In the Van der Waals equation `(P + (a)/(V^(2)))(V-b)` = constant, the unit of a is

Options

A. dyne/`cm^(5)`
B. dyne/`cm^(4)`
C. dyne/`cm^(3)`
D. dyne/`cm^(2)`

Answer

To determine the unit of \( a \) in the Van der Waals equation, we start with the equation itself:

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

### Step 1: Identify the units of pressure \( P \)
Pressure \( P \) is defined as force per unit area. The unit of force is dyne and the unit of area is square centimeters. Therefore, the unit of pressure \( P \) is:

\[ P = \frac{\text{Force}}{\text{Area}} = \frac{\text{dyne}}{\text{cm}^2} = \text{dyne/cm}^2 \]

**Hint:** Remember that pressure is force divided by area.

### Step 2: Analyze the term \( \frac{a}{V^2} \)
In the equation, \( \frac{a}{V^2} \) must have the same unit as pressure \( P \) for the equation to be dimensionally consistent. Therefore, we can write:

\[ \frac{a}{V^2} \text{ has the same unit as } P \]

### Step 3: Determine the unit of volume \( V \)
The unit of volume \( V \) is cubic centimeters, or:

\[ V = \text{cm}^3 \]

### Step 4: Calculate the unit of \( V^2 \)
Since \( V \) is in cubic centimeters, \( V^2 \) will have the unit:

\[ V^2 = (\text{cm}^3)^2 = \text{cm}^6 \]

### Step 5: Set up the equation for \( a \)
Now, substituting the unit of \( V^2 \) into the equation for \( \frac{a}{V^2} \):

\[ \frac{a}{\text{cm}^6} = \text{dyne/cm}^2 \]

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

\[ a = \text{dyne/cm}^2 \times \text{cm}^6 \]

This simplifies to:

\[ a = \text{dyne} \times \text{cm}^{4} \]

### Final Answer
Thus, the unit of \( a \) is:

\[ \text{dyne} \cdot \text{cm}^4 \]

### Summary
The unit of \( a \) in the Van der Waals equation is \( \text{dyne} \cdot \text{cm}^4 \).

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