The volume occupied by 8.8 g of `CO_(2)` at `31.1^(@)C` and 1 bar pressure ( in L ) is

Explanation

To solve the problem of finding the volume occupied by 8.8 g of CO₂ at 31.1°C and 1 bar pressure, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in bar)
- \( V \) = volume (in liters)
- \( n \) = number of moles of the gas
- \( R \) = ideal gas constant
- \( T \) = temperature (in Kelvin)

### Step 1: Convert the temperature from Celsius to Kelvin
To convert the temperature from Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

For our case:

\[ T = 31.1 + 273.15 = 304.25 \, K \]

### Step 2: Calculate the number of moles of CO₂
To find the number of moles, we use the formula:

\[ n = \frac{mass}{molar \, mass} \]

The molar mass of CO₂ is calculated as follows:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 2 = 32 g/mol
- Therefore, molar mass of CO₂ = 12 + 32 = 44 g/mol

Now, we can calculate the number of moles:

\[ n = \frac{8.8 \, g}{44 \, g/mol} = 0.2 \, mol \]

### Step 3: Use the Ideal Gas Law to find the volume
Now we can rearrange the Ideal Gas Law to solve for volume \( V \):

\[ V = \frac{nRT}{P} \]

Substituting the known values:
- \( n = 0.2 \, mol \)
- \( R = 0.0821 \, L \cdot bar/(K \cdot mol) \)
- \( T = 304.25 \, K \)
- \( P = 1 \, bar \)

Now we can plug in the values:

\[ V = \frac{(0.2 \, mol) \times (0.0821 \, L \cdot bar/(K \cdot mol)) \times (304.25 \, K)}{1 \, bar} \]

Calculating this gives:

\[ V = \frac{(0.2 \times 0.0821 \times 304.25)}{1} \]

\[ V \approx 5.0 \, L \]

### Final Answer:
The volume occupied by 8.8 g of CO₂ at 31.1°C and 1 bar pressure is approximately **5.0 liters**.

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