Air expands from 5 litres to 10 litres at 2 atm pressure. External workdone is

Options

A. 10J
B. 1000J
C. 3000 J
D. 300 J

Answer

To solve the problem of calculating the external work done when air expands from 5 liters to 10 liters at a pressure of 2 atm, we can follow these steps:

### Step 1: Convert Pressure from atm to Pascal
The pressure is given as 2 atm. To convert this to Pascal (Pa), we use the conversion factor:
1 atm = 1.013 × 10^5 Pa.

So,
\[ P = 2 \, \text{atm} = 2 \times 1.013 \times 10^5 \, \text{Pa} = 2.026 \times 10^5 \, \text{Pa} \]

### Step 2: Calculate Change in Volume
The initial volume (V1) is 5 liters and the final volume (V2) is 10 liters. The change in volume (ΔV) can be calculated as:
\[ \Delta V = V2 - V1 = 10 \, \text{liters} - 5 \, \text{liters} = 5 \, \text{liters} \]

We need to convert liters to cubic meters for our calculations:
1 liter = 10^-3 m³, so:
\[ \Delta V = 5 \, \text{liters} = 5 \times 10^{-3} \, \text{m}^3 \]

### Step 3: Calculate Work Done
The work done (W) by the gas during expansion at constant pressure can be calculated using the formula:
\[ W = P \times \Delta V \]

Substituting the values we have:
\[ W = (2.026 \times 10^5 \, \text{Pa}) \times (5 \times 10^{-3} \, \text{m}^3) \]

### Step 4: Perform the Calculation
Now we calculate the work done:
\[ W = 2.026 \times 10^5 \times 5 \times 10^{-3} \]
\[ W = 2.026 \times 5 \times 10^{5 - 3} \]
\[ W = 10.13 \times 10^2 \]
\[ W = 1013 \, \text{Joules} \]

### Final Answer
The external work done during the expansion is approximately:
\[ W \approx 1013 \, \text{J} \]