The equation of state of some gases can be expressed as ,
`(P+(a)/(V^(2)))(V-b)=RY`
where P is the pressure, V the volume, T the absolute temperature and a, b, R are constants. The dimensions of 'a' are -

Options

A. `[ML^(5)T^(-2)]`
B. `[ML^(-1)T^(-2)]`
C. `[L^(3)]`
D. `[L^(6)]`

Explanation

To determine the dimensions of the constant 'a' in the given equation of state, we start with the equation:

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

### Step 1: Analyze the Equation
In this equation, we know that pressure \(P\) and the term \(\frac{a}{V^2}\) are being added together. For two quantities to be added, they must have the same dimensions.

### Step 2: Identify the Dimensions of Pressure \(P\)
Pressure \(P\) can be expressed in terms of force and area:

\[ P = \frac{\text{Force}}{\text{Area}} = \frac{F}{A} \]

Force \(F\) can be expressed as mass times acceleration:

\[ F = m \cdot a = m \cdot \frac{L}{T^2} \]

Thus, the dimensions of force are:

\[ [F] = M \cdot L \cdot T^{-2} \]

Area \(A\) is given by:

\[ A = L^2 \]

So, the dimensions of pressure \(P\) become:

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

### Step 3: Identify the Dimensions of Volume \(V\)
Volume \(V\) is given by:

\[ V = L^3 \]

Thus, the dimensions of \(V^2\) are:

\[ [V^2] = (L^3)^2 = L^6 \]

### Step 4: Set Up the Equation for Dimensions of \(a\)
Since \(P\) and \(\frac{a}{V^2}\) have the same dimensions, we can write:

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

Substituting the dimensions we found:

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

### Step 5: Solve for Dimensions of \(a\)
Rearranging the equation to solve for the dimensions of \(a\):

\[ [a] = [P] \cdot [V^2] = (M \cdot L^{-1} \cdot T^{-2}) \cdot (L^6) \]

This simplifies to:

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

### Conclusion
Thus, the dimensions of the constant \(a\) are:

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