If pressure of `CO_(2)` (real gas ) in a container is given by `P = (RT)/(2V - b) - (a)/(4b^(2))`, then mass of the gas in container is a) 11g b) 22g c) 33g d) 44g

Options

A. 11g
B. 22g
C. 33g
D. 44g

Answer

To solve the problem, we need to find the mass of the carbon dioxide gas in the container using the given pressure equation. The pressure of the gas is given by:

\[ P = \frac{RT}{2V - b} - \frac{a}{4V^2} \]

We will compare this with the Van der Waals equation for real gases:

\[ P = \frac{nRT}{V - nb} - \frac{a n^2}{V^2} \]

### Step 1: Identify the terms in the equations
From the Van der Waals equation, we can identify:
- \( n \) is the number of moles of the gas.
- \( R \) is the universal gas constant.
- \( T \) is the temperature.
- \( V \) is the volume of the gas.
- \( a \) and \( b \) are Van der Waals constants specific to the gas.

### Step 2: Compare the two equations
We can rewrite the given pressure equation in a form that allows us to compare it with the Van der Waals equation. We can express the first part of the equation as:

\[ P = \frac{RT}{2V - b} - \frac{a}{4V^2} \]

Now, we can rewrite the Van der Waals equation as:

\[ P = \frac{nRT}{V - nb} - \frac{a n^2}{V^2} \]

### Step 3: Set the two equations equal
By comparing the two equations, we can set the terms equal to each other:

1. From \( \frac{RT}{2V - b} = \frac{nRT}{V - nb} \)
2. From \( -\frac{a}{4V^2} = -\frac{a n^2}{V^2} \)

### Step 4: Solve for the number of moles \( n \)
From the second equation, we can simplify:

\[ \frac{a}{4V^2} = \frac{a n^2}{V^2} \]

Cancelling \( a \) and \( V^2 \) (assuming \( a \neq 0 \) and \( V \neq 0 \)), we get:

\[ \frac{1}{4} = n^2 \]

Taking the square root:

\[ n = \frac{1}{2} \]

### Step 5: Relate moles to mass
We know that the number of moles \( n \) is related to mass \( m \) and molar mass \( M \) by the equation:

\[ n = \frac{m}{M} \]

For carbon dioxide (CO₂), the molar mass \( M \) is:

\[ M = 12 \, \text{(for C)} + 2 \times 16 \, \text{(for O)} = 44 \, \text{g/mol} \]

### Step 6: Calculate the mass
Now substituting \( n \) into the equation:

\[ \frac{1}{2} = \frac{m}{44} \]

Multiplying both sides by 44:

\[ m = 44 \times \frac{1}{2} = 22 \, \text{g} \]

### Conclusion
Thus, the mass of the gas in the container is:

**22 g** (Option B)

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