## Gas Laws in Physics: Exploring the Fundamentals of Gas Behavior

**Gas Laws in Physics: **Gas is a state of matter that permeates our everyday lives. From the air we breathe to the fuel that powers our vehicles, understanding the behaviour of gases is essential.

In this article, we will learn about general gas laws, which provide the foundational principles governing the properties and behaviour of gases.

Gas laws are a set of mathematical relationships that describe the behaviour of gases under various conditions. They provide a framework for understanding how gases respond to changes in temperature, pressure, volume, and the number of gas molecules.

Therefore, these laws help scientists and engineers to predict and manipulate gas properties, enabling advancements in fields such as chemistry, physics, and engineering.

## 2. The Basics of Gas Behavior

Before diving into specific gas laws, let us establish some foundational concepts about gas behaviour. Gases consist of tiny particles, such as atoms or molecules, that are in constant motion and exhibit negligible intermolecular forces. These particles move randomly and collide with each other and the walls of their container, exerting pressure.

The behaviour of gases can be described using variables such as pressure (P), volume (V), temperature (T), and the number of moles (n) of gas molecules present. Gas laws provide relationships between these variables, allowing us to make predictions and understand how changes in one variable affect others.

### Gas Laws Formula

Here is a table of gas laws formulas along with their inventors, dates of invention, formulas, units, and respective dimensions based on the information available.

Here’s the table:

Gas Law | Inventor | Date of Invention | Formula | Unit | Dimensions |
---|---|---|---|---|---|

Boyle’s Law | Robert Boyle | 1662 | PV = constant | P * V | Pressure * Volume |

Charles’s Law | Jacques Charles | 1787 | V / T = constant | V / T | Volume / Temperature |

Gay-Lussac’s Law | Joseph Gay-Lussac | 1802 | P / T = constant | P / T | Pressure / Temperature |

Avogadro’s Law | Amedeo Avogadro | 1811 | V / n = constant | V / n | Volume / Amount of substance |

Ideal Gas Law | Emil Clapeyron | 1834 (further developed by others later) | PV / nT = constant | P * V / (n * T) | Pressure * Volume / (Amount of substance * Temperature) |

Combined Gas Law | (P_{1} * V_{1}) / (T_{1} * n_{1}) = (P_{2} * V_{2}) / (T_{2} * n_{2}) | Varies | Varies |

Please note that these dates and attributions are based on historical records.

### Properties of Gases

Gas is one of the states of matter. Therefore, it is mostly obtained when a solid is heated to its melting point. Therefore, it’s forces of attraction become weaker and will break into liquid. When we add more heat to the liquid to its boiling point, the liquid will start vaporizing and give up in the form of gases. Here are properties of gases you will need to know:

- The molecules possess a weak cohesive force
- The molecules are negligible.
- They fill the entire container in which they are.
- They have the highest kinetic energy compared to solid and liquid
- The degree of randomness is high.

#### From Kinetic Theory of Gases

According to the kinetic theory of gases, the following also apply:

- Gases are tiny particles that move in a straight line and collide with one another and also with the wall of the container.
- The collision between the gases is perfectly elastic.
- The cohesive force of the gases is also negligible
- The temperature of the gas is the measure of the average kinetic energy of the gas.

## 3. Boyle’s Law: Pressure-Volume Relationship

Boyle’s Law, named after physicist Robert Boyle, **states that the pressure of a gas is inversely proportional to its volume when the temperature and the number of gas molecules are constant**. Mathematically, it can be expressed as: V ∝ (1/P)

**Therefore, Boyle’s Law Formula is:**

P_{1}V_{1} = P_{2}V_{2}

where P_{1} and V_{1} represent the initial pressure and volume, and P_{2} and V_{2} represent the final pressure and volume, respectively.

This law explains why squeezing a gas-filled balloon reduces its volume, causing an increase in pressure.

## 4. Charles’s Law: Temperature-Volume Relationship

Charles’ Law, formulated by physicist Jacques Charles, describes the relationship between the temperature and volume of a gas when the pressure and the number of gas molecules are constant. According to Charles’s Law, as the temperature of a gas increases, its volume also increases, and vice versa.

Therefore, Charles’ Law states that the volume of a fixed gas is directly proportional to its absolute temperature, provided pressure is kept constant.

Mathematically, the charles’ law formula can be stated as: V ∝ T

**Therefore, Charles’s Law Equation:**

V_{1}/T_{1} = V_{2}/T_{2}

where V_{1} and T_{1} represent the initial volume and temperature, and V_{2} and T_{2} represent the final volume and temperature, respectively. As we can see, temperature and volume are involved in the case of Charles’ Law.

This law explains why a helium-filled balloon expands when exposed to heat.

