Gas Laws Practice Problems And Answers

Gas Laws Practice Problems and Answers: A Comprehensive Guide

Understanding gas laws is crucial in chemistry and physics. These laws describe the relationship between pressure, volume, temperature, and the amount of gas in a system. Mastering these concepts requires practice, and that's where solving problems comes in. This comprehensive guide provides a wide range of gas law practice problems with detailed answers and explanations, helping you solidify your understanding and improve your problem-solving skills.

Understanding the Key Gas Laws

Before diving into the problems, let's briefly review the fundamental gas laws:

1. Boyle's Law: This law states that at a constant temperature, the volume of a gas is inversely proportional to its pressure. Mathematically, this is represented as:

P₁V₁ = P₂V₂

where:

• P₁ = initial pressure
• V₁ = initial volume
• P₂ = final pressure
• V₂ = final volume

2. Charles's Law: This law states that at a constant pressure, the volume of a gas is directly proportional to its absolute temperature (in Kelvin). The formula is:

V₁/T₁ = V₂/T₂

where:

• V₁ = initial volume
• T₁ = initial absolute temperature (Kelvin)
• V₂ = final volume
• T₂ = final absolute temperature (Kelvin)

3. Gay-Lussac's Law: This law states that at a constant volume, the pressure of a gas is directly proportional to its absolute temperature. The formula is:

P₁/T₁ = P₂/T₂

where:

• P₁ = initial pressure
• T₁ = initial absolute temperature (Kelvin)
• P₂ = final pressure
• T₂ = final absolute temperature (Kelvin)

4. Avogadro's Law: This law states that at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas. The formula is:

V₁/n₁ = V₂/n₂

where:

• V₁ = initial volume
• n₁ = initial number of moles
• V₂ = final volume
• n₂ = final number of moles

5. The Ideal Gas Law: This law combines Boyle's, Charles's, and Avogadro's laws into a single equation:

PV = nRT

where:

• P = pressure
• V = volume
• n = number of moles
• R = ideal gas constant (0.0821 L·atm/mol·K or 8.314 J/mol·K)
• T = absolute temperature (Kelvin)

This law is applicable under ideal conditions, where gas molecules are assumed to have negligible volume and no intermolecular forces.

Practice Problems with Detailed Answers and Explanations

Let's tackle a variety of problems, progressing from simple applications of individual laws to more complex scenarios involving the ideal gas law and its variations. Remember to always convert units to match the units of the gas constant you are using.

Problem 1 (Boyle's Law): A gas occupies a volume of 5.0 L at a pressure of 1.0 atm. If the pressure is increased to 2.5 atm at constant temperature, what is the new volume?

Answer: Using Boyle's Law: P₁V₁ = P₂V₂

1.0 atm * 5.0 L = 2.5 atm * V₂ V₂ = (1.0 atm * 5.0 L) / 2.5 atm V₂ = 2.0 L

The new volume is 2.0 L.

Problem 2 (Charles's Law): A balloon has a volume of 2.0 L at 25°C. What will be its volume at 50°C if the pressure remains constant? Remember to convert Celsius to Kelvin (K = °C + 273.15).

Answer: Using Charles's Law: V₁/T₁ = V₂/T₂

First, convert Celsius to Kelvin: T₁ = 25°C + 273.15 = 298.15 K T₂ = 50°C + 273.15 = 323.15 K

Now apply Charles's Law: 2.0 L / 298.15 K = V₂ / 323.15 K V₂ = (2.0 L * 323.15 K) / 298.15 K V₂ ≈ 2.16 L

The new volume is approximately 2.16 L.

Problem 3 (Gay-Lussac's Law): A gas in a rigid container has a pressure of 1.5 atm at 20°C. What will be the pressure if the temperature is increased to 80°C at constant volume?

Answer: Using Gay-Lussac's Law: P₁/T₁ = P₂/T₂

Convert Celsius to Kelvin: T₁ = 20°C + 273.15 = 293.15 K T₂ = 80°C + 273.15 = 353.15 K

Apply Gay-Lussac's Law: 1.5 atm / 293.15 K = P₂ / 353.15 K P₂ = (1.5 atm * 353.15 K) / 293.15 K P₂ ≈ 1.81 atm

The new pressure is approximately 1.81 atm.

Problem 4 (Avogadro's Law): A sample of gas occupies 10.0 L at a certain temperature and pressure. If the number of moles of gas is doubled while the temperature and pressure remain constant, what is the new volume?

