2. Two vessels of equal volume are perfectly insulated and contain the gases at the temperatures described in the boxes. They are placed together, and the partition removed. initial: n = 3.0 mol Ar n= 1.0 mol Ne V = 1 L V = 1L T= 358K T= 298 K final: n = 3.0 mol Ar + 1.0 mol Ne V = 2 L Calculate the change in entropy of the system for this whole process. Assume ideal gas behavior, no heat is lost to the surroundings, and that the heat capacity is independent of temperature. HINT 1: notice that the initial T and P of each gas is different. Think about what is happening to the system in terms of T, P and V for each gas when the 2 containers are placed together HINT 2: Remember that entropy is a state function, so AS is independent of which path you chose, so you can split it up into whichever pathway is most convenient for you to calculate.

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### Problem 2: Entropy Change Calculation

Two vessels of equal volume are perfectly insulated and contain gases at the specified temperatures as described in the boxes. The vessels are placed together, and the partition is removed.

#### Initial Conditions:
- **Vessel 1:**
  - Gas: Ar (Argon)
  - Amount: \( n = 3.0 \, \text{mol} \)
  - Volume: \( V = 1\, \text{L} \)
  - Temperature: \( T = 358\, \text{K} \)

- **Vessel 2:**
  - Gas: Ne (Neon)
  - Amount: \( n = 1.0 \, \text{mol} \)
  - Volume: \( V = 1\, \text{L} \)
  - Temperature: \( T = 298\, \text{K} \)

#### Final Conditions:
- Combined contents:
  - Gases: 3.0 mol Ar and 1.0 mol Ne
  - Total Volume: \( V = 2\, \text{L} \)

### Task:
Calculate the change in entropy of the system for the entire process.

### Assumptions:
- Assume ideal gas behavior.
- No heat is lost to the surroundings.
- Heat capacity is independent of temperature.

### Hints:
- **Hint 1:** Notice that the initial temperature and pressure of each gas are different. Consider what happens to the system's temperature (T), pressure (P), and volume (V) for each gas when the two containers are combined.
  
- **Hint 2:** Remember that entropy is a state function, so the change in entropy (\( \Delta S \)) is independent of the path taken. You can split it into any pathway that is most convenient for calculation.
Transcribed Image Text:### Problem 2: Entropy Change Calculation Two vessels of equal volume are perfectly insulated and contain gases at the specified temperatures as described in the boxes. The vessels are placed together, and the partition is removed. #### Initial Conditions: - **Vessel 1:** - Gas: Ar (Argon) - Amount: \( n = 3.0 \, \text{mol} \) - Volume: \( V = 1\, \text{L} \) - Temperature: \( T = 358\, \text{K} \) - **Vessel 2:** - Gas: Ne (Neon) - Amount: \( n = 1.0 \, \text{mol} \) - Volume: \( V = 1\, \text{L} \) - Temperature: \( T = 298\, \text{K} \) #### Final Conditions: - Combined contents: - Gases: 3.0 mol Ar and 1.0 mol Ne - Total Volume: \( V = 2\, \text{L} \) ### Task: Calculate the change in entropy of the system for the entire process. ### Assumptions: - Assume ideal gas behavior. - No heat is lost to the surroundings. - Heat capacity is independent of temperature. ### Hints: - **Hint 1:** Notice that the initial temperature and pressure of each gas are different. Consider what happens to the system's temperature (T), pressure (P), and volume (V) for each gas when the two containers are combined. - **Hint 2:** Remember that entropy is a state function, so the change in entropy (\( \Delta S \)) is independent of the path taken. You can split it into any pathway that is most convenient for calculation.
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