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Step 1: Identify the Known Parameters
• Initial volume, $V_i = 1 \times 10^{-3}\,\text{m}^3$
• Final volume, $V_f = 1 \times 10^{-2}\,\text{m}^3$
• Constant external pressure, $P = 1 \times 10^5\,\text{N}\,\text{m}^{-2}$ (or Pa)
• Temperature, $T = 300\,\text{K}$ (not directly needed for work calculation under constant pressure).
Step 2: Recall the Formula for Work Done by an Expanding Gas
Under an isobaric (constant pressure) process, the work done by the gas is given by:
$$w = -\,P\,\Delta V$$
The negative sign indicates that when the gas expands, it does work on the surroundings and hence the system loses energy.
Step 3: Compute the Change in Volume
$$\Delta V = V_f - V_i = (1 \times 10^{-2}) - (1 \times 10^{-3}) \,\text{m}^3 = 9 \times 10^{-3} \,\text{m}^3$$
Step 4: Substitute the Values into the Work Formula
$$w = -\,P\,\Delta V = -\,\bigl(1 \times 10^5\,\text{N}\,\text{m}^{-2}\bigr)\,\bigl(9 \times 10^{-3}\,\text{m}^3\bigr)$$
Note: $1\,\text{N}\,\text{m}^{-2} = 1\,\text{Pa}$, and $1\,\text{J} = 1\,\text{N}\,\text{m}$.
Step 5: Perform the Multiplication
$$w = - \Bigl(1 \times 10^5 \times 9 \times 10^{-3}\Bigr)\,\text{J} = - (1 \times 9 \times 10^{5 - 3})\,\text{J} = - (9 \times 10^2)\,\text{J}$$
$$w = -900\,\text{J}$$
Step 6: Interpret the Result
The negative sign indicates that the system (gas) is doing work on the surroundings by expanding. The magnitude of the work is $900\,\text{J}$, and since the process is expansion, the sign is negative.
Final Answer
The work done in the process is -900 J.
