The volume occupied by 88 g of `CO_(2)` at `30 .^(@)C` and 1 bar pressure will be

Options

A. 5.05 L
B. 49.8 L
C. 2 L
D. 55 L

Explanation

To find the volume occupied by 88 g of CO₂ at 30 °C and 1 bar pressure, we will use the Ideal Gas Law, which is expressed as:

\[ PV = nRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles of gas
- \( R \) = Universal gas constant
- \( T \) = Temperature in Kelvin

### Step-by-step Solution:

1. **Convert Temperature to Kelvin:**
- The temperature given is 30 °C. To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273.15 \]
- Therefore:
\[ T = 30 + 273.15 = 303.15 \, K \]

2. **Calculate the Number of Moles (n):**
- The number of moles \( n \) can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
- The molar mass of CO₂ (Carbon Dioxide) is calculated as follows:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (and there are 2 oxygen atoms in CO₂)
- Therefore, molar mass of CO₂:
\[ \text{Molar mass of CO₂} = 12 + (2 \times 16) = 12 + 32 = 44 \, g/mol \]
- Now, substituting the values:
\[ n = \frac{88 \, g}{44 \, g/mol} = 2 \, mol \]

3. **Use the Ideal Gas Law to Find Volume (V):**
- Rearranging the Ideal Gas Law to solve for \( V \):
\[ V = \frac{nRT}{P} \]
- Substitute the known values:
- \( n = 2 \, mol \)
- \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \) (Note: 1 bar ≈ 0.9869 atm, but for simplicity, we can use 1 bar directly in calculations)
- \( T = 303.15 \, K \)
- \( P = 1 \, bar \)
- Plugging in the values:
\[ V = \frac{2 \, mol \times 0.0821 \, L \cdot atm/(K \cdot mol) \times 303.15 \, K}{1 \, bar} \]
- Calculate:
\[ V = \frac{2 \times 0.0821 \times 303.15}{1} \]
\[ V \approx 49.75 \, L \]

4. **Final Answer:**
- Rounding off, the volume occupied by 88 g of CO₂ at 30 °C and 1 bar pressure is approximately:
\[ V \approx 49.8 \, L \]