The ratio of Van Der Waal's 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 `mo1^(-2)`

Explanation

To solve the problem of finding the dimension of the ratio of Van der Waals constants \( a \) and \( b \), we will follow these steps:

### Step 1: Understand the Van der Waals equation
The Van der Waals equation is given by:
\[ P + \frac{n^2 a}{V^2} = \frac{nRT}{V - nB} \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = absolute temperature
- \( a \) and \( b \) = Van der Waals constants

### Step 2: Find the dimension of constant \( a \)
From the equation, we can isolate \( a \):
\[ P = \frac{n^2 a}{V^2} \implies a = \frac{PV^2}{n^2} \]
Now, we need to determine the dimensions of \( a \):
- Dimension of pressure \( P \) is \( [P] = \text{ML}^{-1}\text{T}^{-2} \)
- Dimension of volume \( V \) is \( [V] = \text{L}^3 \)
- Dimension of number of moles \( n \) is \( [n] = \text{mol} \)

Substituting these into the equation for \( a \):
\[ [a] = \frac{[\text{ML}^{-1}\text{T}^{-2}][\text{L}^3]^2}{[\text{mol}]^2} = \frac{\text{ML}^{-1}\text{T}^{-2} \cdot \text{L}^6}{\text{mol}^2} = \frac{\text{ML}^{5}}{\text{mol}^2\text{T}^{2}} \]

### Step 3: Find the dimension of constant \( b \)
From the equation, we can isolate \( b \):
\[ V - nB = V \implies nB = V \implies B = \frac{V}{n} \]
Now, we need to determine the dimensions of \( b \):
\[ [b] = \frac{[\text{L}^3]}{[\text{mol}]} = \text{L}^3 \cdot \text{mol}^{-1} \]

### Step 4: Find the ratio \( \frac{a}{b} \)
Now we can find the ratio of \( a \) and \( b \):
\[ \frac{a}{b} = \frac{\frac{\text{ML}^{5}}{\text{mol}^2\text{T}^{2}}}{\text{L}^3 \cdot \text{mol}^{-1}} = \frac{\text{ML}^{5}}{\text{mol}^2\text{T}^{2}} \cdot \frac{\text{mol}}{\text{L}^3} = \frac{\text{ML}^{5}}{\text{L}^3\text{mol}\text{T}^{2}} = \frac{\text{ML}^{2}}{\text{mol}\text{T}^{2}} \]

### Step 5: Final dimension of \( \frac{a}{b} \)
The final dimension simplifies to:
\[ \frac{a}{b} = \text{L}^{2} \cdot \text{atm} \cdot \text{mol}^{-1} \]
This corresponds to the dimension of \( \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \).

### Conclusion
The dimension of the ratio \( \frac{a}{b} \) is:
\[ \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \]