One mole of a gas occupying `3dm^(3)` expands against a constant external pressure of 1 atm to a volume of 13 lit. The workdone is :-

Options

A. `-10` atm `dm^(3)`
B. `-20` atm `dm^(3)`
C. `-39` atm `dm^(3)`
D. `-48` atm `dm^(3)`

Answer

To solve the problem of calculating the work done during the expansion of one mole of gas, we can follow these steps:

### Step 1: Understand the Given Information
We have:
- Initial volume (V1) = 3 dm³
- Final volume (V2) = 13 L (which is equivalent to 13 dm³ since 1 L = 1 dm³)
- External pressure (P_external) = 1 atm

### Step 2: Use the Formula for Work Done
The work done (W) during the expansion of a gas can be calculated using the formula:
\[ W = -P_{\text{external}} \times (V_2 - V_1) \]

### Step 3: Calculate the Change in Volume
Calculate the change in volume:
\[ V_2 - V_1 = 13 \, \text{dm}^3 - 3 \, \text{dm}^3 = 10 \, \text{dm}^3 \]

### Step 4: Substitute Values into the Formula
Now substitute the values into the work done formula:
\[ W = -1 \, \text{atm} \times 10 \, \text{dm}^3 \]
\[ W = -10 \, \text{atm} \cdot \text{dm}^3 \]

### Step 5: State the Final Answer
The work done during the expansion is:
\[ W = -10 \, \text{atm} \cdot \text{dm}^3 \]