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 determine the correct unit for the van der Waals constant \( a \), we need to analyze its role and the units involved in the van der Waals equation of state for real gases. The van der Waals equation is an improvement over the ideal gas law, accounting for intermolecular forces and the finite size 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 accounting for intermolecular forces
• \( b \): van der Waals constant accounting for the finite size of molecules

1. Step 1: Understand the Units of Each Term
   First, let’s identify the units of each variable in the equation:
   • Pressure (\( P \)) has units of atmospheres (atm).
   • Molar volume (\( V_m \)) has units of cubic decimeters per mole (\( \text{dm}^3 \, \text{mol}^{-1} \)).
   • Temperature (\( T \)) has units of Kelvin (K).
   • Gas constant (\( R \)) has units of \( \text{dm}^3 \, \text{atm} \, \text{mol}^{-1} \, \text{K}^{-1} \).
2. Step 2: Analyze the Term Involving \( a \)
   The term \( \frac{a}{V_m^2} \) is added to the pressure \( P \). For dimensional consistency, the units of \( \frac{a}{V_m^2} \) must be the same as those of \( P \), which is atm.
   \[ \frac{a}{V_m^2} \, \text{has units of} \, \text{atm.} \]
3. Step 3: Express \( a \) in Terms of Units
   Given that: \[ \frac{a}{V_m^2} \, \text{has units of} \, \text{atm,} \] we can express the units of \( a \) as: \[ \text{Units of} \, a = \text{atm} \times (\text{dm}^3 \, \text{mol}^{-1})^2 = \text{atm} \times \text{dm}^6 \, \text{mol}^{-2}. \] \[ \text{Units of} \, a = \text{dm}^6 \, \text{atm} \, \text{mol}^{-2}. \]
4. Step 4: Confirm Dimensional Consistency in the van der Waals Equation
   To ensure our derived units for \( a \) are correct, let’s check the dimensional consistency of the entire equation: \[ \left( P + \frac{a}{V_m^2} \right)(V_m - b) = RT \]
   • \( P \) has units of atm.
   • \( \frac{a}{V_m^2} \) has units of atm.
   • \( V_m \) and \( b \) both have units of \( \text{dm}^3 \, \text{mol}^{-1} \).
   • \( RT \) has units of \( \text{dm}^3 \, \text{atm} \, \text{mol}^{-1} \) (since \( R \) is in \( \text{dm}^3 \, \text{atm} \, \text{mol}^{-1} \, \text{K}^{-1} \) and \( T \) is in K).

Multiplying the terms on the left side: \[ \text{atm} \times \text{dm}^3 \, \text{mol}^{-1} = \text{dm}^3 \, \text{atm} \, \text{mol}^{-1}. \] Which matches the units on the right side (\( RT \)), confirming dimensional consistency.
Conclusion:
The van der Waals constant \( a \) must have units of \( \text{dm}^6 \, \text{atm} \, \text{mol}^{-2} \) to ensure that the van der Waals equation remains dimensionally consistent. Therefore, the correct unit for \( a \) is: \[ \boxed{\text{dm}^6 \, \text{atm} \, \text{mol}^{-2}} \]