The equation of state of some gases can be expressed as `(P+ (a)/(V^(2)))(V-b)= RT`, where P is the pressure, V is the volume, T is 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 find the dimensions of the constant 'a' in the equation of state given by:

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

we will follow these steps:

### Step 1: Understand the equation
The equation involves pressure (P), volume (V), temperature (T), and constants (a, b, R). We need to find the dimensions of 'a'.

### Step 2: Identify the dimensions of each variable
- Pressure (P) has dimensions of force per unit area:
\[ [P] = \frac{[F]}{[A]} = \frac{[M][L][T^{-2}]}{[L^2]} = [M][L^{-1}][T^{-2}] \]
- Volume (V) has dimensions:
\[ [V] = [L^3] \]
- Temperature (T) has dimensions:
\[ [T] = [\Theta] \quad (\text{where } \Theta \text{ is the dimension of temperature}) \]
- The constant R (the gas constant) has dimensions:
\[ [R] = \frac{[P][V]}{[T]} = \frac{[M][L^{-1}][T^{-2}][L^3]}{[\Theta]} = [M][L^2][T^{-2}][\Theta^{-1}] \]

### Step 3: Rearranging the equation
From the equation, we can isolate the term involving 'a':
\[ P + \frac{a}{V^2} = \frac{RT}{V - b} \]
Since we are interested in the term \(\frac{a}{V^2}\), we can express it as:
\[ \frac{a}{V^2} = P - \frac{RT}{V - b} \]

### Step 4: Equating dimensions
For the left-hand side to have the same dimensions as the right-hand side, we can equate:
\[ \frac{a}{V^2} \text{ must have dimensions of } [P] \]
Thus, we can write:
\[ \frac{[a]}{[V]^2} = [P] \]

### Step 5: Substitute dimensions
Substituting the dimensions we found:
\[ \frac{[a]}{[L^3]^2} = [M][L^{-1}][T^{-2}] \]
This simplifies to:
\[ \frac{[a]}{[L^6]} = [M][L^{-1}][T^{-2}] \]

### Step 6: Solve for dimensions of 'a'
Now, multiply both sides by \(L^6\):
\[ [a] = [M][L^{-1}][T^{-2}][L^6] = [M][L^{5}][T^{-2}] \]

### Final Answer
Thus, the dimensions of 'a' are:
\[ [a] = [M][L^{5}][T^{-2}] \]