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. `-10^(3) kJ`
B. `10^(2) kJ`
C. `-0.9 kJ`
D. `-900 kJ`

Explanation

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

### Step 1: Understand the Work Done Formula
The work done \( W \) by an ideal gas during expansion against a constant external pressure \( P_{\text{ext}} \) is given by the formula:
\[ W = -P_{\text{ext}} \Delta V \]
where \( \Delta V = V_2 - V_1 \).

### Step 2: Identify the Given Values
From the question, we have:
- Initial volume \( V_1 = 10^{-3} \, m^3 \)
- Final volume \( V_2 = 10^{-2} \, m^3 \)
- External pressure \( P_{\text{ext}} = 10^5 \, N/m^2 \)

### Step 3: Calculate the Change in Volume
Calculate the change in volume \( \Delta V \):
\[ \Delta V = V_2 - V_1 = 10^{-2} \, m^3 - 10^{-3} \, m^3 \]
\[ \Delta V = 10^{-2} \, m^3 - 0.1 \times 10^{-2} \, m^3 = 0.9 \times 10^{-2} \, m^3 = 9 \times 10^{-3} \, m^3 \]

### Step 4: Substitute Values into the Work Done Formula
Now substitute the values into the work done formula:
\[ W = -P_{\text{ext}} \Delta V = -10^5 \, N/m^2 \times 9 \times 10^{-3} \, m^3 \]

### Step 5: Perform the Calculation
Now calculate the work done:
\[ W = -10^5 \times 9 \times 10^{-3} = -900 \, J \]

### Final Answer
The work done by the gas during the expansion is:
\[ W = -900 \, J \]