A gas expands from a volume of `1m^(3)` to a volume of `2m^(3)` against an external pressure of `10^(5)Nm^(-2)`. The work done of the gas will be

Options

A. `10^(5)kJ`
B. `10^(2)kJ`
C. `10^(2)J`
D. `10^(3)J`

Explanation

To find the work done by the gas during its expansion, we can use the formula for work done against an external pressure:

\[ W = P \Delta V \]

where:
- \( W \) is the work done,
- \( P \) is the external pressure,
- \( \Delta V \) is the change in volume.

### Step 1: Identify the initial and final volumes
- Initial volume \( V_1 = 1 \, m^3 \)
- Final volume \( V_2 = 2 \, m^3 \)

### Step 2: Calculate the change in volume \( \Delta V \)
\[ \Delta V = V_2 - V_1 = 2 \, m^3 - 1 \, m^3 = 1 \, m^3 \]

### Step 3: Identify the external pressure
- External pressure \( P = 10^5 \, N/m^2 \)

### Step 4: Substitute the values into the work done formula
\[ W = P \Delta V = (10^5 \, N/m^2) \times (1 \, m^3) \]

### Step 5: Calculate the work done
\[ W = 10^5 \, N \cdot m = 10^5 \, J \]

### Step 6: Convert the work done into kilojoules
\[ W = 10^5 \, J = 10^2 \, kJ = 100 \, kJ \]

### Final Answer
The work done by the gas is \( 10^5 \, J \) or \( 100 \, kJ \).

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