A balloon is filled with 3.00L of helium at a pressure of 765 torr. What is the volume of the balloon at an altitude where the pressure is 530 torr if the temperature remains constant? O 0.231L O2.08L O 1.00L O 4.33L

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### Gas Laws in Action: Solving Volumetric Changes of a Helium Balloon

A balloon is filled with 3.00L of helium at a pressure of 765 torr. What is the volume of the balloon at an altitude where the pressure is 530 torr if the temperature remains constant?

#### Options:
- \( \circ \) 0.231L
- \( \circ \) 2.08L
- \( \circ \) 1.00L
- \( \circ \) 4.33L

#### Explanation:

This problem can be solved using **Boyle's Law**, which states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. Mathematically, it is represented as:

\[ P_1 V_1 = P_2 V_2 \]

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

Given:
- \( P_1 = 765 \, \text{torr} \)
- \( V_1 = 3.00 \, \text{L} \)
- \( P_2 = 530 \, \text{torr} \)

We need to find \( V_2 \), the final volume of the balloon.

Rearranging the equation to solve for \( V_2 \):
\[ V_2 = \frac{P_1 V_1}{P_2} \]

Substitute the known values:
\[ V_2 = \frac{765 \, \text{torr} \times 3.00 \, \text{L}}{530 \, \text{torr}} \]

\[ V_2 = \frac{2295}{530} \]

\[ V_2 \approx 4.33 \, \text{L} \]

Therefore, the correct answer is:
- \( \circ \) 4.33L
Transcribed Image Text:### Gas Laws in Action: Solving Volumetric Changes of a Helium Balloon A balloon is filled with 3.00L of helium at a pressure of 765 torr. What is the volume of the balloon at an altitude where the pressure is 530 torr if the temperature remains constant? #### Options: - \( \circ \) 0.231L - \( \circ \) 2.08L - \( \circ \) 1.00L - \( \circ \) 4.33L #### Explanation: This problem can be solved using **Boyle's Law**, which states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. Mathematically, it is represented as: \[ P_1 V_1 = P_2 V_2 \] Where: - \( P_1 \) is the initial pressure - \( V_1 \) is the initial volume - \( P_2 \) is the final pressure - \( V_2 \) is the final volume Given: - \( P_1 = 765 \, \text{torr} \) - \( V_1 = 3.00 \, \text{L} \) - \( P_2 = 530 \, \text{torr} \) We need to find \( V_2 \), the final volume of the balloon. Rearranging the equation to solve for \( V_2 \): \[ V_2 = \frac{P_1 V_1}{P_2} \] Substitute the known values: \[ V_2 = \frac{765 \, \text{torr} \times 3.00 \, \text{L}}{530 \, \text{torr}} \] \[ V_2 = \frac{2295}{530} \] \[ V_2 \approx 4.33 \, \text{L} \] Therefore, the correct answer is: - \( \circ \) 4.33L
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