Find the temperature at which 3 moles of `SO_2` will occupy a volume of 10 litre at a pressure of 15 atm. a=6.71` atm litre^2mol^(−2)` ;b=0.0564 `litre mol^(−1)` .

Options

A. 623
B. 615
C. 602
D. 603

Explanation

To find the temperature at which 3 moles of \( SO_2 \) will occupy a volume of 10 liters at a pressure of 15 atm, we will use the Van der Waals equation:

\[ \left(P + \frac{a n^2}{V^2}\right)(V - nb) = nRT \]

Where:
- \( P \) = pressure (15 atm)
- \( n \) = number of moles (3 moles)
- \( V \) = volume (10 liters)
- \( a \) = Van der Waals constant for \( SO_2 \) (6.71 atm L²/mol²)
- \( b \) = Van der Waals constant for \( SO_2 \) (0.0564 L/mol)
- \( R \) = ideal gas constant (0.0821 atm L/(K mol))
- \( T \) = temperature (in Kelvin)

### Step 1: Substitute the known values into the Van der Waals equation.

\[ \left(15 + \frac{6.71 \times 3^2}{10^2}\right)(10 - 3 \times 0.0564) = 3 \times 0.0821 \times T \]

### Step 2: Calculate \( \frac{a n^2}{V^2} \).

\[ \frac{6.71 \times 3^2}{10^2} = \frac{6.71 \times 9}{100} = \frac{60.39}{100} = 0.6039 \]

### Step 3: Calculate \( V - nb \).

\[ V - nb = 10 - 3 \times 0.0564 = 10 - 0.1692 = 9.8308 \]

### Step 4: Substitute these values back into the equation.

\[ (15 + 0.6039)(9.8308) = 3 \times 0.0821 \times T \]

### Step 5: Calculate \( 15 + 0.6039 \).

\[ 15 + 0.6039 = 15.6039 \]

### Step 6: Calculate the left side of the equation.

\[ 15.6039 \times 9.8308 \approx 153.398 \]

### Step 7: Set up the equation to solve for \( T \).

\[ 153.398 = 3 \times 0.0821 \times T \]

### Step 8: Calculate \( 3 \times 0.0821 \).

\[ 3 \times 0.0821 = 0.2463 \]

### Step 9: Solve for \( T \).

\[ T = \frac{153.398}{0.2463} \approx 622.813 \text{ K} \]

### Step 10: Round the temperature to the nearest whole number.

\[ T \approx 623 \text{ K} \]

Thus, the temperature at which 3 moles of \( SO_2 \) will occupy a volume of 10 liters at a pressure of 15 atm is approximately **623 K**.