added? Explain. SP6. Suppose the pressure of an ideal gas mixture remains con- stant at $$ 1800 \mathrm{~Pa}\left(1 \mathrm{~Pa}=1 \mathrm{~N} / \mathrm{m}^{2}\right) $$ and the temperature is in- creased from 220 K to 1100 K . a. If the original volume of the gas was $$ 0.17 \mathrm{~m}^{3} $$, what is the final volume? (See example box 10.3 .) b. What is the change in volume $$ \Delta V $$ for this process? c. How much work does the gas do on the surroundings during the expansion? d. If the initial volume was $$ 0.24 \mathrm{~m}^{3} $$ and the same tempera- ture change occurred, would the work done be the same as in the first case? Show by repeating the steps of th first three parts. e. Is the same amount of gas involved in these two situ tions? Explain.

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UpStudy AI · Accepted Answer

Sure, let's solve each part of the problem step by step. We'll use the **Ideal Gas Law**, which is fundamental in thermodynamics for relating pressure ($$ P $$), volume ($$ V $$), temperature ($$ T $$), and the number of moles of gas ($$ n $$):

$$
PV = nRT
$$

Where:
- $$ R $$ is the universal gas constant ($$ R \approx 8.314 \, \text{J/mol·K} $$)
- All units should be consistent (e.g., pressure in Pascals, volume in cubic meters, temperature in Kelvin)

Given:
- Pressure ($$ P $$) = 1800 Pa (constant)
- Initial Temperature ($$ T_1 $$) = 220 K
- Final Temperature ($$ T_2 $$) = 1100 K

### **Part a. Final Volume Calculation**

**Given:**
- Initial Volume ($$ V_1 $$) = 0.17 m³
- Temperature increases from $$ T_1 $$ to $$ T_2 $$ at constant pressure.

**Using the Ideal Gas Law:**
Since pressure is constant, we can relate the initial and final states as follows:

$$
\frac{V_1}{T_1} = \frac{V_2}{T_2}
$$

**Solving for Final Volume ($$ V_2 $$):**
$$
V_2 = V_1 \times \frac{T_2}{T_1} = 0.17 \, \text{m}^3 \times \frac{1100 \, \text{K}}{220 \, \text{K}} = 0.17 \times 5 = 0.85 \, \text{m}^3
$$

**Answer:**
The final volume ($$ V_2 $$) is **0.85 m³**.

---

### **Part b. Change in Volume ($$ \Delta V $$)**

**Calculation:**
$$
\Delta V = V_2 - V_1 = 0.85 \, \text{m}^3 - 0.17 \, \text{m}^3 = 0.68 \, \text{m}^3
$$

**Answer:**
The change in volume ($$ \Delta V $$) is **0.68 m³**.

---

### **Part c. Work Done by the Gas During Expansion**

**Concept:**
Work done ($$ W $$) by an expanding gas at constant pressure is given by:
$$
W = P \times \Delta V
$$

**Given:**
- Pressure ($$ P $$) = 1800 Pa
- Change in Volume ($$ \Delta V $$) = 0.68 m³

**Calculation:**
$$
W = 1800 \, \text{Pa} \times 0.68 \, \text{m}^3 = 1224 \, \text{J}
$$

**Answer:**
The gas does **1224 Joules** of work on the surroundings during the expansion.

---

### **Part d. Work Done with a Different Initial Volume**

**Given:**
- New Initial Volume ($$ V_1' $$) = 0.24 m³
- Same Pressure ($$ P $$) = 1800 Pa
- Temperature change from $$ T_1 $$ = 220 K to $$ T_2 $$ = 1100 K

**Step 1: Calculate Final Volume ($$ V_2' $$)**
$$
V_2' = V_1' \times \frac{T_2}{T_1} = 0.24 \, \text{m}^3 \times \frac{1100 \, \text{K}}{220 \, \text{K}} = 0.24 \times 5 = 1.2 \, \text{m}^3
$$

**Step 2: Change in Volume ($$ \Delta V' $$)**
$$
\Delta V' = V_2' - V_1' = 1.2 \, \text{m}^3 - 0.24 \, \text{m}^3 = 0.96 \, \text{m}^3
$$

**Step 3: Work Done ($$ W' $$)**
$$
W' = P \times \Delta V' = 1800 \, \text{Pa} \times 0.96 \, \text{m}^3 = 1728 \, \text{J}
$$

**Comparison:**
- **First Case Work ($$ W $$)** = 1224 J
- **Second Case Work ($$ W' $$)** = 1728 J

**Conclusion:**
The work done in the second case (**1728 J**) is **greater** than in the first case (**1224 J**).

**Answer:**
No, the work done is not the same. With an initial volume of 0.24 m³, the gas performs **1728 Joules** of work, compared to **1224 Joules** in the first case.

---

### **Part e. Amount of Gas Involved in Both Situations**

**Concept:**
From the Ideal Gas Law:
$$
n = \frac{PV}{RT}
$$

Where $$ n $$ is the number of moles.

**First Situation:**
$$
n_1 = \frac{P V_1}{R T_1} = \frac{1800 \, \text{Pa} \times 0.17 \, \text{m}^3}{R \times 220 \, \text{K}}
$$

**Second Situation:**
$$
n_2 = \frac{P V_1'}{R T_1} = \frac{1800 \, \text{Pa} \times 0.24 \, \text{m}^3}{R \times 220 \, \text{K}}
$$

**Ratio of Moles:**
$$
\frac{n_2}{n_1} = \frac{V_1'}{V_1} = \frac{0.24}{0.17} \approx 1.4118
$$

This indicates that approximately **1.41 times more** moles of gas are present in the second situation compared to the first.

**Conclusion:**
Different amounts of gas are involved in the two situations.

**Answer:**
**No**, the two scenarios involve different amounts of gas. The second situation contains approximately **1.41 times more** gas than the first.
**Part a:** The final volume is 0.85 m³. **Part b:** The change in volume is 0.68 m³. **Part c:** The gas does 1224 Joules of work. **Part d:** With an initial volume of 0.24 m³, the work done is 1728 Joules, which is more than the first case. **Part e:** Different amounts of gas are involved in the two situations.

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