The average kinetic energy of a molecule of a perfect gas at temperature T is given by:
Show 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 \( \frac{3}{2} k_B T \).
Concept:
According to the kinetic theory of gases, the pressure of an ideal gas arises due to collisions of molecules with the container walls. The microscopic motion of gas molecules is related to macroscopic variables such as pressure (P), volume (V), and temperature (T).
The fundamental relation connecting pressure and molecular motion is:
\[
PV = \frac{1}{3} M v_{\text{rms}}^2
\]
where \(M\) is the total mass of gas and \(v_{\text{rms}}\) is the root mean square speed.
Step 1: Expressing kinetic energy in terms of pressure and volume
We rewrite:
\[
PV = \frac{2}{3} \left( \frac{1}{2} M v_{\text{rms}}^2 \right)
\]
The term:
\[
\frac{1}{2} M v_{\text{rms}}^2
\]
represents the total translational kinetic energy of all molecules in the gas.
Thus:
\[
PV = \frac{2}{3} E_{\text{total}}
\Rightarrow E_{\text{total}} = \frac{3}{2} PV
\]
Step 2: Using ideal gas law for one mole
For 1 mole of gas:
\[
PV = RT
\]
So:
\[
E_{\text{total}} = \frac{3}{2} RT
\]
Step 3: Average kinetic energy per molecule
If a mole contains \(N_A\) molecules:
\[
E_{\text{avg}} = \frac{E_{\text{total}}}{N_A}
\]
\[
E_{\text{avg}} = \frac{\frac{3}{2} RT}{N_A}
\]
Using:
\[
k_B = \frac{R}{N_A}
\]
We get:
\[
E_{\text{avg}} = \frac{3}{2} k_B T
\]
Thus, the average kinetic energy of a molecule is:
\[
\boxed{\frac{3}{2} k_B T}
\]
