The equation `(P+a/(V^(2)))(V-b)` constant. The units of `a` are

Options

A. `"Dyne"xxcm^(5)`
B. `"Dyne"xxcm^(4)`
C. `"Dyne"//cm^(3)`
D. `"Dyne"//cm^(2)`

Answer

To find the units of \( a \) in the equation \( (P + \frac{a}{V^2})(V - b) = \text{constant} \), we need to ensure that the terms being added have the same units. Let's break down the steps:

### Step-by-Step Solution:

1. **Identify the Terms**:
The equation consists of pressure \( P \) and the term \( \frac{a}{V^2} \). For these two terms to be added together, they must have the same units.

2. **Determine the Unit of Pressure \( P \)**:
Pressure is defined as force per unit area. In the CGS (centimeter-gram-second) system, the unit of force is the dyne and the unit of area is square centimeters.
\[ \text{Unit of Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{\text{dyne}}{\text{cm}^2} \]

3. **Express the Units of \( \frac{a}{V^2} \)**:
Since \( P \) and \( \frac{a}{V^2} \) must have the same units, we can write:
\[ \frac{a}{V^2} \text{ must also have units of } \frac{\text{dyne}}{\text{cm}^2} \]

4. **Determine the Unit of Volume \( V \)**:
The volume \( V \) is measured in cubic centimeters (cm³). Therefore, \( V^2 \) will have units of:
\[ V^2 = (\text{cm}^3)^2 = \text{cm}^6 \]

5. **Set Up the Equation for Units**:
Now, we can express the unit of \( a \):
\[ \frac{a}{\text{cm}^6} = \frac{\text{dyne}}{\text{cm}^2} \]

6. **Solve for the Units of \( a \)**:
Rearranging the equation gives us:
\[ a = \frac{\text{dyne}}{\text{cm}^2} \times \text{cm}^6 = \text{dyne} \cdot \text{cm}^4 \]

7. **Final Result**:
Thus, the units of \( a \) are:
\[ \text{Unit of } a = \text{dyne} \cdot \text{cm}^4 \]

### Conclusion:
The units of \( a \) are \( \text{dyne} \cdot \text{cm}^4 \).