## 5. Gay-Lussac’s Law: Pressure-Temperature Relationship

Gay-Lussac’s Law, named after chemist Joseph Louis Gay-Lussac, relates the pressure of a gas to its temperature when the volume and the number of gas molecules remain constant. According to this law, as the temperature of a gas increases, its pressure also increases, and vice versa. Mathematically, it can be expressed as:

**Gay-Lussac’s Law Equation:**

P_{1}/T_{1} = P_{2}/T_{2}

where P_{1} and T_{1} represent the initial pressure and temperature, and P_{2} and T_{2} represent the final pressure and temperature, respectively.

This law helps explain why the pressure inside a car tire increases on a hot day.

## 6. The Combined Gas Law

The Combined Gas Law combines Boyle’s, Charles’s, and Gay-Lussac’s Laws into a single equation that describes the relationship between a gas’s pressure, volume, and temperature. Thus, it allows us to calculate the changes in these properties when more than one variable is altered. The equation can be written as:

(P_{1}V_{1})/T_{1} = (P_{2}V_{2})/T_{2}

The Combined Gas Law is particularly useful when studying gases undergoing complex changes.

## 7. Avogadro’s Law: Volume-Amount Relationship

Avogadro’s Law, proposed by the Italian scientist Amedeo Avogadro, states that equal volumes of gases, at the same temperature and pressure, contain an equal number of particles. This law allows us to understand the relationship between the volume and the amount (in moles) of gas. Mathematically, it can be expressed as:

**Avogadro’s Law Equation:**

V_{1}/n_{1} = V_{2}/n_{2}

where V_{1} and n_{1} represent the initial volume and amount of gas, and V_{2} and n_{2} represent the final volume and amount of gas, respectively.

Avogadro’s Law is particularly relevant in stoichiometry calculations and understanding gas reactions.

## 8. Ideal Gas Law: The Grand Unification

The Ideal Gas Law combines all the previously discussed gas laws (Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s Laws) into a single equation that describes the behavior of an ideal gas. The Ideal Gas Law equation is given as:

PV = nRT

where (P) represents pressure, (V) represents volume, (n) represents the amount of gas (in moles), (R) represents the ideal gas constant, and (T) represents temperature in Kelvin.

The Ideal Gas Law allows us to relate the macroscopic properties of a gas to its microscopic behaviour.

## 9. Dalton’s Law of Partial Pressures

Dalton’s Law of Partial Pressures states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of each individual gas. Mathematically, it can be expressed as:

**Dalton’s Law Equation:**

P_{total} = P_{1} + P_{2} + P_{3} + ….

where P_{total} represents the total pressure of the gas mixture, and (P1 + P2 + P3 ) represents the partial pressures of each gas component.

Dalton’s Law is essential when dealing with gas mixtures, such as the composition of the Earth’s atmosphere.

## 10. Graham’s Law of Effusion and Diffusion

Graham’s Law of Effusion and Diffusion describes the rate at which gases effuse (escape through a small opening) or diffuse (mix with other gases).

According to Graham’s Law, the rate of effusion or diffusion of a gas is inversely proportional to the square root of Therefore, understanding these deviations is very important when studying gases in extreme conditions or when high precision is required.

## 12. Applications of Gas Laws in Everyday Life

Gas laws find practical applications in various fields and aspects of everyday life. Some notable examples include:

- Understanding and predicting weather phenomena, such as changes in atmospheric pressure and temperature.
- Designing and operating scuba diving equipment by considering gas solubility and pressure changes at different depths.
- Developing efficient fuel combustion strategies in engines by optimizing air-fuel ratios.
- Analyzing and interpreting gas data in industrial processes, such as chemical reactions and gas storage.

## 13. General Gas Law Formula

The general gas law in physics is the combination of Boyle’s, Charle’s, and Pressure laws for an ideal gas.

We can write the formula as

PV / T = Constant

Where

P = pressure

V = volume

T = Temperature

And because of the constant, we can now write the formula for the general gas law as

P_{1}V_{1} / T_{1} = P_{2}V_{2} / T_{2}

**Similarly, the Ideal Gas Law is written as PV = nRT**

## 14. General Gas Law in Physics Practice Problems

Here are a few examples to help you understand how to apply the formula for general gas law and calculate a problem.

### Problem 1: General Gas Law

A mass of gas at 17^{0}c and 700mmHg has a volume of 1.2m^{3}. Determine its volume at 27^{0}c and pressure at 750 mmHg.