Answer: Using Avogadro's Law: V₁/n₁ = V₂/n₂

Let n₁ represent the initial number of moles. If the number of moles is doubled, n₂ = 2n₁.

10.0 L / n₁ = V₂ / (2n₁) V₂ = 2 * 10.0 L V₂ = 20.0 L

The new volume is 20.0 L.

Problem 5 (Ideal Gas Law): What is the pressure of 2.0 moles of an ideal gas in a 5.0 L container at 27°C?

Answer: Using the Ideal Gas Law: PV = nRT

Convert Celsius to Kelvin: T = 27°C + 273.15 = 300.15 K

Use R = 0.0821 L·atm/mol·K

P * 5.0 L = 2.0 mol * 0.0821 L·atm/mol·K * 300.15 K P = (2.0 mol * 0.0821 L·atm/mol·K * 300.15 K) / 5.0 L P ≈ 9.86 atm

The pressure is approximately 9.86 atm.

Problem 6 (Combined Gas Law): A gas occupies 1.0 L at 25°C and 1.0 atm. What volume will it occupy at 50°C and 2.0 atm?

Answer: The combined gas law combines Boyle's and Charles's laws: (P₁V₁)/T₁ = (P₂V₂)/T₂

Convert Celsius to Kelvin: T₁ = 25°C + 273.15 = 298.15 K T₂ = 50°C + 273.15 = 323.15 K

(1.0 atm * 1.0 L) / 298.15 K = (2.0 atm * V₂) / 323.15 K V₂ = (1.0 atm * 1.0 L * 323.15 K) / (298.15 K * 2.0 atm) V₂ ≈ 0.54 L

The new volume is approximately 0.54 L.

Problem 7 (Ideal Gas Law with Density): What is the density of nitrogen gas (N₂) at 25°C and 1.0 atm? The molar mass of N₂ is 28.02 g/mol.

Answer: We can use the ideal gas law to find the density. Density (ρ) is mass (m) divided by volume (V): ρ = m/V. We can rearrange the ideal gas law to solve for density:

PV = nRT and n = m/M (where M is molar mass)

PV = (m/M)RT PM = mRT/V PM/RT = m/V = ρ

ρ = PM/RT

Convert Celsius to Kelvin: T = 25°C + 273.15 = 298.15 K

ρ = (1.0 atm * 28.02 g/mol) / (0.0821 L·atm/mol·K * 298.15 K) ρ ≈ 1.14 g/L

The density of nitrogen gas is approximately 1.14 g/L.

Problem 8 (Ideal Gas Law with Partial Pressures – Dalton's Law): A container holds a mixture of helium (He) and neon (Ne) gases. The partial pressure of He is 0.60 atm, and the partial pressure of Ne is 0.40 atm. What is the total pressure in the container?

Answer: Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of the individual gases.

Total pressure = Partial pressure of He + Partial pressure of Ne Total pressure = 0.60 atm + 0.40 atm Total pressure = 1.00 atm

The total pressure in the container is 1.00 atm.

Problem 9 (Stoichiometry and Ideal Gas Law): What volume of hydrogen gas (H₂) at STP (Standard Temperature and Pressure: 0°C and 1.0 atm) is produced when 1.0 g of zinc (Zn) reacts completely with excess hydrochloric acid (HCl) according to the following balanced equation: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)? The molar mass of Zn is 65.38 g/mol.

Answer: First, find the moles of Zn:

Moles of Zn = mass / molar mass = 1.0 g / 65.38 g/mol ≈ 0.0153 mol

From the balanced equation, 1 mole of Zn produces 1 mole of H₂. Therefore, 0.0153 moles of Zn produce 0.0153 moles of H₂.

Now, use the ideal gas law at STP (T = 273.15 K, P = 1.0 atm):

PV = nRT V = nRT/P = (0.0153 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1.0 atm V ≈ 0.34 L

The volume of hydrogen gas produced is approximately 0.34 L.

These problems demonstrate the diverse applications of gas laws. Remember that meticulous attention to units and careful application of the appropriate formulas are key to success. Consistent practice with a variety of problem types will build your understanding and confidence in working with gas laws. Further practice can be found in textbooks, online resources, and chemistry problem sets. Remember to always check your work and ensure your answers are reasonable within the context of the problem.