Question

If 0.55 g of a gas dissolves in 1.0 L of water at 20.0 kPa of pressure, how much will dissolve at 110.0 kPa of pressure

Short answer

The solubility of the gas in water at an increased pressure of 110.0 kPa will be approximately 3.025 g/L.

Step-by-step solution

Identify the known variables

First, identify the known variables from the problem statement. - Solubility1 (initial solubility) = 0.55 g/L - Pressure1 (initial pressure) = 20.0 kPa - Pressure2 (final pressure) = 110.0 kPa The unknown variable is Solubility2 (final solubility).

Write down the formula for Henry's Law

According to Henry's Law, the solubility of a gas is directly proportional to its partial pressure. So, we can write the formula as follows: Solubility1 / Pressure1 = Solubility2 / Pressure2

Substitute the known values into the formula and solve for Solubility2

Now substitute the known values into the formula and solve for the unknown variable Solubility2 (final solubility): \( (\frac{0.55\ g/L}{20\ kPa}) = (\frac{Solubility2}{110 kPa}) \) To find the value of Solubility2 (final solubility), first multiply both sides of the equation by 110 kPa: \( Solubility2 = (\frac{0.55 g/L}{20 kPa}) \times 110\ kPa \)

Calculate the final solubility

Now, evaluate the expression to find the final solubility Solubility2: \(Solubility2 = (\frac{0.55}{20}) \times 110 \) \(Solubility2 ≈ 3.025\ g/L\) So, the solubility of the gas in water at an increased pressure of 110.0 kPa will be approximately 3.025 g/L.

Key concepts

Solubility of Gases

Understanding the solubility of gases in liquids is essential for a variety of applications, ranging from industrial processes to the physiological behavior of gases in our bloodstream. Solubility refers to how well a gas can be dissolved in a liquid under certain conditions. This phenomenon is influenced by factors such as temperature, pressure, and the nature of the gas and liquid involved.

Henry's Law provides a valuable way to quantify this relationship, stating that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. This implies that as the pressure of the gas increases, so does its solubility, assuming the temperature remains constant. Thus, under increased pressure conditions, more gas will dissolve until a new equilibrium is achieved. The exercise provided illustrates how the solubility changes with varying pressure, highlighting the practical application of this law.

Partial Pressure

Partial pressure is a concept that can sometimes confuse students, but it is a cornerstone of understanding gas behavior. In essence, the partial pressure of any gas is the pressure it would exert if it alone occupied the entire volume of the mixture at the same temperature. It's a way of describing how the total pressure of a gas mixture, such as air, is the sum of the pressures of each constituent gas.

In the context of Henry's Law, we are concerned with the partial pressure of the gas in contact with the liquid surface, because it is this pressure that drives the gas into solution. The step in the exercise where we consider the changes in partial pressure is crucial: it directly influences the solubility of the gas. This is a key factor when calculating how much gas will dissolve in a liquid under different pressure conditions.

Gas Laws

Gas laws are fundamental principles that describe the behavior of gases and how they react to changes in temperature, volume, and pressure. These laws lay the foundation for understanding various phenomena in chemistry and physics, and Henry's Law is one of these pivotal principles, related specifically to the solubility of gases.

Other relevant gas laws include Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at a constant temperature, and Charles's Law, which shows that volume and temperature are directly related under constant pressure. The Ideal Gas Law combines these relationships and adds in the variable of quantity of gas, giving us a comprehensive equation to describe a gas's state.

The exercise demonstrates how an understanding of these laws is not just theoretical but has real-world applications. It helps explain why gases dissolve more at higher pressures (as stated by Henry's Law) and how the partial pressures impact this process—firmly rooting the behavior of gases within the predictable laws that govern their properties.