Question

In the van der Waals equation of state, the volume correction factor (b) arises from the repulsive interactions between the gas moleculesConsidering the molecules to behave as hard spheres of radius r, b = Y $\times$ V$_{\text{molecule}}$ $\times$ N$_A$The value of Y is _ _ _.

• A. 8
• B. 2
• C. 16
• D. 4

Hint:

In van der Waals equation, excluded volume per mole is four times the actual molecular volume

Answer 1

The Correct Option is D

Step 1: Understand van der Waals constant b.
The constant $b$ represents excluded volume due to finite size of moleculesIt accounts for repulsive interactions when molecules cannot occupy the same space

Step 2: Consider molecules as hard spheres.
Each molecule is treated as a sphere of radius $r$Thus, actual molecular volume is:
\[ V_{\text{molecule}} = \frac{4}{3}\pi r^3 \]

Step 3: Concept of excluded volume.
When two molecules approach, the center of one cannot come closer than a distance $2r$ from the otherThus, excluded volume corresponds to a sphere of radius $2r$

Step 4: Calculate excluded volume per pair.
\[ V_{\text{excluded}} = \frac{4}{3}\pi (2r)^3 = 8 \times \frac{4}{3}\pi r^3 = 8V_{\text{molecule}} \]

Step 5: Correct for double counting.
Since this excluded volume is counted for a pair of molecules, for one molecule it becomes half
\[ \text{Excluded volume per molecule} = \frac{8V_{\text{molecule}}}{2} = 4V_{\text{molecule}} \]

Step 6: Total excluded volume for 1 mole.
\[ b = 4 \times V_{\text{molecule}} \times N_A \]

Step 7: Conclusion.
\[ \boxed{Y = 4} \]