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 the first three parts. e. Is the same amount of gas involved in these two situa- tions? Explain.

Question

UpStudy AI · Accepted Answer

Let's address part **(e)** of the problem:

**e. Is the same amount of gas involved in these two situations? Explain.**

**Answer:**

No, the same amount of gas is **not** involved in the two situations. Here's why:

1. **Ideal Gas Law Relationship:**
   
   The ideal gas law is given by:
   $$
   PV = nRT
   $$
   where:
   - $$ P $$ = pressure (constant at 1800 Pa)
   - $$ V $$ = volume
   - $$ n $$ = number of moles of gas
   - $$ R $$ = universal gas constant
   - $$ T $$ = temperature

2. **Comparing the Two Situations:**

- **First Situation:**
     - Initial Volume ($$ V_{1a} $$) = 0.17 m³
     - Temperature increases from 220 K to 1100 K
   
   - **Second Situation:**
     - Initial Volume ($$ V_{1d} $$) = 0.24 m³
     - Same temperature change from 220 K to 1100 K

3. **Determining the Number of Moles ($$ n $$):**
   
   For both situations, since pressure $$ P $$ and temperature change are the same, the initial number of moles can be compared using the ideal gas law rearranged as:
   $$
   n = \frac{PV}{RT}
   $$
   
   - **First Situation:**
     $$
     n_a = \frac{P V_{1a}}{R T_1} = \frac{1800 \times 0.17}{R \times 220}
     $$
   
   - **Second Situation:**
     $$
     n_d = \frac{P V_{1d}}{R T_1} = \frac{1800 \times 0.24}{R \times 220}
     $$
   
   - **Comparing $$ n_a $$ and $$ n_d $$:**
     $$
     \frac{n_d}{n_a} = \frac{V_{1d}}{V_{1a}} = \frac{0.24}{0.17} \approx 1.4118
     $$
     This means the second situation has approximately 1.41 times more moles of gas than the first situation.

4. **Conclusion:**

Since the initial volumes ($$ V_{1a} $$ and $$ V_{1d} $$) are different while pressure and temperature conditions are the same, the number of moles of gas in each situation must also be different. Therefore, the two situations involve different amounts of gas.

**Summary:**
The two scenarios have different initial volumes under the same pressure and temperature conditions, leading to different numbers of moles of gas. Consequently, the amount of gas involved in the two situations is not the same.
No, the two situations involve different amounts of gas because the initial volumes are different while pressure and temperature remain the same.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Mind Expander

preguntas relacionadas

Latest Physics Questions