An ideal gas expands from `10^(-3) m^(3)` to `10^(-2) m^(3)` at 300 K against a constant pressure of `10^(5) Nm^(-2)`. The workdone is

Options

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

Explanation

To solve the problem of calculating the work done by an ideal gas expanding against a constant pressure, we can follow these steps:

### Step 1: Identify the given values
- Initial volume, \( V_1 = 10^{-3} \, m^3 \)
- Final volume, \( V_2 = 10^{-2} \, m^3 \)
- Constant external pressure, \( P = 10^5 \, N/m^2 \)

### Step 2: Calculate the change in volume
The change in volume (\( \Delta V \)) can be calculated using the formula:
\[ \Delta V = V_2 - V_1 \]
Substituting the values:
\[ \Delta V = 10^{-2} \, m^3 - 10^{-3} \, m^3 = 10^{-2} \, m^3 - 0.1 \times 10^{-2} \, m^3 = 0.9 \times 10^{-2} \, m^3 = 9 \times 10^{-3} \, m^3 \]

### Step 3: Use the formula for work done
The work done (\( W \)) by the gas during expansion against a constant pressure is given by:
\[ W = -P \Delta V \]
Substituting the values:
\[ W = - (10^5 \, N/m^2) \times (9 \times 10^{-3} \, m^3) \]

### Step 4: Calculate the work done
Calculating the above expression:
\[ W = - (10^5) \times (9 \times 10^{-3}) = - 900 \, J \]

### Step 5: Interpret the result
The negative sign indicates that the work is done by the gas during expansion.

### Final Answer
The work done by the gas is \( -900 \, J \).

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