Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

**Given:** - Initial volume (\(V_1\)) = 8.0 dm³ - Final volume (\(V_2\)) = 4.0 dm³ - External pressure (\(P_{\text{ext}}\)) = 43 bar **To find:** The work done (\(W\)) and direction of the work energy flow. **Formulas:** \[W = -P_{\text{ext}} \Delta V = -P_{\text{ext}} (V_2 - V_1)\] **Calculations:** From the formula, \[W = -P_{\text{ext}} \Delta V = -P_{\text{ext}} (V_2 - V_1)\] Therefore, \[W = -43 \, \text{bar} \times (4.0 \, \text{dm}^3 - 8.0 \, \text{dm}^3) = 172 \, \text{dm}^3 \, \text{bar}\] Now, 1 dm³ bar = 100 J Hence, \[172 \, \text{dm}^3 \times \left(\frac{100 \, \text{J}}{1 \, \text{dm}^3 \, \text{bar}}\right) = 17200 \, \text{J} = 17.2 \, \text{kJ}\] Since the work is done on the system, work-energy flows into the system from the surroundings. Therefore: - The work done (\(W\)) = 17.2 kJ - Work energy flows into the system.