If the solubility of a gas is 5.6 L at 505 kPa pressure, what is the solubility of the gas when the pressure is 1010 kPa? Show your work. Use equation editor.
If the solubility of a gas is 5.6 L at 505 kPa pressure, what is the solubility of the gas when the pressure is 1010 kPa? Show your work. Use equation editor.
Chapter6: The Systematic Approach To Equilibria: Solving Many Equations
Section: Chapter Questions
Problem 13P
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![### Problem: Solubility of a Gas under Different Pressures
If the solubility of a gas is 5.6 \( \frac{g}{L} \) at 505 kPa pressure, what is the solubility of the gas when the pressure is 1010 kPa? Show your work. Use equation editor.
#### Solution:
To solve this problem, we can use Henry's Law, which states that the solubility of a gas is directly proportional to its partial pressure. This can be expressed as:
\[ S_1 P_1 = S_2 P_2 \]
where:
- \( S_1 \) is the initial solubility,
- \( P_1 \) is the initial pressure,
- \( S_2 \) is the final solubility,
- \( P_2 \) is the final pressure.
Given:
- \( S_1 = 5.6 \frac{g}{L} \)
- \( P_1 = 505 \text{ kPa} \)
- \( P_2 = 1010 \text{ kPa} \)
We need to find \( S_2 \).
Rearranging Henry’s Law formula to solve for \( S_2 \):
\[ S_2 = \frac{S_1 P_1}{P_2} \]
Substitute the known values:
\[ S_2 = \frac{5.6 \frac{g}{L} \times 1010 \text{ kPa}}{505 \text{ kPa}} \]
\[ S_2 = \frac{5.6 \times 1010}{505} \]
\[ S_2 = 11.2 \frac{g}{L} \]
Therefore, the solubility of the gas at 1010 kPa pressure is \( 11.2 \frac{g}{L} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81d865ca-ed2c-4f6d-a4cc-7f843242dec1%2F902bba9d-b174-437b-b53d-02e0ccdc5588%2Fr4c34hl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem: Solubility of a Gas under Different Pressures
If the solubility of a gas is 5.6 \( \frac{g}{L} \) at 505 kPa pressure, what is the solubility of the gas when the pressure is 1010 kPa? Show your work. Use equation editor.
#### Solution:
To solve this problem, we can use Henry's Law, which states that the solubility of a gas is directly proportional to its partial pressure. This can be expressed as:
\[ S_1 P_1 = S_2 P_2 \]
where:
- \( S_1 \) is the initial solubility,
- \( P_1 \) is the initial pressure,
- \( S_2 \) is the final solubility,
- \( P_2 \) is the final pressure.
Given:
- \( S_1 = 5.6 \frac{g}{L} \)
- \( P_1 = 505 \text{ kPa} \)
- \( P_2 = 1010 \text{ kPa} \)
We need to find \( S_2 \).
Rearranging Henry’s Law formula to solve for \( S_2 \):
\[ S_2 = \frac{S_1 P_1}{P_2} \]
Substitute the known values:
\[ S_2 = \frac{5.6 \frac{g}{L} \times 1010 \text{ kPa}}{505 \text{ kPa}} \]
\[ S_2 = \frac{5.6 \times 1010}{505} \]
\[ S_2 = 11.2 \frac{g}{L} \]
Therefore, the solubility of the gas at 1010 kPa pressure is \( 11.2 \frac{g}{L} \).
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