Understanding Van der Waals Constants and Gas Liquefaction

The van der Waals equation corrects the ideal gas law for real gas behavior:

$(P+\frac{a}{V^2})(V-b)=RT$

Where:

• a accounts for intermolecular attraction (higher a = stronger attraction)
• b accounts for molecular volume (higher b = larger molecular size)

Step 1: Relate Liquefaction to Intermolecular Forces

Gases liquefy more easily when intermolecular forces are stronger. The constant a directly measures these attractive forces.

Step 2: Compare Chlorine (Cl₂) and Ethane (C₂H₆)

Chlorine has larger molecules with greater polarizability, leading to stronger London dispersion forces than ethane. Thus:

$a_{\mathrm{Cl}2}>a_{C2H6}$

Step 3: Consider Molecular Size

Chlorine molecules (Cl₂) are larger than ethane molecules (C₂H₆), so:

$b_{\mathrm{Cl}2}>b_{C2H6}$

Step 4: Evaluate the Options

Both constants are larger for chlorine than for ethane. Therefore, the correct option is:

a and b for Cl₂ > a and b for C₂H₆

Final Answer

Chlorine is more easily liquefied because both van der Waals constants a and b are greater for Cl₂ than for C₂H₆.

Related Topics

• Intermolecular Forces
• Critical Temperature and Pressure
• Real Gas Behavior
• Gas Liquefaction Processes

Key Formulae

Van der Waals Equation: $(P+\frac{a}{V^2})(V-b)=RT$

Critical Temperature: $T_c=\frac{8a}{27Rb}$

Theory: Higher critical temperature (T_c) indicates easier liquefaction. Since T_c ∝ a/b, both larger a and larger b contribute to easier liquefaction, though a has a stronger effect.