 = \frac{3}{2} k_B T \).
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