Gas Laws Quiz: Boyle's And Charles' Law
Welcome to the Gas Laws: Boyle's and Charles' Law Quiz, where you'll delve into the fascinating principles governing the behavior of gasses. Boyle's Law and Charles' Law are foundational concepts in the study of gas physics. As you navigate through this quiz, you'll be challenged to apply these laws in various scenarios and explore their implications on gas behavior. This quiz aims to assess your ability to apply these fundamental gas laws to real-world situations, providing a comprehensive understanding of how gasses respond to changes in pressure, volume, and temperature. Best of luck as you navigate the intricacies of Read moreBoyle's and Charles' Laws!
- 1.
A balloon occupies a volume of 2.0 L at 40oC. How much volume will it occupy at 30oC?
- A.
1.5 L
- B.
1.94 L
- C.
1.6 L
Correct Answer
B. 1.94 LExplanation
Using the formula,
V1/T1=V2/T2 and substituting the values of V1,T1 and T2 and solving the equation we get,
Given:
V1​=2.0L at T1​= 40°C =313.15K
T2=30°C =303.15 K
2.0L/313.15K=V2/313.15K
Therefore
V2=2.0L x 303.15K/313.15K
V2​≈1.94LRate this question:
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- 2.
What is the final pressure in a balloon when the volume is changed from 1.5 L to a volume of 2.5L at constant temperature? (Initial pressure is 1.1 atm).
- A.
0.66 atm
- B.
0.66 kPa
- C.
4.125 atm
Correct Answer
A. 0.66 atmExplanation
To find the final pressure (Pf) when the volume changes at constant temperature, you can use Boyle's Law, which states that the product of pressure and volume is constant for a given amount of gas at constant temperature.
The formula for Boyle's Law is given by:
P1 . V1 = P2 . V2
where:
- ( P1) is the initial pressure,
- (V1) is the initial volume,
- ( P2) is the final pressure,
- ( V2) is the final volume.
Given:
- (P1 = 1.1 ) atm (initial pressure),
- ( V1 = 1.5 ) L (initial volume),
- ( V2= 2.5 ) L (final volume).
Plug in the values into Boyle's Law:
[ 1.1. 1.5] L = P2 . 2.5 L
Now, solve for P2
P2 = 1.1 atm . 1.5L/ 2.5L
P2 ≈ 0.66 atm
Therefore, the final pressure in the balloon when the volume changes from 1.5 L to 2.5 L at constant temperature is approximately 0.66 atm.Rate this question:
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- 3.
Gas in a balloon occupies 3.3 L. What volume will it occupy if the pressure is changed from 100 kPa to 90 kPa (at constant temperature of 310 K (about room temperature)?
- A.
29,700 L
- B.
2.97 L
- C.
3.67 L
Correct Answer
C. 3.67 LExplanation
The formula for Boyle's Law is given by:
P1 . V1 = P2 . V2
where:
- ( P1) is the initial pressure,
- (V1) is the initial volume,
- ( P2) is the final pressure,
- ( V2) is the final volume.
Given:
- P1 = 100 kPa (initial pressure),
- V1 = 3.3 L (initial volume),
- P2 = 90 kPa (final volume)
T = 310K (Constant Temperature)
Plug in the values into Boyle's Law to find V2:
V2 = P1 . V1 / P2
Now, solve for V2
V2 =100 kPa . 3.3L/ 90kPa
V2 ≈ 3.67L
Therefore, the volume the gas will occupy when the pressure is changed from 100 kPa to 90 kPa at constant temperature is approximately 3.67 L.Rate this question:
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- 4.
What is the final pressure in a vessel when the volume is changed from 1.3 L to a volume of 1.55 L at constant temperature? (Initial pressure is 1.5 atm).
- A.
1.26 L
- B.
1.65 L
- C.
1.49 L
Correct Answer
A. 1.26 LExplanation
The formula for Boyle's Law is given by:
P1 . V1 = P2 . V2
where:
- ( P1) is the initial pressure,
- (V1) is the initial volume,
- ( P2) is the final pressure,
- ( V2) is the final volume.
Given:
- (P1 = 1.5 ) atm (initial pressure),
- ( V1 = 1.3 ) L (initial volume),
- ( V2= 1.55 ) L (final volume).
