An ideal gas expands in volume from `1 xx 10^(3) m^(3)` at 300 K against a constant pressure of `1 xx 10^(5)` N `m^(-2)`.The work done is

Options

A. `-900` J
B. `-900` kJ
C. `270` kJ
D. `900` kJ

Explanation

To find the work done by an ideal gas during expansion at constant pressure, we can use the formula for work done (W) in thermodynamics:

\[ W = -P \Delta V \]

where:
- \( P \) is the constant pressure,
- \( \Delta V \) is the change in volume (final volume - initial volume).

### Step 1: Identify the given values
- Initial volume (\( V_1 \)) = \( 1 \times 10^3 \, m^3 \)
- Final volume (\( V_2 \)) = \( 2 \times 10^3 \, m^3 \) (assuming the final volume is double the initial volume for this example)
- Constant pressure (\( P \)) = \( 1 \times 10^5 \, N/m^2 \)

### Step 2: Calculate the change in volume (\( \Delta V \))
\[ \Delta V = V_2 - V_1 \]
\[ \Delta V = (2 \times 10^3 \, m^3) - (1 \times 10^3 \, m^3) \]
\[ \Delta V = 1 \times 10^3 \, m^3 \]

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

### Step 4: Perform the multiplication
\[ W = - (1 \times 10^5 \times 1 \times 10^3) \]
\[ W = - 1 \times 10^8 \, J \]

### Step 5: Write the final answer
The work done by the gas during expansion is:
\[ W = -1 \times 10^8 \, J \]