2L of gas is enclosed in a vessel at a pressure of 760 mm Hg. If temperature remain constant , calculate pressure when volume changes to 3 L .

Options

A. 506.6 mm Hg
B. 606 mm Hg
C. 459 mm Hg
D. 485 mm Hg

Explanation

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 a 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:

**Step 1: Identify the known values.**
- Initial pressure, \( P_1 = 760 \, \text{mm Hg} \)
- Initial volume, \( V_1 = 2 \, \text{L} \)
- Final volume, \( V_2 = 3 \, \text{L} \)

**Step 2: Use Boyle's Law to find the final pressure \( P_2 \).**
We can rearrange the equation to solve for \( P_2 \):

\[ P_2 = \frac{P_1 \times V_1}{V_2} \]

**Step 3: Substitute the known values into the equation.**
\[ P_2 = \frac{760 \, \text{mm Hg} \times 2 \, \text{L}}{3 \, \text{L}} \]

**Step 4: Calculate the value.**
\[ P_2 = \frac{1520 \, \text{mm Hg}}{3} = 506.67 \, \text{mm Hg} \]

**Step 5: Round the answer.**
The final pressure \( P_2 \) is approximately \( 506.6 \, \text{mm Hg} \).

### Final Answer:
The pressure when the volume changes to 3 L is approximately **506.6 mm Hg**.

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