Let's **solve question (2) with step-by-step explanation**: --- ## **Question (2) Restated** A semi-flexible vessel (4 L air at 1 atm) is heated to 100°C with the cap off, then sealed and cooled to 20°C. The vessel is partially crushed (volume decreases but not to zero) as it cools, equalizing some pressure with the atmosphere. **Will the final pressure inside be the same, less than, or greater than 1 atm? Also, compare this to the case of a rigid (fixed-volume) can.** --- ## **Step 1: Initial Conditions** - Initial volume \( V_1 = 4 \) L - Initial pressure \( P_1 = 1 \) atm - Initial temperature \( T_1 = 100^\circ \text{C} = 373 \) K (after heating and just before sealing) When the vessel is open and heated to 100°C, the air inside is at **atmospheric pressure** (1 atm), but at 100°C. **The cap is then sealed at this condition:** - \( P_{\text{seal}} = 1 \) atm - \( T_{\text{seal}} = 373 \) K - \( V_{\text{seal}} = V_1 \) --- ## **Step 2: Cooling with a Semi-Flexible Vessel** - Final temperature: \( T_2 = 20^\circ \text{C} = 293 \) K - The vessel is *semi-flexible*: as the air cools and its pressure drops below 1 atm, the vessel is partially crushed by the outside atmosphere, reducing its volume until the pressure inside matches the outside. ### **Key Point:** The vessel cannot maintain a pressure below atmospheric pressure (1 atm), because the external pressure will crush it until the pressures are equal! --- ### **What happens as it cools?** - The air inside **wants** to reach a lower pressure (by the ideal gas law), but - The vessel collapses until \( P_{\text{inside}} = P_{\text{outside}} = 1 \) atm. --- ## **Step 3: Final Pressure in the Semi-Flexible Vessel** **The final pressure, assuming the can does not leak, will be:** \[ \boxed{1 \text{ atm}} \] because the vessel will be crushed by the atmosphere until the pressure inside matches the outside pressure. --- ## **Step 4: Comparison with the Rigid Vessel** ### **Rigid Vessel (Fixed Volume):** Apply the **ideal gas law** (\( PV = nRT \)), with the number of moles \( n \) fixed. Let’s use ratios: \[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \] \[ P_2 = P_1 \times \frac{T_2}{T_1} \] \[ P_2 = 1 \text{ atm} \times \frac{293\text{ K}}{373\text{ K}} = .785 \text{ atm} \] So, for a **rigid can**, the final pressure would be **less than 1 atm** (specifically about .79 atm). --- ## **Step 5: Summary Table** | Case | Final Pressure | |-----------------------|----------------------------| | **Semi-flexible can** | **1 atm** | | **Rigid can** | **.79 atm (less than 1)** | --- ## **Final Answers** - **Semi-flexible can:** Final pressure is **the same as** the external pressure (1 atm). - **Rigid can:** Final pressure is **less than** 1 atm. - **Comparison:** The final pressure in the semi-flexible can is **greater than** the pressure in the rigid can. --- ### **Conceptual Explanation** - **Semi-flexible:** The can collapses until the pressure inside rises to match the atmosphere, so the final pressure can't go below 1 atm. - **Rigid:** The volume can't change, so the pressure drops as the temperature drops. --- ### **Final Boxed Answers** 1. **Final pressure in the semi-flexible can:** **Same as 1 atm** 2. **Compared to rigid can:** **Greater than the pressure in the rigid can (which is .79 atm)** --- Let me know if you'd like a diagram or further clarification!