Problem 37 - Chemistry Matter and Change

Question

A gas has a solubility of 0.66 \(\mathrm{g} / \mathrm{L}\) at 10.0 atm of pressure. What is the pressure on a \(1.0-\mathrm{L}\) sample that contains 1.5 \(\mathrm{g}\) of gas?

Step-by-Step Solution

Answer

The pressure is 22.73 atm.

Step 1: Understanding the Initial Conditions

We are given that a gas has a solubility of 0.66 g/L at a pressure of 10.0 atm. This means that at 10.0 atm, 0.66 grams of the gas can dissolve in 1 liter of solution.

Step 2: Determining Solubility under Different Conditions

Using Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid (S = kP), we can find the solubility at other pressures. Here, \[ k = \frac{S_1}{P_1} = \frac{0.66 \, \mathrm{g/L}}{10.0 \, \mathrm{atm}} = 0.066 \, \mathrm{g/(L \, atm)} \]

Step 3: Solving for the New Pressure

We want to find the pressure at which 1.5 grams of the gas dissolves in 1 liter. This becomes: \[ S_2 = \frac{1.5 \, \mathrm{g}}{1.0 \, \mathrm{L}} = 1.5 \, \mathrm{g/L} \] Using Henry's Law, solve for the new pressure, \( P_2 \): \[ k = \frac{S_2}{P_2} \rightarrow P_2 = \frac{S_2}{k} = \frac{1.5 \, \mathrm{g/L}}{0.066 \, \mathrm{g/(L \, atm)}} = 22.73 \, \mathrm{atm} \]

Step 4: Conclusion and Result

Therefore, at 22.73 atm, 1.5 grams of the gas would be soluble in 1 liter of the solution.

Key Concepts

Gas Solubility

Solubility is a measure of how much of a substance can dissolve in a liquid to form a homogeneous solution. For gases, solubility is often expressed as grams of gas per liter of liquid, indicating how much gas can be dissolved in a given volume of liquid at a specific condition.

When dealing with gases, the solubility can be influenced by several factors such as temperature and pressure of the liquid. In general, the solubility of most gases in water decreases with an increase in temperature. This is because higher temperatures give gas molecules more energy to escape from the liquid form back into the gas phase.

To find a gas's solubility, consider its ability to dissolve until no more gas can be accepted by the liquid without forming bubbles or escaping. In the original problem, the solubility of the gas is given as 0.66 g/L at 10 atm, meaning that this is the maximum amount of gas that can remain dissolved without escaping at this pressure and temperature.

Pressure and Solubility Relationship

Henry's Law describes the direct relationship between the pressure of a gas and its solubility in a liquid. Simply put, according to Henry's Law:

• \( S = kP \)

where \( S \) is the solubility, \( k \) is Henry's Law constant, and \( P \) is the pressure. This equation tells us that when pressure increases, solubility also increases, assuming the temperature remains constant.

This principle is why carbonated beverages are bottled under high pressure; more carbon dioxide can dissolve in the drink than would be possible under normal atmospheric pressure. When you open a bottle, the pressure decreases, and gas bubbles escape, resulting in the fizzy sensation.

In the original exercise, the pressure needed to dissolve 1.5 grams of gas in 1 liter of liquid is higher than the initial 10 atm required for 0.66 g/L, as calculated using Henry's Law. This demonstrates the direct relationship where increasing gas content in a solution necessitates higher pressure.

Chemistry Mathematical Problems

Solving chemistry problems often involves applying formulas and concepts to find unknown values based on given data. Understanding the problem, identifying what's given and what's needed are crucial first steps.

In the original exercise, we started with a known solubility and pressure, using these to find a constant (\( k \)) that can be applied to find unknown quantities. The problem involved algebraic manipulation to solve for the new pressure (\( P_2 \)) required to dissolve a different amount of gas.

To solve such problems:

• First, understand the formula: Here it was \( S = kP \).
• Exchange known values to find the constant \( k \)
• Substitute into the formula to find unknowns (\( P_2 \)).

These steps are essential in approaching chemistry mathematical problems effectively. By breaking down the problem and using known principles like Henry's Law, solving complex questions becomes more approachable.