# Step-by-Step Solution: Work Done by Gas at Constant Pressure 1. Understanding the Problem - **Initial volume** \( V_1 = .03~\mathrm{m}^3 \) - **Final volume,** \( V_2 = .06~\mathrm{m}^3 \) - **Constant pressure,** \( P = 200~\mathrm{kN/m}^2 \) We need to **find the work done** by the gas during this expansion. --- ## 2. Work Done at Constant Pressure The formula for work done by a gas expanding at constant pressure is: \[ W = P \Delta V \] where - \( W \) = work done (in Joules) - \( P \) = pressure (in Pascals, \( \mathrm{N/m}^2 \)) - \( \Delta V = V_2 - V_1 \) --- ## 3. Substitute the Values ### Convert Pressure The pressure is given as \( 200~\mathrm{kN/m}^2 \). - \( 1~\mathrm{kN} = 100~\mathrm{N} \) - \( 200~\mathrm{kN/m}^2 = 200 \times 100 = 200,000~\mathrm{N/m}^2 \) So, \( P = 200,000~\mathrm{N/m}^2 \). ### Calculate \(\Delta V\): \[ \Delta V = V_2 - V_1 = .06~\mathrm{m}^3 - .03~\mathrm{m}^3 = .03~\mathrm{m}^3 \] --- ## 4. Calculate the Work \[ W = P \Delta V = 200,000~\mathrm{N/m}^2 \times .03~\mathrm{m}^3 \] \[ W = 6,000~\mathrm{J} \] --- ## 5. Final Answer **The work done by the gas is \(\boxed{6,000~\mathrm{J}}\) (Joules).** --- ### Summary Table | Quantity | Value | |---------------|---------------------| | Initial Volume (\(V_1\)) | \(.03~\mathrm{m}^3\) | | Final Volume (\(V_2\)) | \(.06~\mathrm{m}^3\) | | Pressure (\(P\)) | \(200,000~\mathrm{N/m}^2\) | | Work Done (\(W\)) | \(6,000~\mathrm{J}\) |