This guide provides comprehensive answers and explanations for common Combined Gas Law problems found in chemistry worksheets. The Combined Gas Law, which combines Boyle's, Charles's, and Gay-Lussac's laws, is a crucial concept in understanding the behavior of gases. We'll cover various scenarios and illustrate how to effectively apply the law to solve them. Remember to always double-check your units before starting any calculation!
The Combined Gas Law is expressed as:
(P₁V₁)/T₁ = (P₂V₂)/T₂
Where:
- P = Pressure (typically in atmospheres (atm), kilopascals (kPa), or millimeters of mercury (mmHg))
- V = Volume (typically in liters (L))
- T = Temperature (always in Kelvin (K)! Remember to convert Celsius to Kelvin using K = °C + 273.15)
Understanding the Combined Gas Law
Before diving into specific worksheet problems, let's revisit the fundamental principles:
- Boyle's Law: At constant temperature, the volume of a gas is inversely proportional to its pressure (P₁V₁ = P₂V₂).
- Charles's Law: At constant pressure, the volume of a gas is directly proportional to its temperature (V₁/T₁ = V₂/T₂).
- Gay-Lussac's Law: At constant volume, the pressure of a gas is directly proportional to its temperature (P₁/T₁ = P₂/T₂).
The Combined Gas Law elegantly combines these three laws, allowing us to solve problems where pressure, volume, and temperature are all changing simultaneously.
Common Worksheet Problems & Solutions
Let's tackle some typical Combined Gas Law worksheet questions:
1. A gas occupies 5.0 L at 20°C and 1.0 atm. What volume will it occupy at 40°C and 1.5 atm?
Solution:
-
Convert temperatures to Kelvin: 20°C + 273.15 = 293.15 K; 40°C + 273.15 = 313.15 K
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Apply the Combined Gas Law: (P₁V₁)/T₁ = (P₂V₂)/T₂
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Plug in the values: (1.0 atm * 5.0 L) / 293.15 K = (1.5 atm * V₂) / 313.15 K
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Solve for V₂: V₂ = [(1.0 atm * 5.0 L * 313.15 K) / (293.15 K * 1.5 atm)] ≈ 3.56 L
2. A balloon filled with helium has a volume of 10.0 L at 25°C and 101.3 kPa. If the balloon is heated to 50°C, what is its new volume assuming the pressure remains constant?
Solution:
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Convert temperatures to Kelvin: 25°C + 273.15 = 298.15 K; 50°C + 273.15 = 323.15 K
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Since pressure is constant, we can simplify to Charles's Law: V₁/T₁ = V₂/T₂
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Plug in the values: 10.0 L / 298.15 K = V₂ / 323.15 K
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Solve for V₂: V₂ = (10.0 L * 323.15 K) / 298.15 K ≈ 10.84 L
3. A sample of gas at 2.0 atm and 273 K occupies 10.0 L. If the gas is compressed to 5.0 L at constant temperature, what is the new pressure?
Solution:
-
Temperature remains constant, so we use Boyle's Law: P₁V₁ = P₂V₂
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Plug in the values: 2.0 atm * 10.0 L = P₂ * 5.0 L
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Solve for P₂: P₂ = (2.0 atm * 10.0 L) / 5.0 L = 4.0 atm
Frequently Asked Questions (FAQs)
This section addresses common queries regarding the Combined Gas Law:
What are the units for each variable in the Combined Gas Law?
Pressure (P) can be in atmospheres (atm), kilopascals (kPa), or millimeters of mercury (mmHg). Volume (V) is typically in liters (L). Crucially, temperature (T) must be in Kelvin (K).
Why is it essential to convert Celsius to Kelvin?
The Kelvin scale is an absolute temperature scale, meaning it starts at absolute zero (0 K), where all molecular motion ceases. Using Celsius would lead to incorrect calculations because Celsius uses an arbitrary zero point.
Can I use the Combined Gas Law for all gases?
The Combined Gas Law is most accurate for ideal gases. Real gases deviate from ideal behavior at high pressures and low temperatures.
What if one of the variables remains constant?
If a variable (pressure, volume, or temperature) remains constant, you can simplify the Combined Gas Law to use either Boyle's, Charles's, or Gay-Lussac's Law.
By understanding the Combined Gas Law and practicing with various problems, you can effectively tackle any worksheet questions related to gas behavior. Remember to always clearly define your variables, convert units appropriately, and meticulously check your work for accuracy.
