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:** $$W=-P_{\text{ext}}\mathrm{\Delta }V=-P_{\text{ext}}(V_2-V_1)$$ **Calculations:** From the formula, $$W=-P_{\text{ext}}\mathrm{\Delta }V=-P_{\text{ext}}(V_2-V_1)$$ 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.