Question

The unit of van der Waals constant ‘a’ is

• A. dm³ mol^-1
• B. dm⁶ atm mol^-2
• C. dm⁶ mol^-1
• D. dm² atm mol^-1

Hint:

When working with the van der Waals equation, ensure dimensional consistency by analyzing the units of each term. The constant a accounts for intermolecular attractions and has units derived from the pressure and volume terms to maintain the equation’s balance.

Answer 1

The Correct Option is B

To find the correct units for the van der Waals constant \( a \), we examine its role in the van der Waals equation of state for real gases. This equation refines the ideal gas law by considering intermolecular forces and the volume of gas molecules: \[ \left( P + \frac{a}{V_m^2} \right)(V_m - b) = RT \] Where:

• \( P \): Pressure
• \( V_m \): Molar volume
• \( T \): Temperature
• \( R \): Universal gas constant
• \( a \): van der Waals constant for intermolecular forces
• \( b \): van der Waals constant for molecular volume

1. Step 1: Identify Variable Units
   First, we list the units for each variable:
   • Pressure (\( P \)): atmospheres (atm).
   • Molar volume (\( V_m \)): cubic decimeters per mole (\( \text{dm}^3 \, \text{mol}^{-1} \)).
   • Temperature (\( T \)): Kelvin (K).
   • Gas constant (\( R \)): \( \text{dm}^3 \, \text{atm} \, \text{mol}^{-1} \, \text{K}^{-1} \).
2. Step 2: Analyze the \( a \) Term
   The term \( \frac{a}{V_m^2} \) is added to pressure \( P \), so it must have the same units as \( P \), which is atm.
   \[ \frac{a}{V_m^2} \, \text{units} = \text{atm.} \]
3. Step 3: Determine Units of \( a \)
   Since \[ \frac{a}{V_m^2} \, \text{units} = \text{atm,} \] the units of \( a \) are: \[ \text{Units of} \, a = \text{atm} \times (\text{dm}^3 \, \text{mol}^{-1})^2 = \text{atm} \times \text{dm}^6 \, \text{mol}^{-2}. \] \[ \text{Units of} \, a = \text{dm}^6 \, \text{atm} \, \text{mol}^{-2}. \]
4. Step 4: Verify Dimensional Consistency
   To confirm the units for \( a \), we check the dimensional consistency of the van der Waals equation: \[ \left( P + \frac{a}{V_m^2} \right)(V_m - b) = RT \]
   • \( P \) units: atm.
   • \( \frac{a}{V_m^2} \) units: atm.
   • \( V_m \) and \( b \) units: \( \text{dm}^3 \, \text{mol}^{-1} \).
   • \( RT \) units: \( \text{dm}^3 \, \text{atm} \, \text{mol}^{-1} \) (as \( R \) is in \( \text{dm}^3 \, \text{atm} \, \text{mol}^{-1} \, \text{K}^{-1} \) and \( T \) is in K).

Checking the left side: \[ \text{atm} \times \text{dm}^3 \, \text{mol}^{-1} = \text{dm}^3 \, \text{atm} \, \text{mol}^{-1}. \] This matches the units on the right side (\( RT \)), confirming consistency.
Conclusion:
The van der Waals constant \( a \) has units of \( \text{dm}^6 \, \text{atm} \, \text{mol}^{-2} \) to keep the van der Waals equation dimensionally consistent. Therefore, the unit for \( a \) is: \[ \boxed{\text{dm}^6 \, \text{atm} \, \text{mol}^{-2}} \]