Boyle's Law And Charles Law Worksheet

Boyle's Law and Charles's Law Worksheet: A Comprehensive Guide

This worksheet will delve into the fascinating world of gas laws, specifically Boyle's Law and Charles's Law. We'll explore their principles, equations, and applications through a series of exercises designed to enhance your understanding. Remember, mastering these laws is crucial for understanding the behavior of gases in various contexts, from everyday phenomena to complex scientific applications. Let's begin!

Understanding Boyle's Law: The Inverse Relationship

Boyle's Law states that for a fixed amount of gas at a constant temperature, the volume of the gas is inversely proportional to its pressure. In simpler terms, if you increase the pressure on a gas, its volume will decrease proportionally, and vice versa. This relationship can be mathematically represented as:

P₁V₁ = P₂V₂

Where:

• P₁ is the initial pressure
• V₁ is the initial volume
• P₂ is the final pressure
• V₂ is the final volume

This equation holds true as long as the temperature and the amount of gas remain constant. The units for pressure and volume can vary (atmospheres, Pascals, liters, cubic centimeters, etc.), but they must be consistent throughout the calculation.

Applying Boyle's Law: Example Problems

Let's work through a couple of example problems to solidify your understanding:

Problem 1: 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, what will be the new volume of the gas, assuming the temperature remains constant?

Solution: We can use Boyle's Law equation: P₁V₁ = P₂V₂

• P₁ = 1.0 atm
• V₁ = 5.0 L
• P₂ = 2.5 atm
• V₂ = ?

Substituting the values into the equation: (1.0 atm)(5.0 L) = (2.5 atm)(V₂)

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

Therefore, the new volume of the gas will be 2.0 L.

Problem 2: A balloon filled with helium has a volume of 2.0 L at a pressure of 1.5 atm. If the balloon is released and ascends to a higher altitude where the pressure drops to 0.8 atm, what will be the new volume of the balloon, assuming the temperature remains constant?

Solution: Again, using Boyle's Law: P₁V₁ = P₂V₂

• P₁ = 1.5 atm
• V₁ = 2.0 L
• P₂ = 0.8 atm
• V₂ = ?

Substituting and solving: (1.5 atm)(2.0 L) = (0.8 atm)(V₂)

V₂ = (1.5 atm * 2.0 L) / 0.8 atm = 3.75 L

The new volume of the balloon at the higher altitude will be 3.75 L.

Understanding Charles's Law: The Direct Relationship

Charles's Law describes the relationship between the volume and temperature of a fixed amount of gas at constant pressure. It states that the volume of a gas is directly proportional to its absolute temperature (measured in Kelvin). This means that if you increase the temperature of a gas, its volume will increase proportionally, and vice versa. The mathematical representation is:

V₁/T₁ = V₂/T₂

Where:

• V₁ is the initial volume
• T₁ is the initial absolute temperature (in Kelvin)
• V₂ is the final volume
• T₂ is the final absolute temperature (in Kelvin)

Remember to always convert Celsius temperatures to Kelvin using the formula: K = °C + 273.15

Applying Charles's Law: Example Problems

Let's tackle some problems involving Charles's Law:

Problem 3: A gas occupies a volume of 3.0 L at 25°C. What will be its volume if the temperature is increased to 50°C, assuming the pressure remains constant?

Solution: First, convert Celsius temperatures to Kelvin:

• T₁ = 25°C + 273.15 = 298.15 K
• T₂ = 50°C + 273.15 = 323.15 K

Now, use Charles's Law: V₁/T₁ = V₂/T₂

• V₁ = 3.0 L
• T₁ = 298.15 K
• V₂ = ?
• T₂ = 323.15 K

Solving for V₂: V₂ = (V₁ * T₂) / T₁ = (3.0 L * 323.15 K) / 298.15 K ≈ 3.25 L

The new volume of the gas will be approximately 3.25 L.

Problem 4: A sample of gas has a volume of 100 mL at 27°C. If the gas is cooled to -73°C at constant pressure, what will be its new volume?

