Let's solve **Question (3)** **step by step**, with clear explanations: --- ## **Given** - Vessel: Semi-flexible (volume can decrease if pressure inside is less than outside) - Initial: 1 L water + 3 L air at 1 atm (atmospheric pressure) - Water is boiled (at 100°C, atmospheric pressure: 1 atm = 101.3 kPa) - While boiling, cap is closed (only steam inside; all air is expelled, so inside is pure H₂O vapor) - Vessel cooled to 20°C - The can is partially crushed (volume decreases as pressure drops inside) - **Assume**: Cap does not leak, vessel is partially flexible --- ## **Step 1: What is inside when the cap is closed?** - At the moment of sealing: **Only water vapor (steam) is inside** at 100°C and 1 atm (101.3 kPa). - Volume is the vessel's original volume, but with only steam (no air). --- ## **Step 2: What happens as the vessel cools to 20°C?** - As it cools, the **vapor will condense** because the saturation pressure of water vapor at 20°C is only **2.3 kPa** (from the FYI). - The can is flexible, so as pressure drops inside, the can is crushed by the atmosphere, reducing its volume until pressure inside rises to match the vapor pressure at 20°C **or** until all vapor condenses and only a little vapor remains in equilibrium with liquid water. --- ## **Step 3: What is the final pressure inside?** - At 20°C, **any water vapor above liquid water will be at the saturation pressure at 20°C = 2.3 kPa** (as long as there is any liquid left). - The pressure **cannot drop lower** than the vapor pressure because any lower, more liquid would evaporate to restore equilibrium. **So, the final pressure inside = 2.3 kPa** --- ## **Step 4: What if the vessel is rigid (fixed volume)?** - In a rigid vessel, the volume stays the same. - After sealing at 100°C, the vessel contains only steam at 101.3 kPa. - Cooling to 20°C: - Most steam will condense, but the remaining vapor will occupy the same volume. - The final pressure is still the **saturation pressure at 20°C = 2.3 kPa** (as long as there's still liquid present). **So, the final pressure (rigid vessel) = 2.3 kPa** --- ## **Step 5: Compare pressures (semi-flexible vs. rigid)** - **Both cases:** Final pressure = **2.3 kPa** (the saturation pressure of water at 20°C) - **Why?**: The system always reaches equilibrium at the saturation pressure of water vapor at the final temperature, as long as there is liquid water present. --- ## **Step 6: Summary Table** | Case | Final Pressure (kPa) | Reason | |-------------------|---------------------|-----------------------------------------| | Semi-flexible | 2.3 | Saturation vapor pressure at 20°C | | Rigid (fixed V) | 2.3 | Saturation vapor pressure at 20°C | --- ## **Final Answers** > ### **Will the final pressure be the same, less than, or greater than the pressure assuming the can is rigid (i.e., fixed volume)?** - **The final pressure will be the SAME** in both the semi-flexible and rigid cases. - **Numerical value:** 2.3 kPa (the saturation vapor pressure of water at 20°C) --- **If you need a calculation for the amount of condensed water or volume, let me know!**