Question

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)

• A. \( \frac{3}{2} mv^2 = \frac{3}{2} k_B T \)
• B. \( \frac{1}{2} mv^2 = \frac{3}{2} k_B T \)
• C. \( \frac{1}{2} mv^2 = \frac{3}{2} k_B T \)
• D. \( \frac{1}{2} mv_x^2 = \frac{3}{2} k_B T \)

Hint:

The total kinetic energy of an ideal gas is directly related to the temperature. The equation \( \langle \frac{1}{2} mv^2 \rangle = \frac{3}{2} k_B T \) gives this relationship.

Answer 1

The Correct Option is B

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) \( \frac{1}{2} mv^2 = \frac{3}{2} k_B T \).