The volume occupied by 4.0 g of oxygen at 100 kPa and 300 K, given that the value of universal gas constant is 8.31 kPa `dm^(3) mol^(-1)` is nearly

Options

A. `1.4 dm^(3)`
B. `2.0 dm^(3)`
C. `3.1 dm^(3)`
D. `4.3 dm^(3)`

Answer

To find the volume occupied by 4.0 g of oxygen at 100 kPa and 300 K, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in kPa)
- \( V \) = volume (in dm³)
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant (8.31 kPa dm³ mol⁻¹ K⁻¹)
- \( T \) = temperature (in K)

### Step 1: Identify the given values
- Mass of oxygen (\( m \)) = 4.0 g
- Pressure (\( P \)) = 100 kPa
- Temperature (\( T \)) = 300 K
- Universal gas constant (\( R \)) = 8.31 kPa dm³ mol⁻¹ K⁻¹

### Step 2: Calculate the molar mass of oxygen
The molecular formula for oxygen is \( O_2 \). The molar mass of oxygen is calculated as:
\[ \text{Molar mass of } O_2 = 2 \times 16 \text{ g/mol} = 32 \text{ g/mol} \]

### Step 3: Calculate the number of moles of oxygen (\( n \))
Using the formula:
\[ n = \frac{m}{\text{Molar mass}} \]
Substituting the values:
\[ n = \frac{4.0 \text{ g}}{32 \text{ g/mol}} = 0.125 \text{ mol} \]

### Step 4: Substitute the values into the Ideal Gas Law equation
We need to find the volume (\( V \)):
\[ V = \frac{nRT}{P} \]

Substituting the known values:
\[ V = \frac{(0.125 \text{ mol}) \times (8.31 \text{ kPa dm}^3 \text{ mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K})}{100 \text{ kPa}} \]

### Step 5: Calculate the volume
Calculating the numerator:
\[ 0.125 \times 8.31 \times 300 = 312.375 \text{ kPa dm}^3 \]

Now, divide by the pressure:
\[ V = \frac{312.375 \text{ kPa dm}^3}{100 \text{ kPa}} = 3.12375 \text{ dm}^3 \]

Rounding this to two decimal places, we get:
\[ V \approx 3.1 \text{ dm}^3 \]

### Final Answer
The volume occupied by 4.0 g of oxygen at 100 kPa and 300 K is approximately **3.1 dm³**.

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