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. 8.3 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 SO₃ and He
1. **Moles of SO₃**:
- Given mass of SO₃ = 40 g
- Molar mass of SO₃ = 80 g/mol
- Moles of SO₃ = Mass / Molar mass = 40 g / 80 g/mol = 0.5 moles

2. **Moles of He**:
- Given mass of He = 4 g
- Molar mass of He = 4 g/mol
- Moles of He = Mass / Molar mass = 4 g / 4 g/mol = 1 mole

### Step 2: Calculate the total number of moles
- Total moles (n_total) = Moles of SO₃ + Moles of He = 0.5 moles + 1 mole = 1.5 moles

### Step 3: Calculate the mole fraction of helium
- Mole fraction of He (X_He) = Moles of He / Total moles = 1 mole / 1.5 moles = 2/3

### Step 4: Use the partial pressure of helium to find the total pressure
- Given partial pressure of He (P_He) = 2.5 atm
- Using the relation: P_He = X_He * P_total
- Rearranging gives: P_total = P_He / X_He = 2.5 atm / (2/3) = 2.5 * (3/2) = 3.75 atm

### Step 5: Use the Ideal Gas Law to find the volume
- The Ideal Gas Law is given by: PV = nRT
- Rearranging gives: V = nRT / P
- Substituting the values:
- n = 1.5 moles
- R = 0.0821 L·atm/(K·mol)
- T = 375 K
- P = 3.75 atm

Calculating the volume:
\[ V = \frac{(1.5 \text{ moles}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (375 \text{ K})}{3.75 \text{ atm}} \]

Calculating:
\[ V = \frac{(1.5) \times (0.0821) \times (375)}{3.75} \]
\[ V = \frac{46.1875}{3.75} \]
\[ V = 12.3 \text{ L} \]

### Final Answer:
The volume of the container is **12.3 L**.

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