To calculate the work done during the expansion of an ideal gas against a constant external pressure, we use the formula $w = -p_{ex} \Delta V$ .

1. Identify Given Values: Initial volume ($V_{i}$) = $1 \times 10^{-3}$ m$^{3}$. Final volume ($V_{f}$) = $1 \times 10^{-2}$ m$^{3}$.

• Constant external pressure ($p_{ex}$) = $1 \times 10^{5}$ N m$^{-2}$ (which is equal to $1 \times 10^{5}$ Pa) .

2. Calculate the change in volume ($\Delta V$): $\Delta V = V_{f} - V_{i} = (1 \times 10^{-2}) - (1 \times 10^{-3})$ m$^{3}$. $\Delta V = (10 \times 10^{-3}) - (1 \times 10^{-3}) = 9 \times 10^{-3}$ m$^{3}$.

3. Calculate Work ($w$): $w = -(1 \times 10^{5} \text{ N m}^{-2}) \times (9 \times 10^{-3} \text{ m}^{3})$. $w = -9 \times 10^{2} \text{ N m} = -900$ J .

According to IUPAC sign conventions, the negative sign indicates that work is done by the system during expansion . Therefore, the work done is –900 J, matching Option C.