Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution
4. Answer the following questions.
xiii. Calculate the work done during synthesis of NH3 in which volume changes from 8.0 dm3 to 4.0 dm3 at a constant external pressure of 43 bar. In what direction the work energy flows?
Ans. : (17.2 kJ, work energy flows into system)
**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.
Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution