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 calculate 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 1: Identify the given data
- Initial volume (V1) = 50 liters
- Final volume (V2) = 150 liters
- Pressure (P) = 2 atmospheres
- Conversion factor: 1 atmosphere = \(10^5 \, \text{N/m}^2\)

### Step 2: Calculate the change in volume
The change in volume (\(\Delta V\)) can be calculated using the formula:
\[ \Delta V = V2 - V1 \]
Substituting the values:
\[ \Delta V = 150 \, \text{liters} - 50 \, \text{liters} = 100 \, \text{liters} \]

### Step 3: Convert the change in volume to cubic meters
Since we need to express the volume in cubic meters, we convert liters to cubic meters using the conversion factor \(1 \, \text{liter} = 0.001 \, \text{m}^3\):
\[ \Delta V = 100 \, \text{liters} \times 0.001 \, \text{m}^3/\text{liter} = 0.1 \, \text{m}^3 \]

### Step 4: Convert the pressure from atmospheres to N/m²
Now, we convert the pressure from atmospheres to N/m²:
\[ P = 2 \, \text{atmospheres} \times 10^5 \, \text{N/m}^2/\text{atmosphere} = 2 \times 10^5 \, \text{N/m}^2 \]

### Step 5: Calculate the external work done
The external work done (W) can be calculated using the formula:
\[ W = P \Delta V \]
Substituting the values we have:
\[ W = (2 \times 10^5 \, \text{N/m}^2) \times (0.1 \, \text{m}^3) \]
Calculating this gives:
\[ W = 2 \times 10^5 \times 0.1 = 2 \times 10^4 \, \text{J} \]

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

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