An ideal gas expands in volume from $$1 \times 10^{-3}$$ m$$^3$$ to $$1 \times 10^{-2}$$ m$$^3$$ at $$300$$ K against a constant pressure of $$1 \times 10^5$$ N m$$^{-2}$$. The work done is

Irreversible Expansion Work

When a gas expands against a constant external pressure ($$P_{\text{ext}}$$), the process is thermodynamically irreversible. The work done ($$W$$) on or by the system is given by the formula:

$$W = -P_{\text{ext}} \cdot \Delta V$$

Where $$\Delta V$$ is the change in volume ($$V_{\text{final}} - V_{\text{initial}}$$). The negative sign follows the IUPAC convention, indicating that work is done by the system on the surroundings, resulting in a loss of energy.

Step-by-Step Calculation:

• Step 1: Identify the given values

  • Constant External Pressure ($$P_{\text{ext}}$$) = $$1 \times 10^5 \text{ N m}^{-2}$$
  • Initial Volume ($$V_{\text{initial}}$$) = $$1 \times 10^{-3} \text{ m}^3$$
  • Final Volume ($$V_{\text{final}}$$) = $$1 \times 10^{-2} \text{ m}^3$$



• Step 2: Calculate the change in volume ($$\Delta V$$)

  $$\Delta V = V_{\text{final}} - V_{\text{initial}}$$

  $$\Delta V = (1 \times 10^{-2}) - (1 \times 10^{-3}) \text{ m}^3$$

  $$\Delta V = (10 \times 10^{-3}) - (1 \times 10^{-3}) \text{ m}^3 = 9 \times 10^{-3} \text{ m}^3$$




• Step 3: Substitute the values into the work formula

  $$W = -(1 \times 10^5 \text{ N m}^{-2}) \times (9 \times 10^{-3} \text{ m}^3)$$

  $$W = -9 \times 10^2 \text{ N m}$$

  Since $$1 \text{ N m} = 1 \text{ Joule (J)}$$:

  $$W = -900 \text{ J}$$

Conclusion:

The total work done by the expanding gas is $$-900 \text{ J}$$.

Answer: Option A $$\rightarrow$$ $$-900$$ J