Air is expanded from 50 litres to 150 litres at 2 atomospheric pressure . The external work done is (Give , 1 atm=`10^(5)N^(-)2)`

Options

A. `2xx10^(-8)J`
B. `2xx10^(4)J`
C. `200 J`
D. `2000 J`

Explanation

To solve the problem of finding the external work done when air is expanded from 50 liters to 150 liters at a pressure of 2 atmospheres, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial volume (V1) = 50 liters
- Final volume (V2) = 150 liters
- Pressure (P) = 2 atmospheres

2. **Convert the Units**:
- Convert the volumes from liters to cubic meters:
\[ V1 = 50 \, \text{liters} = 50 \times 10^{-3} \, \text{m}^3 = 0.05 \, \text{m}^3 \]
\[ V2 = 150 \, \text{liters} = 150 \times 10^{-3} \, \text{m}^3 = 0.15 \, \text{m}^3 \]

3. **Convert Pressure to SI Units**:
- Convert pressure from atmospheres to pascals (N/m²):
\[ 1 \, \text{atm} = 10^5 \, \text{N/m}^2 \]
\[ P = 2 \, \text{atm} = 2 \times 10^5 \, \text{N/m}^2 \]

4. **Calculate the Change in Volume**:
- The change in volume (ΔV) is given by:
\[ ΔV = V2 - V1 = 0.15 \, \text{m}^3 - 0.05 \, \text{m}^3 = 0.10 \, \text{m}^3 \]

5. **Calculate the Work Done**:
- The work done (W) during the expansion at constant pressure is given by:
\[ W = P \times ΔV \]
- Substituting the values:
\[ W = (2 \times 10^5 \, \text{N/m}^2) \times (0.10 \, \text{m}^3) \]
\[ W = 2 \times 10^5 \times 0.10 = 2 \times 10^4 \, \text{J} \]

### Final Answer:
The external work done is \( W = 2 \times 10^4 \, \text{J} \).

---