`16kg` 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. 271 kcal

Answer

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

### Step 1: Understand the Conditions at STP
At Standard Temperature and Pressure (STP), the temperature is 0°C (273 K) and the pressure is 1 atm.

### Step 2: Determine the Molar Mass of Oxygen
The molar mass of oxygen (O₂) is calculated as follows:
- The atomic mass of oxygen is 16 g/mol.
- Therefore, the molar mass of O₂ = 16 g/mol × 2 = 32 g/mol.

### Step 3: Calculate the Number of Moles of Oxygen
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{16,000 \text{ g}}{32 \text{ g/mol}} = 500 \text{ moles} \]

### Step 4: Calculate the Initial Volume of Oxygen
At STP, 1 mole of gas occupies 22.4 liters. Therefore, the volume occupied by 500 moles is:
\[ V_1 = 500 \text{ moles} \times 22.4 \text{ L/mole} = 11,200 \text{ L} \]

### Step 5: Determine the Final Volume After Expansion
The problem states that the gas expands to occupy double its initial volume:
\[ V_2 = 2 \times V_1 = 2 \times 11,200 \text{ L} = 22,400 \text{ L} \]

### Step 6: Calculate the Change in Volume
The change in volume (dV) is:
\[ dV = V_2 - V_1 = 22,400 \text{ L} - 11,200 \text{ L} = 11,200 \text{ L} \]

### Step 7: Calculate the Work Done
The work done (W) during the expansion at constant external pressure is given by:
\[ W = -P_{\text{external}} \times dV \]
Substituting the values:
\[ W = -1 \text{ atm} \times 11,200 \text{ L} = -11,200 \text{ atm L} \]

### Step 8: Convert Work Done to Kilocalories
To convert atm L to kilocalories, we use the conversion factor:
\[ 1 \text{ atm L} = 24.22 \times 10^{-3} \text{ kcal} \]
Thus,
\[ W = -11,200 \text{ atm L} \times 24.22 \times 10^{-3} \text{ kcal/atm L} \]
Calculating this gives:
\[ W = -271.3 \text{ kcal} \]

### Final Answer
The work done during the process is:
\[ \text{Work Done} = -271.3 \text{ kcal} \]