7L of gas is enclosed in a vessel at a pressure of 760 mm Hg. If temperature remain constant , calculate pressure when volume changes to 9 L .

Options

A. 606 mm Hg
B. 760 mm Hg
C. 591 mm Hg
D. 740 mm Hg

Answer

To solve the problem, we will use Boyle's Law, which states that for a given mass of gas at constant temperature, the product of pressure and volume is constant. This can be expressed mathematically as:

\[ P_1 V_1 = P_2 V_2 \]

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( P_2 \) = final pressure
- \( V_2 \) = final volume

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial pressure, \( P_1 = 760 \, \text{mm Hg} \)
- Initial volume, \( V_1 = 7 \, \text{L} \)
- Final volume, \( V_2 = 9 \, \text{L} \)

2. **Set Up the Equation Using Boyle's Law:**
Since we need to find the final pressure \( P_2 \), we rearrange the equation:
\[ P_2 = \frac{P_1 \times V_1}{V_2} \]

3. **Substitute the Known Values:**
\[ P_2 = \frac{760 \, \text{mm Hg} \times 7 \, \text{L}}{9 \, \text{L}} \]

4. **Calculate the Numerator:**
\[ 760 \times 7 = 5320 \, \text{mm Hg} \cdot \text{L} \]

5. **Divide by the Final Volume:**
\[ P_2 = \frac{5320 \, \text{mm Hg} \cdot \text{L}}{9 \, \text{L}} = 591.11 \, \text{mm Hg} \]

6. **Round to Appropriate Significant Figures:**
The final pressure \( P_2 \) is approximately:
\[ P_2 \approx 591 \, \text{mm Hg} \]

### Final Answer:
The pressure when the volume changes to 9 L is approximately **591 mm Hg**.