In vanderwaal equation at const temperature 300K, `a=1.4 atm "litre"^(2)"mole"^(2),v=100 ml, n=1` mole, what is pressure of gas:

Options

A. 42 atm
B. 210 atm
C. 500 atm
D. 106 atm

Explanation

To solve the problem using the Van der Waals equation, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Van der Waals Equation**:
The Van der Waals equation for one mole of gas is given by:
\[ \left( P + \frac{a n^2}{V^2} \right)(V - nb) = nRT \]
For this problem, we will assume \( b \) is negligible since it is not provided.

2. **Identify Given Values**:
- Temperature (T) = 300 K
- \( a \) = 1.4 atm L²/mol²
- Volume (V) = 100 mL = 0.1 L (convert mL to L)
- Number of moles (n) = 1 mole
- Ideal gas constant (R) = 0.082 L·atm/(K·mol)

3. **Substitute Values into the Equation**:
Since \( b \) is not provided, we can simplify the equation to:
\[ P + \frac{a n^2}{V^2} = \frac{nRT}{V} \]
Substituting the known values:
\[ P + \frac{1.4 \times 1^2}{(0.1)^2} = \frac{1 \times 0.082 \times 300}{0.1} \]

4. **Calculate the Left Side**:
- Calculate \( \frac{1.4 \times 1^2}{(0.1)^2} \):
\[ \frac{1.4}{0.01} = 140 \]
- So, the equation becomes:
\[ P + 140 = \frac{24.6}{0.1} \]

5. **Calculate the Right Side**:
- Calculate \( \frac{1 \times 0.082 \times 300}{0.1} \):
\[ \frac{24.6}{0.1} = 246 \]
- Now, the equation is:
\[ P + 140 = 246 \]

6. **Solve for Pressure (P)**:
- Rearranging gives:
\[ P = 246 - 140 = 106 \text{ atm} \]

### Final Answer:
The pressure of the gas is **106 atm**.