For `H_2O`
`T_C = 647 K`,
`P_C = 218 atm`
`V_C = 0.057 L mol^(-1)`
What is the value of .a. for `H_2O`?

Options

A. `2.12 L^2 atm mol^(-2)`
B. `0.019 L^2 atm mol^(-2)`
C. `8.31 L^2 atm mol^(-2)`
D. `0.082 L^2 atm mol^(-2)`

Explanation

To find the value of the Van der Waals constant \( a \) for water (\( H_2O \)), we can use the relationships between critical temperature (\( T_C \)), critical pressure (\( P_C \)), and critical volume (\( V_C \)) with the Van der Waals constants. Here are the steps to solve the problem:

### Step 1: Understand the Relationships
The relationships we need are:
1. \( T_C = \frac{8a}{27bR} \)
2. \( P_C = \frac{a}{27b^2} \)
3. \( V_C = 3b \)

Where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol)).
- \( a \) is the Van der Waals constant related to the attraction between particles.
- \( b \) is the Van der Waals constant related to the volume occupied by the gas particles.

### Step 2: Express \( b \) in terms of \( V_C \)
From the equation \( V_C = 3b \), we can express \( b \) as:
\[ b = \frac{V_C}{3} \]

Given \( V_C = 0.057 \, \text{L/mol} \):
\[ b = \frac{0.057}{3} = 0.019 \, \text{L/mol} \]

### Step 3: Substitute \( b \) into the \( P_C \) equation
Now substitute \( b \) into the equation for \( P_C \):
\[ P_C = \frac{a}{27b^2} \]
Rearranging gives:
\[ a = 27 P_C b^2 \]

### Step 4: Substitute known values
We know:
- \( P_C = 218 \, \text{atm} \)
- \( b = 0.019 \, \text{L/mol} \)

Now substitute these values into the equation:
\[ a = 27 \times 218 \times (0.019)^2 \]

### Step 5: Calculate \( a \)
First, calculate \( (0.019)^2 \):
\[ (0.019)^2 = 0.000361 \]
Now substitute this value back into the equation:
\[ a = 27 \times 218 \times 0.000361 \]
Calculating this step by step:
1. \( 27 \times 218 = 5896 \)
2. \( 5896 \times 0.000361 \approx 2.125 \)

Thus, the value of \( a \) is approximately:
\[ a \approx 2.125 \, \text{atm} \cdot \text{L}^2/\text{mol}^2 \]

### Final Answer
The value of \( a \) for \( H_2O \) is approximately \( 2.125 \, \text{atm} \cdot \text{L}^2/\text{mol}^2 \).

---