`16g` oxygen gas expands at `STP` to occupy double of its oxygen volume. The work done during the process is:

Options

A. `260 kcal`
B. `180 kcal`
C. `130 kcal`
D. `272.84` kcal

Explanation

To solve the problem of calculating the work done during the expansion of 16 g of oxygen gas at STP, we can follow these steps:

### Step 1: Calculate the number of moles of oxygen gas (O2).
- The molecular weight of O2 is 32 g/mol.
- Given mass of O2 = 16 g.

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molecular weight}} = \frac{16 \, \text{g}}{32 \, \text{g/mol}} = 0.5 \, \text{moles} \]

### Step 2: Determine the initial volume (V1) of the gas at STP.
- At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters.
- Therefore, the volume occupied by 0.5 moles is:

\[ V_1 = 0.5 \, \text{moles} \times 22.4 \, \text{L/mole} = 11.2 \, \text{L} \]

### Step 3: Determine the final volume (V2) after expansion.
- The problem states that the gas expands to occupy double its initial volume.

\[ V_2 = 2 \times V_1 = 2 \times 11.2 \, \text{L} = 22.4 \, \text{L} \]

### Step 4: Calculate the work done (W) during the expansion.
- The formula for work done at constant pressure is:

\[ W = P \times (V_2 - V_1) \]

- At STP, the pressure (P) is 1 atm.
- Substitute the values into the formula:

\[ W = 1 \, \text{atm} \times (22.4 \, \text{L} - 11.2 \, \text{L}) = 1 \, \text{atm} \times 11.2 \, \text{L} = 11.2 \, \text{L atm} \]

### Step 5: Convert the work done from L atm to calories.
- To convert L atm to calories, we use the conversion factor:
1 L atm = 0.0821 L atm / 1 cal.

\[ W = 11.2 \, \text{L atm} \times \frac{1 \, \text{cal}}{0.0821 \, \text{L atm}} \approx 272.84 \, \text{cal} \]

### Final Answer:
The work done during the process is approximately **272.84 calories**.

---