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The combined gas law describes the relationship between the pressure (P), volume (V), and temperature (T) of a gas. It is derived by combining Boyle’s law, Charles’s law, and Gay-Lussac’s law into a single expression. The combined gas law states that the ratio of the initial pressure, volume, and temperature of a gas to the final pressure, volume, and temperature is constant, provided the amount of gas and its state (solid, liquid, or gas) remain constant.
Mathematically, the combined gas law is expressed as P1V1/T1 = P2V2/T2, where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively. The combined gas law is particularly useful in understanding how changes in pressure, volume, and temperature affect the behavior of gases. It is commonly used in various scientific and engineering applications, such as in the design of gas storage and distribution systems, as well as in the study of gas behavior in chemical reactions.
Formula
The combined gas law formula, expressed as P1V1/T1 = P2V2/T2, describes the relationship between pressure (P), volume (V), and temperature (T) of a given quantity of gas. In this formula, P1, V1, and T1 represent the initial pressure, volume, and temperature of the gas, while P2, V2, and T2 represent the final pressure, volume, and temperature, respectively. This formula is used to calculate the changes in pressure, volume, or temperature of a gas under different conditions while keeping the amount of gas constant.
Practice problems
Problem #1
A cylinder initially filled with 4 L of argon gas at 24 ℃ temperature has a pressure of 2 atm. What will be the final volume of the gas when the pressure increases to 8 atm and the temperature increases to 30 ℃?
Solution
Given data:
- Initial volume of the gas, V1 = 4 L
- Initial temperature of the gas, T1 = 24 ℃ = 297 K
- Initial pressure of the gas, P1 = 2 atm
- Final volume of the gas, V2 = ?
- Final pressure of the gas, P2 = 8 atm
- Final temperature of the gas, T2 = 30 ℃ = 303 K
Applying the formula:
- P1V1/T1 = P2V2/T2
- V2 = (P1 × V1 × T2) ÷ (T1 × P2)
- V2 = (2 × 4 × 303) ÷ (297 × 8)
- V2 = 303 ÷ 297
- V2 = 1.02 L
Therefore, the final volume of the gas is 1.02 L.
Problem #2
A balloon initially containing 12 L of helium gas at 20 ℃ and 5 atm pressure is heated to 40 ℃, causing the volume to increase to 60 L. What is the final pressure of the gas?
Solution
Given data:
- Initial volume of the gas, V1 = 12 L
- Initial temperature of the gas, T1 = 20 ℃ = 293 K
- Initial pressure of the gas, P1 = 5 atm
- Final temperature of the gas, T2 = 40 ℃ = 313 K
- Final volume of the gas, V2 = 60 L
- Final pressure of the gas, P2 = ?
Applying the formula:
P1V1/T1 = P2V2/T2
P2 = (P1 × V1 × T2) ÷ (T1 × V2)
P2 = (5 × 12 × 313) ÷ (293 × 60)
P2 = 313 ÷ 293
P2 = 1.06 atm
Therefore, the final pressure of the gas is 1.06 atm.
Problem #3
A rubber balloon is taken out of cold storage, and 8 L of helium gas is released into it at a temperature of 70 ℃ and pressure of 6 atm. If the initial temperature and pressure of the gas were 25 ℃ and 3 atm, respectively, what was the initial volume of the gas?
Solution
Given data:
- Final volume of the gas, V2 = 8 L
- Final temperature of the gas, T2 = 70 ℃ = 343 K
- Final pressure of the gas, P2 = 6 atm
- Initial temperature of the gas, T1 = 25 ℃ = 298 K
- Initial pressure of the gas, P1 = 3 atm
- Initial volume of the gas, V1 = ?
Applying the formula:
- P1V1/T1 = P2V2/T2
- V1 = (P2 × V2 × T1) ÷ (T2 × P1)
- V1 = (6 × 8 × 298) ÷ (343 × 3)
- V1 = 14304 ÷ 1029
- V1 = 13.90 L
Therefore, the initial volume of the gas is 13.90 L.
Problem #4
At an initial temperature of 10 ℃, the container holds 6 L of neon gas, exerting a certain pressure on the container walls. When the temperature is raised to 80 ℃ and the volume of the gas is increased to 12 L, the pressure exerted by the gas on the container walls becomes 12 atm. What was the initial pressure of the gas in the container?
Solution
Given data:
- Initial temperature of the gas, T1 = 10 ℃ = 283 K
- Initial volume of the gas, V1 = 6 L
- Final temperature of the gas, T2 = 80 ℃ = 353 K
- Final volume of the gas, V2 = 12 L
- Final pressure of the gas, P2 = 12 atm
- Initial pressure of the gas, P1 = ?
Applying the formula:
- P1V1/T1 = P2V2/T2
- P1 = (P2 × V2 × T1) ÷ (T2 × V1)
- P1 = (12 × 12 × 283) ÷ (353 × 6)
- P1 = 40752 ÷ 2118
- P1 = 19.24 atm
Therefore, the initial pressure of the gas is 19.24 atm.
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Related
More topics
- Boyle’s law
- Charles’s law
- Gay-Lussac’s law
- Avogadro’s law
- Ideal gas law
- Dalton’s law
- Henry’s law
- Combined gas law
- Graham’s law
External links
- Combined Gas Law | Definition, Formula & Example – Study.com
- Combined Gas Law Definition and Examples – ThoughtCo
- 14.6: Combined Gas Law – Chemistry LibreTexts
- Combined Gas Law – ChemTalk
- Combined gas law – Wikipedia
- Gas Laws – Florida State University
- Combined Gas Law – CK-12
- Combined Gas Law — Overview & Calculations – Expii
- Combined Gas Law Ten Examples – ChemTeam
- Combined & Ideal Gas Laws – EdTech Books
- The Combined Gas Law – Florida State College
- Combined Gas Law – The University of Texas at Austin
- Combined Gas Law – Chemistry – Socratic
- Combined Gas Law – Problems 1 – 15 – ChemTeam
- High School Chemistry : Using the Ideal Gas Law and Combined Gas Law – Varsity Tutors
- Combined Gas Law Calculator – SensorsONE
- Combined gas law Facts for Kids – Kids encyclopedia facts
- Combined Gas Equation Chemistry Tutorial – AUS-e-TUTE
- Why is the combined gas law useful? How is it applied? – Quora
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