A sample of an ideal gas at 1.00 atm and a volume of 1.67 L was placed in a weighted balloon and dropped into the ocean. As the sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 70.0 atm, what was the volume of the sample? Assume that the temperature was held constant. V = 41.9 L

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Chapter1: Chemical Foundations
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### Problem Statement

A sample of an ideal gas at 1.00 atm and a volume of 1.67 L was placed in a weighted balloon and dropped into the ocean. As the sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 70.0 atm, what was the volume of the sample? Assume that the temperature was held constant.

### Solution

To solve this problem, we can use Boyle's Law, which states that for a given mass of an ideal gas kept at a constant temperature, the product of the pressure and volume is constant:

\[ P_1 \times V_1 = P_2 \times V_2 \]

Where:
- \( P_1 = 1.00 \) atm (initial pressure)
- \( V_1 = 1.67 \) L (initial volume)
- \( P_2 = 70.0 \) atm (final pressure)
- \( V_2 = ? \) L (final volume)

Solving for \( V_2 \):

\[ V_2 = \frac{P_1 \times V_1}{P_2} \]
\[ V_2 = \frac{1.00 \text{ atm} \times 1.67 \text{ L}}{70.0 \text{ atm}} \]
\[ V_2 = \frac{1.67}{70.0} \]
\[ V_2 = 0.0238571 \text{ L} \]

Therefore,

\[ V = 0.024 \text{ L} \]

However, the provided answer in the box is \( 41.9 \) L, which does not align with the expected solution of the equation given. This suggests there might be an error in either the data provided, the answer expected, or the calculations. Please cross-verify all values and calculations to ensure accuracy.
Transcribed Image Text:### Problem Statement A sample of an ideal gas at 1.00 atm and a volume of 1.67 L was placed in a weighted balloon and dropped into the ocean. As the sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 70.0 atm, what was the volume of the sample? Assume that the temperature was held constant. ### Solution To solve this problem, we can use Boyle's Law, which states that for a given mass of an ideal gas kept at a constant temperature, the product of the pressure and volume is constant: \[ P_1 \times V_1 = P_2 \times V_2 \] Where: - \( P_1 = 1.00 \) atm (initial pressure) - \( V_1 = 1.67 \) L (initial volume) - \( P_2 = 70.0 \) atm (final pressure) - \( V_2 = ? \) L (final volume) Solving for \( V_2 \): \[ V_2 = \frac{P_1 \times V_1}{P_2} \] \[ V_2 = \frac{1.00 \text{ atm} \times 1.67 \text{ L}}{70.0 \text{ atm}} \] \[ V_2 = \frac{1.67}{70.0} \] \[ V_2 = 0.0238571 \text{ L} \] Therefore, \[ V = 0.024 \text{ L} \] However, the provided answer in the box is \( 41.9 \) L, which does not align with the expected solution of the equation given. This suggests there might be an error in either the data provided, the answer expected, or the calculations. Please cross-verify all values and calculations to ensure accuracy.
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