Question

The unit of the van der Waals gas equation parameter 'a' in \(\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\) is:

• A. kg m s\(^{-2}\)
• B. atm dm\(^6\) mol\(^{-2}\)
• C. dm\(^3\) mol\(^{-1}\)
• D. kg m s\(^{-1}\)

Hint:

To find the units of constants in physical equations (like 'a' and 'b' in the van der Waals equation), always use the principle of dimensional homogeneity. The pressure correction term \((an^2/V^2)\) must have units of pressure, and the volume correction term \((nb)\) must have units of volume.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the units of the van der Waals constant 'a' based on the van der Waals equation.
Step 2: Key Formula or Approach:
The principle of dimensional homogeneity states that quantities can be added or subtracted only if they have the same physical dimensions. In the van der Waals equation, the term \(\frac{an^2}{V^2}\) is added to the pressure \(P\). Therefore, the units of \(\frac{an^2}{V^2}\) must be the same as the units of pressure \(P\).
\[ \text{Unit of } P = \text{Unit of } \left(\frac{an^2}{V^2}\right) \] Step 3: Detailed Explanation:
Let's rearrange the equation to solve for the units of 'a'.
\[ \text{Unit of } a = \frac{(\text{Unit of } P) \times (\text{Unit of } V^2)}{(\text{Unit of } n^2)} \] We are given the units in one of the options as atm, dm\(^3\) (since dm\(^3\) = L, a common unit for volume), and mol. Let's use these units.

Unit of Pressure (P) = atm
Unit of Volume (V) = dm\(^3\)
Unit of moles (n) = mol
Now, substitute these into the equation for the units of 'a'.
\[ \text{Unit of } a = \frac{(\text{atm}) \times (\text{dm}^3)^2}{(\text{mol})^2} = \frac{\text{atm} \cdot \text{dm}^6}{\text{mol}^2} = \text{atm dm}^6 \text{mol}^{-2} \] Step 4: Final Answer:
The unit of the parameter 'a' is atm dm\(^6\) mol\(^{-2}\), which matches option (B).