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 Initial and Final Volumes
The initial volume of the gas is given as 3 dm³. Since 1 dm³ is equivalent to 1 liter, we can convert this to liters:
- Initial Volume (V_i) = 3 dm³ = 3 liters

The final volume after expansion is given as:
- Final Volume (V_f) = 13 liters

### Step 2: Calculate the Change in Volume
The change in volume (ΔV) can be calculated using the formula:
\[ \Delta V = V_f - V_i \]
Substituting the values:
\[ \Delta V = 13 \, \text{liters} - 3 \, \text{liters} = 10 \, \text{liters} \]

### Step 3: Use the Formula for Work Done
The work done (W) by the gas during expansion against a constant external pressure (P) can be calculated using the formula:
\[ W = -P \Delta V \]
Here, the external pressure (P) is given as 1 atm. Thus, we can substitute the values into the formula:
\[ W = -1 \, \text{atm} \times 10 \, \text{liters} \]

### Step 4: Calculate the Work Done
Calculating the work done:
\[ W = -10 \, \text{atm} \cdot \text{liters} \]
This can also be expressed in terms of decimeters cubed (since 1 liter = 1 dm³):
\[ W = -10 \, \text{atm} \cdot \text{dm}^3 \]

### Conclusion
The work done during the expansion of the gas is:
\[ W = -10 \, \text{atm} \cdot \text{dm}^3 \]