To determine the temperature at which 192 g of SO[tex]\(_2\)[/tex] would occupy a volume of 6 dm[tex]\(^3\)[/tex] at a pressure of 15 atm using the Van der Waals equation, follow these steps:
### Step 1: Understand the Given Data
1. Van der Waals constants for SO[tex]\(_2\)[/tex]:
- [tex]\(a = 6.865\)[/tex] atm·L[tex]\(^2\)[/tex]/mol[tex]\(^2\)[/tex]
- [tex]\(b = 0.05679\)[/tex] L/mol
2. Given values:
- Pressure, [tex]\(P = 15\)[/tex] atm
- Volume, [tex]\(V = 6\)[/tex] dm[tex]\(^3\)[/tex]
- Mass of SO[tex]\(_2\)[/tex], [tex]\( \text{mass}_{\text{SO}_2} = 192\)[/tex] g
3. Molar mass of SO[tex]\(_2\)[/tex]:
- [tex]\( M_{\text{SO}_2} = 64.066 \)[/tex] g/mol
### Step 2: Conversion of Volume
- Volume needs to be in Liters for the calculation:
- [tex]\( V = 6 \)[/tex] dm[tex]\(^3\)[/tex] = 6 L
### Step 3: Calculate the Number of Moles of SO[tex]\(_2\)[/tex]
- Use the molar mass of SO[tex]\(_2\)[/tex] to find the number of moles, [tex]\( n \)[/tex]:
[tex]\[
n = \frac{\text{mass}_{\text{SO}_2}}{M_{\text{SO}_2}} = \frac{192 \text{ g}}{64.066 \text{ g/mol}} \approx 2.9969 \text{ moles}
\][/tex]
### Step 4: Use the Van der Waals Equation
The Van der Waals equation for a real gas is:
[tex]\[
\left(P + a \left( \frac{n}{V} \right)^2 \right) \left( V - n b \right) = n R T
\][/tex]
Where [tex]\( R \)[/tex] is the universal gas constant, [tex]\( R = 0.0821 \)[/tex] L·atm/K·mol.
### Step 5: Solve for Temperature [tex]\( T \)[/tex]
Rearrange the Van der Waals equation to solve for [tex]\( T \)[/tex]:
[tex]\[
T = \frac{\left(P + a \left( \frac{n}{V} \right)^2 \right) \left( V - n b \right)}{n R}
\][/tex]
Plug in the known values:
[tex]\[
T = \frac{\left(15 + 6.865 \left( \frac{2.9969}{6} \right)^2 \right) \left(6 - 2.9969 \times 0.05679 \right)}{2.9969 \times 0.0821}
\][/tex]
### Step 6: Calculation
Perform the calculations step-by-step:
1. Calculate [tex]\(\left( \frac{n}{V} \right)^2 \)[/tex]:
[tex]\[
\left( \frac{2.9969}{6} \right)^2 \approx 0.2497
\][/tex]
2. Calculate the pressure factor:
[tex]\[
15 + 6.865 \times 0.2497 \approx 16.7137
\][/tex]
3. Calculate the volume factor:
[tex]\[
6 - 2.9969 \times 0.05679 \approx 5.8298
\][/tex]
4. Combine these into the formula for [tex]\( T \)[/tex]:
[tex]\[
T = \frac{16.7137 \times 5.8298}{2.9969 \times 0.0821} \approx \frac{97.4504}{0.2461} \approx 395.99
\][/tex]
### Step 7: Final Result
The temperature at which 192 g of SO[tex]\(_2\)[/tex] would occupy a volume of 6 dm[tex]\(^3\)[/tex] at 15 atm pressure is approximately 396 Kelvin.