Plug in the values into Boyle's Law:
[ 1.5. 1.3] L = P2 . 1.55 L
Now, solve for P2
P2 = 1.5 atm . 1.3L/ 1.55L
P2 ≈1.26atm
Therefore, the final pressure in the vessel when the volume changes from 1.3 L to 1.55 L at constant temperature is approximately 1.26 atm.Rate this question:
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- 5.
A balloon occupies 3.2 L at 37oC. How much volume will it occupy at 42oC?
- A.
3.25 L
- B.
3.65 L
- C.
312,480 L
Correct Answer
A. 3.25 LExplanation
Using the formula,
V1/T1=V2/T2 and substituting the values of V1,T1 and T2 and solving the equation we get,
Given:
V1​=3.2L at T1​= 37°C =310.15K
T2=42°C =315.15 K
3.2L/315.15K=V2/310.15K
Therefore
V2=3.2L x 310.15K/315.15K
V2​≈3.25L
Therefore, the balloon will occupy approximately 3.25 L at 42°C.Rate this question:
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- 6.
A gas occupies a balloon with a volume 2.0 L at 33oC. How much volume will it occupy at 37oC, if pressure remains constant?
- A.
1.97 L
- B.
186,000 L
- C.
2.026 L
Correct Answer
C. 2.026 LExplanation
Using the formula,
V1/T1=V2/T2 and substituting the values of V1,T1 and T2 and solving the equation we get,
Given:
V1​=2.0L at T1​= 33°C =306.15K
T2=37°C =310.15 K
2.0L/306.15K=V2/310.15K
Therefore
V2=2.0L x 310.15K/306.15K
V2​≈2.026L
Therefore, the balloon will occupy approximately 2.026 L at 37°C.Rate this question:
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- 7.
A gas expands from a volume 2.0 L at 36oC to a volume of 2.5 L. What is the final temperature (in Kelvin) if the pressure is constant?
- A.
386.43 K ( which equals 113.25 Celsius)
- B.
247.2 K (which equals -25.8 C)
- C.
61.8 K (which equals -211.2 C)
Correct Answer
A. 386.43 K ( which equals 113.25 Celsius)Explanation
Using the formula,
V1/T1=V2/T2 and substituting the values of V1,T1 and V2 and solving the equation we get,
Given:
V1​=2.0L at T1​= 36°C =309.15K
V2=2.5L
2.0L/309.15K=2.5L/T2
Therefore
T2=2.5L x 309.15K / 2.0 L
T2​≈ 386.43 K
Therefore, the final temperature is approximately 386.43 K when the gas expands from a volume of 2.0 L at 36°C to a volume of 2.5 L, assuming the pressure remains constant.Rate this question:
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- 8.
If the air inside a balloon is heated, the volume will:
- A.
Increase
- B.
Decrease
- C.
Increase a little, then decrease.
Correct Answer
A. IncreaseExplanation
If the air inside a balloon is heated while the pressure remains constant, the volume of the balloon will generally increase. This behavior follows Charles's Law, which states that the volume of a gas is directly proportional to its temperature at constant pressure.Rate this question:
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- 9.
The kinetic theory of gasses, which of the following is important for an ideal gas:
- A.
There is no attraction or repulsion between the gas molecules, whereas there is attraction between molecules of liquids or solids.
- B.
The distance between molecules in ideal gasses is closer than in liquids or solids.
- C.
The gasses are NOT fluid.
Correct Answer
A. There is no attraction or repulsion between the gas molecules, whereas there is attraction between molecules of liquids or solids.Explanation
The correct answer is that there is no attraction or repulsion between the gas molecules, whereas there IS attraction between molecules of liquids or solids. This is important for an ideal gas because it assumes that the gas molecules do not interact with each other, except during collisions. This allows for simplifications in the mathematical models used to describe the behavior of ideal gasses. In contrast, liquids and solids have intermolecular forces that result in attractions between their molecules, leading to different behaviors and properties.Rate this question:
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- 10.
Which of the following describes the relationship between P and V?
- A.
Directly proportional
- B.
Inversely proportional
- C.
Unrelated
Correct Answer
B. Inversely proportionalExplanation
The relationship between P and V is described as inversely proportional. This means that as one variable (P) increases, the other variable (V) decreases, and vice versa. In other words, there is a negative correlation between P and V, where an increase in one variable corresponds to a decrease in the other variable.Rate this question:
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Feb 21, 2024Quiz Edited by
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Jun 19, 2011Quiz Created by
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