According to Vander Wall's equation pressure`(P)`, volume `(V)` and temperature`(T)` are related as `(P+(a)/(V^(2)))(V-b)=RT` [for `1` mole of gas] Then dimension of `(ab)/(V^(2))` is equivalent to `:-`

Options

A. force
B. energy
C. power
D. Pressure

Explanation

To find the dimension of \(\frac{ab}{V^2}\) from the Van der Waals equation, we will follow these steps:

### Step 1: Identify the dimensions of \(V\)
The volume \(V\) has the dimension of length cubed. Therefore,
\[ [V] = L^3 \]

### Step 2: Determine the dimensions of \(b\)
Since \(b\) is subtracted from \(V\) in the equation, \(b\) must have the same dimensions as \(V\). Thus,
\[ [b] = L^3 \]

### Step 3: Determine the dimensions of \(a\)
From the Van der Waals equation, \(P + \frac{a}{V^2}\) implies that \(\frac{a}{V^2}\) must have the same dimensions as pressure \(P\).

The dimension of pressure \(P\) is given by:
\[ [P] = \frac{Force}{Area} = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

Now, we need to find the dimension of \(a\):
\[ \frac{a}{V^2} \Rightarrow [a] = [P] \cdot [V^2] \]
Since \(V^2\) has dimensions:
\[ [V^2] = (L^3)^2 = L^6 \]
Thus,
\[ [a] = [P] \cdot [V^2] = (M L^{-1} T^{-2}) \cdot (L^6) = M L^{5} T^{-2} \]

### Step 4: Calculate the dimensions of \(\frac{ab}{V^2}\)
Now we can find the dimensions of \(\frac{ab}{V^2}\):
\[ [ab] = [a] \cdot [b] = (M L^{5} T^{-2}) \cdot (L^{3}) = M L^{8} T^{-2} \]
Now divide by \(V^2\):
\[ \frac{ab}{V^2} = \frac{M L^{8} T^{-2}}{L^{6}} = M L^{2} T^{-2} \]

### Step 5: Final Result
The dimension of \(\frac{ab}{V^2}\) is:
\[ [M L^{2} T^{-2}] \]

This is equivalent to the dimension of energy (Joules).