In equation `(P+(a)/(V^(2)))(V-b)=RT`, the dimensional formula of a is

Options

A. `[ML^(3)T^(-2)]`
B. `[ML^(-5)T^(-2)]`
C. `[ML^(5)T^(-2)]`
D. `[ML^(2)T^(-2)]`

Explanation

To find the dimensional formula of \( a \) in the equation

\[ (P + \frac{a}{V^2})(V - b) = RT \]

we can follow these steps:

### Step 1: Understand the equation
The equation represents a modified form of the Van der Waals equation for real gases. Here, \( P \) is pressure, \( a \) is a constant related to the attractive forces between molecules, \( V \) is volume, \( b \) is the volume occupied by the gas molecules, \( R \) is the universal gas constant, and \( T \) is temperature.

### Step 2: Identify the dimensions of pressure \( P \)
Pressure \( P \) is defined as force per unit area. The dimensional formula for force is:

\[ \text{Force} = \text{mass} \times \text{acceleration} = M L T^{-2} \]

Area has the dimensional formula:

\[ \text{Area} = L^2 \]

Thus, the dimensional formula for pressure \( P \) is:

\[ [P] = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

### Step 3: Identify the dimensions of volume \( V \)
Volume \( V \) has the dimensional formula:

\[ [V] = L^3 \]

### Step 4: Write the expression for \( a \)
From the equation, we can isolate \( a \):

\[ P + \frac{a}{V^2} = \frac{RT}{V - b} \]

Since \( P \) and \( \frac{a}{V^2} \) must have the same dimensions, we can set:

\[ [P] = \left[\frac{a}{V^2}\right] \]

### Step 5: Substitute the dimensions
Substituting the dimensions we have:

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

This simplifies to:

\[ M L^{-1} T^{-2} = \frac{[a]}{L^6} \]

### Step 6: Solve for the dimensions of \( a \)
Rearranging gives:

\[ [a] = M L^{-1} T^{-2} \cdot L^6 = M L^{5} T^{-2} \]

### Conclusion
Thus, the dimensional formula of \( a \) is:

\[ [a] = M L^{5} T^{-2} \]