A closed cylinder contains 40 g of `SO_3` and 4 g of helium gas at 375 K. If the partial pressure of helium gas is 2.5 atm, then the volume of the container is (R = 0.082 L-atm `K^-1` `mol^-1`)

Options

A. 18.45 L
B. 2.5 L
C. 15.5 L
D. 12.3 L

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of \( SO_3 \) and helium gas.

1. **Given data:**
- Mass of \( SO_3 = 40 \, \text{g} \)
- Mass of helium \( He = 4 \, \text{g} \)
- Molar mass of \( SO_3 = 80 \, \text{g/mol} \) (calculated as \( 32 \, \text{g/mol} \) for S + \( 48 \, \text{g/mol} \) for O)
- Molar mass of helium \( He = 4 \, \text{g/mol} \)

2. **Calculate moles of \( SO_3 \):**
\[ \text{Moles of } SO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{40 \, \text{g}}{80 \, \text{g/mol}} = 0.5 \, \text{mol} \]

3. **Calculate moles of helium:**
\[ \text{Moles of } He = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \, \text{g}}{4 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 2: Calculate the total number of moles in the container.

\[ \text{Total moles} = \text{Moles of } SO_3 + \text{Moles of } He = 0.5 \, \text{mol} + 1 \, \text{mol} = 1.5 \, \text{mol} \]

### Step 3: Use the ideal gas law to find the volume of the container.

1. **Ideal gas equation:**
\[ PV = nRT \]
Rearranging for volume \( V \):
\[ V = \frac{nRT}{P} \]

2. **Substituting known values:**
- \( n = 1.5 \, \text{mol} \)
- \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 375 \, \text{K} \)
- \( P = 2.5 \, \text{atm} \)

\[ V = \frac{(1.5 \, \text{mol}) \times (0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (375 \, \text{K})}{2.5 \, \text{atm}} \]

### Step 4: Calculate the volume.

\[ V = \frac{(1.5 \times 0.082 \times 375)}{2.5} \]

Calculating the numerator:
\[ 1.5 \times 0.082 \times 375 = 46.125 \]

Now, divide by the pressure:
\[ V = \frac{46.125}{2.5} = 18.45 \, \text{L} \]

### Final Answer:
The volume of the container is \( 18.45 \, \text{L} \).

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