`1` mole of gas occupying `3` litre volume is expanded against a constant external pressure of `1` atm to a volume of `15` litre. The work done by the system is:

Options

A. `1.215xx10^(3) J`
B. `12.15xx10^(3) J`
C. `121.5xx10^(3) J`
D. None of these

Explanation

To solve the problem of calculating the work done by the system during the expansion of 1 mole of gas from 3 liters to 15 liters against a constant external pressure of 1 atm, we can follow these steps:

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Initial volume (V1) = 3 liters
- Final volume (V2) = 15 liters
- External pressure (P_external) = 1 atm

**Step 2: Use the formula for work done in an irreversible process.**
The work done (W) by the system during expansion against a constant external pressure is given by the formula:
\[ W = -P_{\text{external}} \times (V_2 - V_1) \]

**Step 3: Calculate the change in volume (V2 - V1).**
\[ V_2 - V_1 = 15 \, \text{liters} - 3 \, \text{liters} = 12 \, \text{liters} \]

**Step 4: Substitute the values into the work formula.**
\[ W = -1 \, \text{atm} \times 12 \, \text{liters} \]
\[ W = -12 \, \text{liters} \cdot \text{atm} \]

**Step 5: Convert the work from liters-atm to joules.**
We know that:
1 liter-atm = 101.3 joules
Thus,
\[ W = -12 \, \text{liters} \cdot \text{atm} \times 101.3 \, \text{joules/liter-atm} \]
\[ W = -1215.6 \, \text{joules} \]

**Step 6: Round the answer appropriately.**
The work done by the system is approximately:
\[ W \approx -1.215 \times 10^3 \, \text{joules} \]

### Final Answer:
The work done by the system is \(-1.215 \times 10^3\) joules.

---