# Combined Gas Law Online Quiz

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| By Jcosh
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Jcosh
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Quizzes Created: 1 | Total Attempts: 6,310
Questions: 10 | Attempts: 6,320

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• 1.

### If the pressure exerted on a gas is tripled, what will happen to the volume (assuming the temperature and amount of gas remains constant)?

• A.

X2

• B.

X3

• C.

X 1/2

• D.

X 1/3

D. X 1/3
Explanation
When the pressure exerted on a gas is tripled, the volume of the gas will decrease by a factor of 1/3. This is because according to Boyle's Law, there is an inverse relationship between the pressure and volume of a gas when the temperature and amount of gas remain constant. As the pressure increases, the volume decreases proportionally. Therefore, if the pressure is tripled, the volume will decrease by a factor of 1/3.

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• 2.

### Carbon dioxide occupies a 2.54 L container at STP. What will be the volume when the pressure is 150 KPa and 26 C?

• A.

1.89 L

• B.

42300 L

• C.

0.163 L

• D.

14.1 L

A. 1.89 L
Explanation
At STP (Standard Temperature and Pressure), the volume of carbon dioxide in a 2.54 L container is given. To find the volume at a different pressure and temperature, we can use the ideal gas law equation, PV = nRT. Since the amount of gas (n) and the gas constant (R) are constant, we can rearrange the equation to solve for the new volume. By plugging in the given values for pressure, temperature, and the initial volume, we can calculate the new volume, which is 1.89 L.

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• 3.

### In which of the following sets of conditions would a gas most act as a real gas?

• A.

High temperature and low pressure

• B.

High temperature and high pressure

• C.

Low temperature and high pressure

• D.

Low temperature and low pressure

C. Low temperature and high pressure
Explanation
At low temperature and high pressure, the gas particles are closer together and have less kinetic energy. This means that the intermolecular forces between the particles become more significant, causing the gas to deviate from ideal gas behavior and behave more like a real gas. In contrast, at high temperature and low pressure, the gas particles have greater kinetic energy and are more spread out, minimizing the effect of intermolecular forces and promoting ideal gas behavior.

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• 4.

### Which of the following terms is used to describe the average kinetic energy of the molecules in a sample?

• A.

Manometer

• B.

Temperature

• C.

Pressure

• D.

Volume

B. Temperature
Explanation
Temperature is the correct answer because it is the term used to describe the average kinetic energy of the molecules in a sample. Temperature is a measure of the average speed and energy of the particles in a substance. It is directly related to the kinetic energy of the molecules, with higher temperatures corresponding to higher average kinetic energies.

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• 5.

### Which of the following is not a standard pressure measurement?

• A.

101.3 kPa

• B.

1 atm

• C.

760 mm Hg

• D.

273 K

D. 273 K
Explanation
The given options consist of different units used to measure pressure. 101.3 kPa, 1 atm, and 760 mm Hg are all standard pressure measurements commonly used in various fields. However, 273 K is not a standard pressure measurement. It is actually a standard temperature measurement in Kelvin. Pressure and temperature are two different physical properties, and therefore, 273 K cannot be considered a standard pressure measurement.

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• 6.

### The volume of a gas is increased from 150.0 mL to 350.0 mL by heating it. If the original temperature of the gas was 25.0 °C?

• A.

-146 C

• B.

10.7 C

• C.

426.85 C

• D.

150 C

C. 426.85 C
Explanation
To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure and the number of moles of gas are constant.
The equation for Charles's Law is:
V₁/T₁ = V₂/T₂
Where: V₁ = Initial volume T₁ = Initial temperature V₂ = Final volume T₂ = Final temperature
Given: V₁ = 150.0 mL T₁ = 25.0 °C = 25.0 + 273.15 K = 298.15 K (converting Celsius to Kelvin) V₂ = 350.0 mL
We need to find T₂.
Using Charles's Law:
150.0 mL / 298.15 K = 350.0 mL / T₂
Now, let's solve for T₂:
T₂ = (350.0 mL * 298.15 K) / 150.0 mL T₂ = 700.0 K
Now, we need to convert the temperature back to Celsius:
T₂ = 700.0 K - 273.15 T₂ ≈ 426.85 °C

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• 7.

### What is the name of the phase change that exists when a solid turns into a liquid?

• A.

Depositon

• B.

Melting

• C.

Sublimation

• D.

Freezing

B. Melting
Explanation
Melting is the correct answer because it refers to the phase change that occurs when a solid substance is heated and transforms into a liquid state. This process involves the breaking of intermolecular forces within the solid, allowing the particles to move more freely and take on the shape of their container.

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• 8.

### A balloon is inflated to a volume of 130 ml at a pressure of 690 mmHg. If the pressure is increased to 1000 mmHg, what will the new volume be?

• A.

188 ml

• B.

98.8 ml

• C.

89.7 ml

• D.

40300 ml

C. 89.7 ml
Explanation
When the pressure of a gas increases, its volume decreases, assuming the temperature and amount of gas remain constant. This relationship is described by Boyle's Law. In this question, the initial volume of the balloon is 130 ml at a pressure of 690 mmHg. When the pressure is increased to 1000 mmHg, the new volume can be calculated using Boyle's Law equation: P1V1 = P2V2. Plugging in the values, we get (690 mmHg)(130 ml) = (1000 mmHg)(V2). Solving for V2, we find that the new volume is 89.7 ml.

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• 9.

### What type of relationship is shown between temperature and pressure?

• A.

Direct

• B.

Indirect (inverse)

• C.

Exponential

• D.

Logarithmic

A. Direct
Explanation
The relationship between temperature and pressure is direct, meaning that as temperature increases, pressure also increases. This is because as temperature rises, the average kinetic energy of the gas molecules increases, leading to more frequent and forceful collisions with the container walls, resulting in an increase in pressure.

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• 10.

### How many mm Hg is equal to 121 kPa?

• A.

1.19

• B.

16.1

• C.

636

• D.

908

D. 908
Explanation
121 kPa is equal to 908 mm Hg. This conversion is based on the fact that 1 kPa is equal to 7.5 mm Hg. Therefore, to convert kPa to mm Hg, you multiply the value in kPa by 7.5. In this case, 121 multiplied by 7.5 equals 908 mm Hg.

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• Current Version
• Mar 13, 2024
Quiz Edited by
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• Mar 01, 2012
Quiz Created by
Jcosh

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