The equation of state for real gas is given by `((p + (a)/(V^(2))(V - b) = RT`. The dimension of the constant `a` is ………………. .

Options

A. `ML^5T^-2`
B. `ML^(-1)T^(-2)`
C. `M^0L^3T^0`
D. `M^0L^6T^0`

Explanation

To determine the dimension 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 can follow these steps:

### Step 1: Understand the Equation
The equation consists of pressure \( p \), the constant \( a \), volume \( V \), the gas constant \( R \), and temperature \( T \). The left-hand side of the equation involves pressure and a term that includes \( a \).

### Step 2: Identify Dimensions
To find the dimension of \( a \), we first need to recognize that the terms being added on the left side of the equation must have the same dimensions. Therefore, we can equate the dimension of pressure \( p \) with the dimension of \( \frac{a}{V^2} \).

### Step 3: Write the Dimension of Pressure
The dimension of pressure \( p \) is given by the formula:

\[ p = \frac{F}{A}, \]

where \( F \) is force and \( A \) is area. The dimension of force \( F \) is \( [M L T^{-2}] \) and the dimension of area \( A \) is \( [L^2] \). Thus, the dimension of pressure is:

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

### Step 4: Write the Dimension of Volume
The dimension of volume \( V \) is:

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

### Step 5: Substitute Dimensions into the Equation
Now, substituting the dimensions into the equation:

\[ \text{Dimension of } \frac{a}{V^2} = [M L^{-1} T^{-2}]. \]

Since \( V^2 = (L^3)^2 = L^6 \), we have:

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

### Step 6: Solve for the Dimension of \( a \)
To find the dimension of \( a \), we can rearrange the equation:

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

### Conclusion
Thus, the dimension of the constant \( a \) is:

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