Consider the apparatus shown below in which the valve is closed. If the temperature is kept constant and the valve opened, what will be the pressure inside the apparatus? H₂ Vacuum Valve 9.00 L 3.00 L 1.00 atm 0.250 atm 1.00 atm 0.500 atm cannot be determined

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Chapter13: Gases
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### Understanding Gas Laws: An Example Problem

Consider the apparatus shown below:

- A container filled with hydrogen gas (H₂) is connected to a vacuum chamber.
- The volume of the H₂ container is given as 3.00 L and its pressure is 1.00 atm.
- The volume of the vacuum chamber is given as 9.00 L.

**Question:** If the temperature remains constant and the valve between the H₂ container and the vacuum chamber is opened, what will be the final pressure inside the entire apparatus?

![Apparatus Diagram](yourimageurl)

**Details of the Apparatus:**
- **H₂ Container:** 3.00 L, 1.00 atm
- **Vacuum Chamber:** 9.00 L, initially at 0.00 atm (since it’s a vacuum)

**Answer Choices:**
- ⓐ 0.250 atm
- ⓑ 1.00 atm
- ⓒ 0.500 atm
- ⓓ cannot be determined

**Solution Explanation:**

To determine the final pressure inside the combined apparatus, apply the principle of conservation of mass (moles of gas) and Boyle’s Law:

Boyle’s Law states: \( P_1V_1 = P_2V_2 \)

Here:
- Initial Pressure, \( P_1 \) = 1.00 atm
- Initial Volume, \( V_1 \) = 3.00 L
- Final Volume, \( V_2 \) = (3.00 L + 9.00 L) = 12.00 L
- Final Pressure, \( P_2 \) = ?

Using the formula:

\[ P_1 \times V_1 = P_2 \times V_2 \]
\[ 1.00 \, \text{atm} \times 3.00 \, \text{L} = P_2 \times 12.00 \, \text{L} \]
\[ P_2 = \frac{1.00 \, \text{atm} \times 3.00 \, \text{L}}{12.00 \, \text{L}} \]
\[ P_2 = 0.250 \, \text{atm} \]

**Correct Answer: ⓐ 0.250 atm**

### Conclusion
Opening the valve allows the hydrogen gas
Transcribed Image Text:### Understanding Gas Laws: An Example Problem Consider the apparatus shown below: - A container filled with hydrogen gas (H₂) is connected to a vacuum chamber. - The volume of the H₂ container is given as 3.00 L and its pressure is 1.00 atm. - The volume of the vacuum chamber is given as 9.00 L. **Question:** If the temperature remains constant and the valve between the H₂ container and the vacuum chamber is opened, what will be the final pressure inside the entire apparatus? ![Apparatus Diagram](yourimageurl) **Details of the Apparatus:** - **H₂ Container:** 3.00 L, 1.00 atm - **Vacuum Chamber:** 9.00 L, initially at 0.00 atm (since it’s a vacuum) **Answer Choices:** - ⓐ 0.250 atm - ⓑ 1.00 atm - ⓒ 0.500 atm - ⓓ cannot be determined **Solution Explanation:** To determine the final pressure inside the combined apparatus, apply the principle of conservation of mass (moles of gas) and Boyle’s Law: Boyle’s Law states: \( P_1V_1 = P_2V_2 \) Here: - Initial Pressure, \( P_1 \) = 1.00 atm - Initial Volume, \( V_1 \) = 3.00 L - Final Volume, \( V_2 \) = (3.00 L + 9.00 L) = 12.00 L - Final Pressure, \( P_2 \) = ? Using the formula: \[ P_1 \times V_1 = P_2 \times V_2 \] \[ 1.00 \, \text{atm} \times 3.00 \, \text{L} = P_2 \times 12.00 \, \text{L} \] \[ P_2 = \frac{1.00 \, \text{atm} \times 3.00 \, \text{L}}{12.00 \, \text{L}} \] \[ P_2 = 0.250 \, \text{atm} \] **Correct Answer: ⓐ 0.250 atm** ### Conclusion Opening the valve allows the hydrogen gas
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