Two moles of ammonia were found to occupy a volume of 5 L at `27^(@)C`. Calculate the pressure using van der Waals equation `(a="4.17 bar L"^(2)"mol"^(-2), b="0.0371 L mol"^(-1)).`

Explanation

To calculate the pressure of ammonia using the van der Waals equation, we will follow these steps:

### Step 1: Identify the given values
- Number of moles (N) = 2 moles
- Volume (V) = 5 L
- Temperature (T) = 27°C = 27 + 273 = 300 K
- Van der Waals constants:
- \( a = 4.17 \, \text{bar L}^2 \text{mol}^{-2} \)
- \( b = 0.0371 \, \text{L mol}^{-1} \)
- Gas constant (R) = 0.082 L bar K\(^{-1}\) mol\(^{-1}\)

### Step 2: Write the van der Waals equation
The van der Waals equation is given by:
\[ P + \frac{aN^2}{V^2} \left(V - Nb\right) = NRT \]

### Step 3: Rearrange the equation to find pressure (P)
Rearranging the equation to solve for P:
\[ P = \frac{NRT}{V - Nb} - \frac{aN^2}{V^2} \]

### Step 4: Substitute the known values into the equation
1. Calculate \( V - Nb \):
\[ Nb = 2 \, \text{moles} \times 0.0371 \, \text{L mol}^{-1} = 0.0742 \, \text{L} \]
\[ V - Nb = 5 \, \text{L} - 0.0742 \, \text{L} = 4.9258 \, \text{L} \]

2. Calculate \( \frac{NRT}{V - Nb} \):
\[ NRT = 2 \, \text{moles} \times 0.082 \, \text{L bar K}^{-1} \text{mol}^{-1} \times 300 \, \text{K} = 49.2 \, \text{L bar} \]
\[ \frac{NRT}{V - Nb} = \frac{49.2 \, \text{L bar}}{4.9258 \, \text{L}} \approx 10.0 \, \text{bar} \]

3. Calculate \( \frac{aN^2}{V^2} \):
\[ aN^2 = 4.17 \, \text{bar L}^2 \text{mol}^{-2} \times (2 \, \text{moles})^2 = 4.17 \times 4 = 16.68 \, \text{bar L}^2 \]
\[ V^2 = (5 \, \text{L})^2 = 25 \, \text{L}^2 \]
\[ \frac{aN^2}{V^2} = \frac{16.68 \, \text{bar L}^2}{25 \, \text{L}^2} \approx 0.6672 \, \text{bar} \]

### Step 5: Calculate the final pressure (P)
Now substitute the values back into the equation for P:
\[ P = 10.0 \, \text{bar} - 0.6672 \, \text{bar} \approx 9.3328 \, \text{bar} \]

### Final Answer
The pressure of the ammonia gas is approximately **9.33 bar**.

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