A sample of helium gas has a volume of 3.50 L at a pressure of 4.00 atm . The temperature and the number of gas particles do not change. Note: Reference the Conversion factors for non-SI units table for additional information. Part 1 of 2 What is the volume of gas (in L) at 11.6 atm ? Be sure your answer has the correct number of significant figures. What is the volume of gas (in L ) at $$ 590 . \mathrm{mmHg} $$ ? Be sure your answer has the correct number of significant figures. Part 2 of 2

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To solve these problems, we'll use **Boyle's Law**, which states that for a given amount of gas at constant temperature, the pressure and volume are inversely proportional. Mathematically, it's expressed as:

$$ P_1 V_1 = P_2 V_2 $$

Where:
- $$ P_1 $$ and $$ V_1 $$ are the initial pressure and volume.
- $$ P_2 $$ and $$ V_2 $$ are the final pressure and volume.

### Part 1: Volume at 11.6 atm

**Given:**
- $$ P_1 = 4.00 \, \text{atm} $$
- $$ V_1 = 3.50 \, \text{L} $$
- $$ P_2 = 11.6 \, \text{atm} $$

**Calculation:**
$$
V_2 = \frac{P_1 V_1}{P_2} = \frac{4.00 \, \text{atm} \times 3.50 \, \text{L}}{11.6 \, \text{atm}} = \frac{14.00 \, \text{atm·L}}{11.6 \, \text{atm}} \approx 1.2069 \, \text{L}
$$

**Significant Figures:**
All given values have **3 significant figures**, so the final answer should also have **3 significant figures**.

**Final Answer:**
$$
V_2 \approx 1.21 \, \text{L}
$$

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### Part 2: Volume at 590 mmHg

**Given:**
- $$ P_1 = 4.00 \, \text{atm} $$
- $$ V_1 = 3.50 \, \text{L} $$
- $$ P_2 = 590 \, \text{mmHg} $$

**Conversion:**
First, convert $$ 590 \, \text{mmHg} $$ to atmospheres (atm) using the conversion factor $$ 1 \, \text{atm} = 760 \, \text{mmHg} $$:

$$
P_2 = \frac{590 \, \text{mmHg}}{760 \, \text{mmHg/atm}} \approx 0.776 \, \text{atm}
$$

**Calculation:**
$$
V_2 = \frac{P_1 V_1}{P_2} = \frac{4.00 \, \text{atm} \times 3.50 \, \text{L}}{0.776 \, \text{atm}} \approx \frac{14.00 \, \text{atm·L}}{0.776 \, \text{atm}} \approx 18.05 \, \text{L}
$$

**Significant Figures:**
All given values (4.00 atm, 3.50 L, 590 mmHg) have **3 significant figures**, so the final answer should also have **3 significant figures**.

**Final Answer:**
$$
V_2 \approx 18.0 \, \text{L}
$$

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**Summary of Answers:**
1. **At 11.6 atm:** $$ V_2 = \mathbf{1.21 \, \text{L}} $$
2. **At 590 mmHg:** $$ V_2 = \mathbf{18.0 \, \text{L}} $$
At 11.6 atm, the volume is 1.21 L. At 590 mmHg, the volume is 18.0 L.

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