Calculate the pressure exerted by 7 mole `CO_2` ​of gas at 273 K, if the van der Waal's constat "a"=3.592`dm^6atm^(-2)` . Assume that the volume occupied by `CO_2` molecules is negligible.

Options

A. 6.65
B. 8.75
C. 8
D. 0.65

Explanation

To calculate the pressure exerted by 7 moles of \( CO_2 \) gas at 273 K using the Van der Waals equation, we 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 of the gas
- \( a \) = Van der Waals constant
- \( n \) = number of moles
- \( V \) = volume of the gas
- \( b \) = volume occupied by gas molecules (assumed to be negligible, hence \( b = 0 \))
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Simplify the equation
Since the volume occupied by the gas molecules is negligible, we can set \( b = 0 \). Thus, the equation simplifies to:
\[ P + \frac{a n^2}{V^2} = \frac{nRT}{V} \]
Rearranging gives:
\[ P = \frac{nRT}{V} - \frac{a n^2}{V^2} \]

### Step 3: Identify the values
From the problem statement:
- \( n = 7 \) moles
- \( T = 273 \) K
- \( a = 3.592 \, \text{dm}^6 \text{atm}^{-2} \)
- The volume \( V \) at STP (Standard Temperature and Pressure) for 1 mole of gas is approximately \( 22.4 \, \text{L} \) or \( 22.4 \, \text{dm}^3 \). Therefore, for 7 moles:
\[ V = 7 \times 22.4 \, \text{L} = 156.8 \, \text{L} = 156.8 \, \text{dm}^3 \]
- The universal gas constant \( R = 0.0821 \, \text{atm L K}^{-1} \text{mol}^{-1} \)

### Step 4: Substitute the values into the equation
Now we can substitute the values into the simplified Van der Waals equation:
\[ P = \frac{7 \times 0.0821 \times 273}{156.8} - \frac{3.592 \times 7^2}{156.8^2} \]

### Step 5: Calculate the first term
Calculating the first term:
\[ \frac{7 \times 0.0821 \times 273}{156.8} = \frac{159.6651}{156.8} \approx 1.016 \]

### Step 6: Calculate the second term
Calculating the second term:
\[ \frac{3.592 \times 49}{156.8^2} = \frac{176.008}{24625.44} \approx 0.00715 \]

### Step 7: Combine the results
Now, substituting back into the equation for pressure:
\[ P \approx 1.016 - 0.00715 \approx 1.00885 \, \text{atm} \]

### Step 8: Final answer
Thus, the pressure exerted by 7 moles of \( CO_2 \) at 273 K is approximately:
\[ P \approx 1.00885 \, \text{atm} \]