Chapter 14: Problem 38
Challenge The solubility of a gas at 7 atm of pressure is 0.52 \(\mathrm{g} / \mathrm{L}\) . How many grams of the gas would be dissolved per 1 \(\mathrm{L}\) if the pressure was raised to 10 \(\mathrm{atm}\) ?
Short Answer
Expert verified
The solubility of the gas at 10 atm is 0.364 g/L. Thus, 0.364 grams of the gas would be dissolved per 1 L if the pressure was raised to 10 atm.
Step by step solution
01
List down the given values
We are given the following values:
- The solubility of the gas (\(C_1\)) at 7 atm (\(P_1\)) is 0.52 g/L
- The pressure (\(P_2\)) raised to 10 atm
We need to determine the solubility of the gas (\(C_2\)) at 10 atm.
02
Write down the Henry's Law formula
According to Henry's Law, the relationship between the solubility and pressure can be written as:
\(C_1P_1 = C_2P_2\)
03
Plug in the given values and solve for \(C_2\)
Substitute the given values in the formula:
\(C_1P_1 = C_2P_2\)
\((0.52\, \mathrm{g/L})(7\, \mathrm{atm}) = C_2(10\, \mathrm{atm})\)
Now solve for \(C_2\):
\(C_2 = \frac{(0.52\, \mathrm{g/L})(7\, \mathrm{atm})}{10\, \mathrm{atm}}\)
04
Calculate the solubility at 10 atm
Now, just perform the calculations:
\(C_2 = \frac{(0.52)(7)}{10}\)
\(C_2 = \frac{3.64}{10}\)
\(C_2 = 0.364\, \mathrm{g/L}\)
The solubility of the gas at 10 atm is 0.364 g/L. Thus, 0.364 grams of the gas would be dissolved per 1 L if the pressure was raised to 10 atm.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gas Solubility
Gas solubility refers to how much of a gas can dissolve in a liquid under certain conditions. It is measured as the amount of gas (in grams) dissolved in a specific volume of liquid (usually 1 liter). The higher the solubility, the more gas can be dissolved in the liquid. For gases, solubility is directly related to the temperature and pressure of the system.
When considering solubility, one must know that different gases have varying solubility levels depending on their nature. For instance, carbon dioxide is quite soluble in water, whereas oxygen is less so. Solubility can change when the temperature or pressure changes, which is critical in many industrial and natural processes.
When considering solubility, one must know that different gases have varying solubility levels depending on their nature. For instance, carbon dioxide is quite soluble in water, whereas oxygen is less so. Solubility can change when the temperature or pressure changes, which is critical in many industrial and natural processes.
- Temperature Dependency: Unlike solids, gases become less soluble as the temperature increases because gases tend to expand and escape from liquids.
- Pressure Dependency: According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.
Pressure Changes
Pressure changes significantly affect the solubility of gases in liquids. According to Henry's Law, an increase in the pressure of a gas above a liquid will increase the solubility of the gas in that liquid. This principle is crucial in understanding the behavior of gases in various environments.
Henry's Law is formulated as:
\[ C = kP \]Where:
Henry's Law is formulated as:
\[ C = kP \]Where:
- \( C \) is the solubility of the gas
- \( k \) is Henry's Law constant
- \( P \) is the partial pressure of the gas
Solubility Calculation
Solubility calculation often involves using formulas derived from Henry's Law. Calculating solubility helps us predict how much of a gas can dissolve in a liquid at specific conditions. The exercise provided is a classic example of calculating solubility changes due to pressure adjustments.
To perform such calculations, we set up an equation using the initial and final conditions of the gas pressure and its solubility:
\[ C_1P_1 = C_2P_2 \]Given:
\[ C_2 = \frac{(0.52 \, \text{g/L})(7 \, \text{atm})}{10 \, \text{atm}} = 0.364 \, \text{g/L} \]This straightforward calculation shows how increased pressure changes the solubility of a gas, giving a practical application of Henry's Law.
To perform such calculations, we set up an equation using the initial and final conditions of the gas pressure and its solubility:
\[ C_1P_1 = C_2P_2 \]Given:
- \( C_1 = 0.52 \, \text{g/L} \) at \( P_1 = 7 \, \text{atm} \)
- \( P_2 = 10 \, \text{atm} \)
\[ C_2 = \frac{(0.52 \, \text{g/L})(7 \, \text{atm})}{10 \, \text{atm}} = 0.364 \, \text{g/L} \]This straightforward calculation shows how increased pressure changes the solubility of a gas, giving a practical application of Henry's Law.
Chemistry Problems
Chemistry problems involving gas solubility often test your understanding of core concepts like Henry's Law, pressure impacts, and solubility relationships. Solving these problems requires a systematic approach, starting with identifying what is known and what needs to be found.
Typical steps include:
Typical steps include:
- Identify Variables: Note the initial conditions, such as initial pressure and solubility, and the target conditions.
- Use Formulas Appropriately: Apply the correct formulas, being careful to follow units and conversion factors.
- Check for Consistency: After calculations, review results to ensure they make physical sense.