What is the unit of 'a' in `(p+(n^2a)/V^2)(V-nb)` = nRT

Answer

To find the unit of 'a' in the equation \((p + \frac{n^2 a}{V^2})(V - nb) = nRT\), we can follow these steps:

### Step 1: Understand the equation
The equation represents a modified ideal gas law. Here, \(p\) is pressure, \(n\) is the number of moles, \(V\) is volume, \(R\) is the ideal gas constant, and \(T\) is temperature.

### Step 2: Identify the units of each term
1. **Pressure (p)**: The unit of pressure is typically atm (atmospheres) or Pa (Pascals). For our calculations, we will use atm.
2. **Volume (V)**: The unit of volume is typically liters (L).
3. **Number of moles (n)**: The unit of the number of moles is moles (mol).
4. **Gas constant (R)**: The unit of the ideal gas constant \(R\) is \(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\).
5. **Temperature (T)**: The unit of temperature is Kelvin (K).

### Step 3: Rearranging the equation
We can rearrange the equation to isolate the term involving 'a':
\[ p + \frac{n^2 a}{V^2} = \frac{nRT}{V - nb} \]
This implies:
\[ \frac{n^2 a}{V^2} = \frac{nRT}{V - nb} - p \]

### Step 4: Analyzing the left side
The left side of the equation is \(\frac{n^2 a}{V^2}\). The units of this term can be expressed as:
\[ \text{Units of } \frac{n^2 a}{V^2} = \frac{(\text{mol})^2 \cdot [a]}{(\text{L})^2} \]
This simplifies to:
\[ \frac{\text{mol}^2 \cdot [a]}{\text{L}^2} \]

### Step 5: Analyzing the right side
The right side of the equation, \(\frac{nRT}{V - nb} - p\), must also have units of pressure (atm). Therefore, we need to ensure that the left side equals the unit of pressure.

### Step 6: Setting the units equal
We know that the unit of pressure is atm. Thus, we can set the left side equal to atm:
\[ \frac{\text{mol}^2 \cdot [a]}{\text{L}^2} = \text{atm} \]

### Step 7: Solving for the unit of 'a'
Rearranging gives us:
\[ [a] = \frac{\text{atm} \cdot \text{L}^2}{\text{mol}^2} \]

### Conclusion
Thus, the unit of 'a' is:
\[ \text{atm} \cdot \text{L}^2 \cdot \text{mol}^{-2} \]