The equation of state of a real gas is given by `(P+a/V^(2)) (V-b)=RT`
where P, V and T are pressure, volume and temperature resperature and R is the universal gas constant. The dimensions of the constant a in the above equation is

Explanation

To find the dimensions of the constant \( a \) in the equation of state for a real gas given by

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

we will follow these steps:

### Step 1: Identify the terms in the equation
The equation consists of pressure \( P \), volume \( V \), temperature \( T \), and the universal gas constant \( R \). The term \( a \) is added to pressure, which implies that \( \frac{a}{V^2} \) must also have the dimensions of pressure.

### Step 2: Understand the dimensions of pressure
Pressure \( P \) is defined as force per unit area. The dimensions of force are given by:

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

The area has dimensions of \( L^2 \). Therefore, the dimensions of pressure \( P \) are:

\[ [P] = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}. \]

### Step 3: Set up the equation for \( a \)
Since \( \frac{a}{V^2} \) must have the same dimensions as pressure, we can write:

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

### Step 4: Determine the dimensions of volume
The dimensions of volume \( V \) are:

\[ [V] = L^3. \]

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

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

### Step 5: Substitute the dimensions into the equation
Now substituting the dimensions into the equation, we have:

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

### Step 6: Solve for the dimensions of \( a \)
Multiplying both sides by \( L^6 \) gives:

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

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

\[ [a] = ML^{5}T^{-2}. \]