Problem 36 - Chemistry Matter and Change

Question

If 0.55 g of a gas dissolves in 1.0 \(\mathrm{L}\) of water at 20.0 \(\mathrm{kPa}\) of pressure, how much will dissolve at 110.0 \(\mathrm{kPa}\) of pressure?

Step-by-Step Solution

Answer

At 110 kPa, 3.025 g of the gas will dissolve in 1.0 L of water.

Step 1: Understand Henry's Law

Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the liquid. Mathematically, it can be expressed as \( S_1/P_1 = S_2/P_2 \), where \( S_1 \) and \( S_2 \) are the solubilities of the gas at pressures \( P_1 \) and \( P_2 \), respectively.

Step 2: Identify Known Values

From the problem statement, we know: \( S_1 = 0.55 \, \text{g/L} \), \( P_1 = 20.0 \, \text{kPa} \), and \( P_2 = 110.0 \, \text{kPa} \). We need to find \( S_2 \), the solubility at \( P_2 \).

Step 3: Set Up the Equation

Using Henry's Law formula, set up the equation: \( \frac{0.55}{20} = \frac{S_2}{110} \).

Step 4: Solve for S2

To find \( S_2 \), rearrange the equation: \( S_2 = \frac{0.55 \times 110}{20} \).

Step 5: Calculate S2

Perform the calculation: \( S_2 = \frac{0.55 \times 110}{20} = 3.025 \, \text{g/L} \).

Key Concepts

Solubility of Gases

The solubility of gases refers to the measure of how much gas can dissolve in a liquid under specific conditions. Imagine adding sugar to tea — similar to that, gases can dissolve into liquids up to a certain limit, known as the solubility limit. For gases, this solubility depends greatly on the temperature and pressure of the surroundings. In general, as the temperature increases, gas solubility decreases. That's why you may notice carbonated drinks losing their fizz faster in the summer heat. With gases, though, pressure plays an equally crucial role, specifically illustrated by Henry's Law.

Pressure and Solubility

Pressure has a direct relationship with the solubility of gases in liquids. According to Henry's Law, if you increase the pressure over a liquid, more gas molecules are 'pushed' into the liquid, which increases its solubility. Let's connect this to the exercise: initially, at 20 kPa, 0.55 g of gas dissolves in water. When pressure is increased to 110 kPa, more gas dissolves, because higher pressure means more gas molecules are available above the liquid, leading to greater solubility. Key takeaways include: - More pressure results in greater solubility of the gas. - The relationship is linear, meaning doubling the pressure doubles the solubility, holding everything else constant.

Chemistry Problem Solving

Solving chemistry problems like this one involves understanding and applying the basic principles of chemistry, such as Henry's Law. Breaking down the problem step by step helps tackle such exercises efficiently:

• Identify what information you know. For example, in this problem, the knowns were solubility at specific pressures.
• Apply the correct formula. Use Henry's Law here, which can be symbolized as \( S_1/P_1 = S_2/P_2 \).
• Rearrange the formula appropriately to isolate the unknown variable, like rearranging to solve for \( S_2 \).
• Perform the necessary calculations. Calculations step is where all setup work bears fruit. Here, \( S_2 = \frac{0.55 \times 110}{20} = 3.025 \text{ g/L} \).

By tackling chemistry problems in a methodical way, you'll ensure that even complex problems become manageable. Linking real-world intuition with formulas will enrich your understanding and help you solve even trickier problems efficiently.