The ratio of van der Waals constants a and b, `((a)/(b))` has the dimension of :

Options

A. atm `L^(-1)`
B. L atm `m o1^(-1)`
C. `L m o1^(-2)`
D. atm L m `o1^(-2)`

Explanation

To solve the problem of determining the dimension of the ratio of van der Waals constants \( \frac{a}{b} \), we will follow these steps:

### Step 1: Understand the van der Waals equation
The van der Waals equation is given by:

\[ \left(P + \frac{aN^2}{V^2}\right)(V - Nb) = NRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( N \) = Number of moles
- \( R \) = Universal gas constant
- \( T \) = Temperature
- \( a \) = Constant representing intermolecular forces
- \( b \) = Constant representing excluded volume

### Step 2: Identify the dimensions of constants \( a \) and \( b \)
1. **Finding the dimension of \( a \)**:
- The term \( \frac{aN^2}{V^2} \) has the same dimension as pressure \( P \).
- The dimension of pressure \( P \) is \( [P] = \frac{ML^{-1}T^{-2}} \).
- The dimension of \( N^2 \) (number of moles squared) is \( [N^2] = [n]^2 = (mol)^2 \).
- The dimension of \( V^2 \) (volume squared) is \( [V^2] = [L^3]^2 = L^6 \).
- Therefore, we can write:

\[ \frac{a \cdot (mol)^2}{(L^3)^2} = \frac{ML^{-1}T^{-2}}{1} \]

- Rearranging gives:

\[ a = \frac{ML^{-1}T^{-2} \cdot L^6}{(mol)^2} = \frac{ML^5T^{-2}}{(mol)^2} \]

2. **Finding the dimension of \( b \)**:
- The dimension of \( b \) is derived from the excluded volume per mole.
- The dimension of \( b \) is:

\[ b = \frac{V}{N} = \frac{L^3}{mol} = \frac{L^3}{(mol)} \]

### Step 3: Calculate the ratio \( \frac{a}{b} \)
Now we can find the ratio \( \frac{a}{b} \):

\[ \frac{a}{b} = \frac{\frac{ML^5T^{-2}}{(mol)^2}}{\frac{L^3}{(mol)}} = \frac{ML^5T^{-2}}{(mol)^2} \cdot \frac{(mol)}{L^3} \]

This simplifies to:

\[ \frac{a}{b} = \frac{ML^5T^{-2}}{(mol)} \cdot \frac{1}{L^3} = \frac{ML^2T^{-2}}{(mol)} \]

### Step 4: Final dimensions
Thus, the dimension of the ratio \( \frac{a}{b} \) is:

\[ \frac{ML^2T^{-2}}{(mol)} \]