Pressure exerted by 1 mole of methane in a 0.25 litre container at 300K using Vander Waal’s equation : (Given : `a = 2.253 "atm" l^(2) "mol"^(-2)` and `b = 0.0428 l "mol"^(-1)`) is

Options

A. 82.82 atm
B. 152.51 atm
C. 190.52 atm
D. 70.52 atm

Answer

To solve the problem of finding the pressure exerted by 1 mole of methane in a 0.25 litre container at 300K using Van der Waals equation, we will follow these steps:

### Step 1: Write down the Van der Waals equation
The Van der Waals equation is given by:
\[ \left( P + \frac{a n^2}{V^2} \right) (V - nb) = nRT \]
Where:
- \( P \) = pressure
- \( a \) = Van der Waals constant for attraction
- \( b \) = Van der Waals constant for volume
- \( n \) = number of moles
- \( V \) = volume
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Substitute the known values
Given:
- \( n = 1 \) mole
- \( V = 0.25 \) L
- \( T = 300 \) K
- \( a = 2.253 \, \text{atm} \, \text{L}^2 \, \text{mol}^{-2} \)
- \( b = 0.0428 \, \text{L} \, \text{mol}^{-1} \)
- \( R = 0.0821 \, \text{L} \, \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1} \)

Substituting these values into the equation:
\[ \left( P + \frac{2.253 \cdot 1^2}{(0.25)^2} \right) \left( 0.25 - 1 \cdot 0.0428 \right) = 1 \cdot 0.0821 \cdot 300 \]

### Step 3: Calculate the terms
First, calculate \( \frac{a n^2}{V^2} \):
\[ \frac{2.253}{(0.25)^2} = \frac{2.253}{0.0625} = 36.048 \]

Next, calculate \( V - nb \):
\[ 0.25 - 0.0428 = 0.2072 \]

Now calculate \( nRT \):
\[ 1 \cdot 0.0821 \cdot 300 = 24.63 \]

### Step 4: Substitute back into the equation
Now we substitute these calculated values back into the equation:
\[ \left( P + 36.048 \right) (0.2072) = 24.63 \]

### Step 5: Expand and solve for \( P \)
Expanding the equation:
\[ P \cdot 0.2072 + 36.048 \cdot 0.2072 = 24.63 \]
Calculating \( 36.048 \cdot 0.2072 \):
\[ 36.048 \cdot 0.2072 = 7.471 \]
So the equation becomes:
\[ P \cdot 0.2072 + 7.471 = 24.63 \]

Now, isolate \( P \):
\[ P \cdot 0.2072 = 24.63 - 7.471 \]
Calculating the right side:
\[ 24.63 - 7.471 = 17.159 \]
Now divide by \( 0.2072 \):
\[ P = \frac{17.159}{0.2072} \approx 82.82 \, \text{atm} \]

### Final Answer
The pressure exerted by 1 mole of methane in a 0.25 litre container at 300K is approximately **82.82 atm**.