What is the pressure of 2 mole of `NH_(3)` at `27^(@)C` when its volume is 5 lit. in van der Waal's equation ? `(a = 4.17, b = 0.03711)`

Options

A. 10.33 atm
B. 9.333 atm
C. 9.74 atm
D. 9.2 atm

Explanation

To find the pressure of 2 moles of ammonia (NH₃) at 27°C in a volume of 5 liters using Van der Waals equation, we can 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
- \( n \) = number of moles
- \( V \) = volume
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin
- \( a \) and \( b \) are Van der Waals constants.

### Step 2: Convert temperature to Kelvin
The temperature given is 27°C. To convert it to Kelvin:
\[ T(K) = T(°C) + 273.15 = 27 + 273.15 = 300.15 \approx 300 \text{ K} \]

### Step 3: Substitute values into the equation
Given:
- \( n = 2 \) moles
- \( V = 5 \) liters
- \( a = 4.17 \)
- \( b = 0.03711 \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)

Substituting these values into the Van der Waals equation:
\[ \left( P + \frac{4.17 \cdot 2^2}{5^2} \right) (5 - 2 \cdot 0.03711) = 2 \cdot 0.0821 \cdot 300 \]

### Step 4: Calculate the left-hand side
First, calculate \( \frac{4.17 \cdot 2^2}{5^2} \):
\[ \frac{4.17 \cdot 4}{25} = \frac{16.68}{25} = 0.6672 \]

Next, calculate \( 5 - 2 \cdot 0.03711 \):
\[ 5 - 0.07422 = 4.92578 \approx 4.93 \]

Now substitute these values:
\[ \left( P + 0.6672 \right) \cdot 4.93 = 2 \cdot 0.0821 \cdot 300 \]

Calculate the right-hand side:
\[ 2 \cdot 0.0821 \cdot 300 = 49.26 \]

### Step 5: Solve for P
Now we have:
\[ (P + 0.6672) \cdot 4.93 = 49.26 \]

Dividing both sides by 4.93:
\[ P + 0.6672 = \frac{49.26}{4.93} \approx 10 \]

Now, isolate \( P \):
\[ P = 10 - 0.6672 \approx 9.3328 \approx 9.33 \text{ atm} \]

### Final Answer
The pressure of 2 moles of NH₃ at 27°C in a volume of 5 liters is approximately:
\[ \boxed{9.33 \text{ atm}} \]