Compare the temperature of 3 molof `SO_(2)` at 15 bar occupying a volume of 10 L obtained by the ideal gas equation and van der Waals equation `(a="6.7 bar L"^(2)"mol"^(-2), b="0.0564 L mol"^(-1))`

Explanation

To solve the problem, we will calculate the temperature of 3 moles of SO₂ at 15 bar occupying a volume of 10 L using both the Ideal Gas Law and the Van der Waals equation.

### Step 1: Calculate Temperature using Ideal Gas Law

The Ideal Gas Law is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in bar)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0831 L·bar/(K·mol))
- \( T \) = temperature (in Kelvin)

Given:
- \( P = 15 \, \text{bar} \)
- \( V = 10 \, \text{L} \)
- \( n = 3 \, \text{mol} \)

Rearranging the Ideal Gas Law to solve for \( T \):

\[ T = \frac{PV}{nR} \]

Substituting the values:

\[ T = \frac{15 \, \text{bar} \times 10 \, \text{L}}{3 \, \text{mol} \times 0.0831 \, \text{L·bar/(K·mol)}} \]

Calculating:

\[ T = \frac{150}{0.2493} \approx 601.68 \, \text{K} \]

### Step 2: Calculate Temperature using 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:
- \( a = 6.7 \, \text{bar L}^2/\text{mol}^2 \)
- \( b = 0.0564 \, \text{L/mol} \)

Substituting the known values into the Van der Waals equation:

1. Calculate \( P + \frac{a n^2}{V^2} \):

\[ P + \frac{6.7 \times 3^2}{10^2} = 15 + \frac{6.7 \times 9}{100} = 15 + 0.603 = 15.603 \, \text{bar} \]

2. Calculate \( V - nb \):

\[ V - nb = 10 - 3 \times 0.0564 = 10 - 0.1692 = 9.8308 \, \text{L} \]

Now substituting these into the Van der Waals equation:

\[ (15.603)(9.8308) = 3 \times 0.0831 \times T \]

Calculating the left side:

\[ 153.033 = 0.2493T \]

Now solve for \( T \):

\[ T = \frac{153.033}{0.2493} \approx 613.23 \, \text{K} \]

### Step 3: Compare the Temperatures

- Temperature from Ideal Gas Law: \( T_{\text{ideal}} \approx 601.68 \, \text{K} \)
- Temperature from Van der Waals Equation: \( T_{\text{real}} \approx 613.23 \, \text{K} \)

### Conclusion

The temperature calculated using the Ideal Gas Law is lower than that calculated using the Van der Waals equation, indicating that real gases tend to have higher temperatures than predicted by the Ideal Gas Law under the same conditions.

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