Question

The average kinetic energy of a molecule of a perfect gas at temperature T is given by:

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

Hint:

By equipartition theorem, each translational degree of freedom contributes \( \frac{1}{2} k_B T \). A monoatomic gas has 3 degrees of freedom, hence total energy is \( \frac{3}{2} k_B T \).

Answer 1

The Correct Option is B

Step 1: Understand what is asked.
We want the average kinetic energy of a single molecule of an ideal gas at temperature $T$. This is a basic result of the kinetic theory of gases.

Step 2: Start from the gas pressure relation.
Kinetic theory gives the pressure of a gas in terms of molecular motion: \[ PV = \frac{1}{3} N m v_{rms}^2 \] Here $N$ is the number of molecules, $m$ is the mass of one molecule and $v_{rms}$ is the root mean square speed.

Step 3: Bring in temperature.
For an ideal gas, the pressure relation can also be written as $PV = N k_B T$ for $N$ molecules. Comparing the two expressions links speed with temperature.

Step 4: Match the two forms.
Setting the pressure relations equal gives: \[ \frac{1}{3} N m v_{rms}^2 = N k_B T \] So $\frac{1}{3} m v_{rms}^2 = k_B T$.

Step 5: Form the kinetic energy.
The average kinetic energy of one molecule is $\frac{1}{2} m v_{rms}^2$. Multiply the previous result by $\frac{3}{2}$: \[ \frac{1}{2} m v_{rms}^2 = \frac{3}{2} k_B T \]

Step 6: State the answer.
So the average kinetic energy of a molecule depends only on temperature: \[ \boxed{\dfrac{3}{2} k_B T} \]