Solution: Convert to Kelvin:

• T₁ = 27°C + 273.15 = 300.15 K
• T₂ = -73°C + 273.15 = 200.15 K

Using Charles's Law: V₁/T₁ = V₂/T₂

• V₁ = 100 mL
• T₁ = 300.15 K
• V₂ = ?
• T₂ = 200.15 K

Solving for V₂: V₂ = (V₁ * T₂) / T₁ = (100 mL * 200.15 K) / 300.15 K ≈ 66.7 mL

The new volume will be approximately 66.7 mL.

Combining Boyle's Law and Charles's Law: The Combined Gas Law

For situations where both pressure and temperature change, we use the Combined Gas Law:

(P₁V₁)/T₁ = (P₂V₂)/T₂

This equation incorporates both Boyle's Law and Charles's Law to describe the relationship between pressure, volume, and temperature for a fixed amount of gas.

Applying the Combined Gas Law: Example Problem

Problem 5: A gas occupies a volume of 4.0 L at 20°C and 1.0 atm. If the temperature is increased to 40°C and the pressure to 1.5 atm, what will be the new volume of the gas?

Solution: Convert to Kelvin:

• T₁ = 20°C + 273.15 = 293.15 K
• T₂ = 40°C + 273.15 = 313.15 K

Use the Combined Gas Law: (P₁V₁)/T₁ = (P₂V₂)/T₂

• P₁ = 1.0 atm
• V₁ = 4.0 L
• T₁ = 293.15 K
• P₂ = 1.5 atm
• V₂ = ?
• T₂ = 313.15 K

Solving for V₂: V₂ = (P₁V₁T₂) / (T₁P₂) = (1.0 atm * 4.0 L * 313.15 K) / (293.15 K * 1.5 atm) ≈ 2.84 L

The new volume will be approximately 2.84 L.

Advanced Applications and Considerations

The principles of Boyle's Law and Charles's Law are fundamental to understanding numerous phenomena, including:

• Weather patterns: Changes in atmospheric pressure and temperature directly influence air volume and weather systems.
• Respiratory function: Boyle's Law governs the mechanics of breathing, as the volume changes in the lungs lead to corresponding pressure changes.
• Engine performance: The compression and expansion of gases in internal combustion engines rely heavily on these gas laws.
• Deep-sea diving: Understanding how pressure affects gas volume is crucial for diver safety.

It's important to note that these laws are ideal gas laws and work best for gases at relatively low pressures and high temperatures. Real gases deviate from ideal behavior at high pressures and low temperatures due to intermolecular forces. However, for many practical applications, these laws provide a very accurate description of gas behavior.

Worksheet Exercises

Now, let's test your understanding with some practice problems:

Section 1: Boyle's Law

1. A gas has a volume of 6.0 L at a pressure of 2.0 atm. What will be its volume if the pressure is reduced to 1.0 atm (constant temperature)?
2. A scuba tank contains 12 L of air at a pressure of 200 atm. If this air is released into a large balloon at 1 atm, what will be the volume of the air in the balloon (constant temperature)?
3. A syringe contains 5.0 mL of air at atmospheric pressure (1.0 atm). If the plunger is pushed to reduce the volume to 2.5 mL, what is the new pressure of the air inside the syringe (constant temperature)?

Section 2: Charles's Law

4. A balloon has a volume of 10.0 L at 20°C. What will its volume be if the temperature is increased to 40°C (constant pressure)?
5. A sample of gas occupies 250 mL at 27°C. To what temperature (in Celsius) must the gas be cooled to reduce its volume to 200 mL (constant pressure)?
6. A hot air balloon has a volume of 2000 m³ at 100°C. If the temperature decreases to 80°C, what is its new volume (constant pressure)?

Section 3: Combined Gas Law

7. A gas occupies 5.0 L at 1.0 atm and 25°C. What will be its volume if the pressure is increased to 2.0 atm and the temperature is increased to 50°C?
8. A weather balloon has a volume of 10 m³ at 20°C and 1.0 atm. What will be its volume at an altitude where the pressure is 0.5 atm and the temperature is -20°C?
9. A sample of gas has a volume of 100 mL at 27°C and 1 atm pressure. If it is heated to 127°C and compressed to 50 mL, what is the final pressure of the gas?

This comprehensive worksheet should provide a solid foundation in understanding Boyle's Law and Charles's Law. Remember to always convert Celsius to Kelvin before calculations, and choose the appropriate formula based on the variables that remain constant. Good luck with your studies!