#### Answer

Solution: To calculate the volume from general gas law, we need to first extract our data from the question

Data:

Initial Pressure = P_{1 }= 700mmHg

Final Pressure = P_{2 }= 750mmHg

Initial Temperature = T_{1 }= 17^{0}c = 273 + 17 = 290

Final Temperature = T_{2 }= 27^{0}c = 273 + 27 = 300

Additionally, initial Volume = V_{1 }= 1.2m^{3}

Also, Final Volume = V_{2 }= ?

Using the general gas law,

( P_{1 }V_{1}) / ( T` _{1}` )= ( P

_{2 }V

_{2}) / ( T

_{2})

By Making V_{2} subject of the formula,we have

_{ }Thus, V_{2} = ( P_{1 }V_{1}T_{2}) / (P_{2} T` _{1}` )

by substituting the above expression with our data, we have:

Final volume, V_{2} = ( 700 x 1.2 x 300 ) / ( 750 x 290 )

Therefore, V_{2} = ( 252,000 ) / ( 217,500 )

This will give us our final answer as V_{2} = 1.2 m^{3}

**Therefore, the final volume is 1.2 m ^{3}**

### Problem 2: General Gas Law

The volume and pressure of a given mass of gas at 27^{0}C are 76 cm^{3} and 80 cm of mercury respectively. Calculate its volume at standard temperature and pressure (s.t.p)

#### Answer

**Data**

The original volume, V_{1} = 76 cm^{3}

Initial pressure, P_{1 } = 80 cmHg

Final pressure, P_{2} = 760 cmHg [Because of the standard temperature and pressure]

The initial temperature, T` _{1}` = 27

^{0}C = 27 + 273 = 300 k

Final temperature, T` _{2}` = 273k [Because of the standard temperature and pressure]

**Unknown**

The final volume, V_{2} = ?

**Formula**

We will now apply our formula to solve the problem

( P_{1 }V_{1}) / ( T` _{1}` )= ( P

_{2 }V

_{2}) / ( T

_{2})

after inserting our data, we will now have

(80 x 76) / (300) = (76 x V_{2}) / 273

By making V_{2} subject of the formula, we will obtain our answer as

**V _{2} = 72.8 cm^{3} **

### Problem 3: General Gas Law

A given mass of an ideal gas occupies a volume V at a temperature T and under a pressure P. If the pressure is increased to 2P and the temperature reduced to (1/2)T, then what is the percentage change in the volume of the gas?

#### Answer

**The final answer to the above question is 75%**

**Explanation**

Since we have a formula that says P_{1}V_{1} / T_{1} = P_{2}V_{2} / T_{2}

and

V_{1} = V

P_{1} = P

P_{2} = 2P

T_{1} = T

T_{2} = (1/2)T

V_{2} = ?

We can make V_{2} subject of the formula and interpret it as

V_{2} = (P x V x (1/2)T ) / (2P x T)

And we will have

V_{2} = (1/2)V / 2 = 0.25 V

The percentage change in volume will now become

[(V_{1} – V_{2}) / V_{1} ] x 100%

Which will now be

[(V – 0.25V) / V ] x 100%

hence

[0.75V / V ] x 100% = 0.75 x 100% = 75%

**Therefore, the percentage change in the volume of the gas is 75%**

### Problem 4: General Gas Law

The pressure of a given mass of a gas changes from 300 Nm^{-2} to 120 Nm^{-2} while the temperature drops from 127^{0}C to -73^{0}C. What is the ratio of the final volume to the initial volume?

#### Answer

**The final answer to the question is V _{1} : V_{2} = 5 : 4**

**Explanation**

P_{1} = 300 Nm^{-2}

and P_{2} = 120 Nm^{-2}

T_{1} = 127 + 273 = 400 K

we have T_{2} = -73 + 273 = 200 K

**Formula**

The formula we will use is P_{1}V_{1} / T_{1} = P_{2}V_{2} / T_{2}

**Solution**

We will now insert our data into the formula

(300 x V_{1}) / 400 = (120 x V_{2}) / 200

To have a ratio of V_{1} to V_{2}

Hence,

**V _{2} : V_{1} = 60000 / 48000**

This implies:

60 / 48 = V_{2} / V_{1}

When we divide both 60 and 48 by 12, we will obtain

**V _{2} / V_{1} = 5 / 4**

**Therefore, the ratio of V _{2} to V_{1} is 5 to 4**

### Problem 5: Boyle’s Law

The volume of a fixed mass of a gas is 200 cm^{3} at 27^{0}C. Calculate its temperature when the volume is reduced to (1/2) its original at the same pressure.

#### Answer

**The final answer to the above question is -123 ^{0}C**

**Explanation**

**Data:**

V_{1} = 200 cm^{3}

T_{1} = 273 + 27 = 300K

V_{2} = (1/2)V_{1} = 0.5 x 200 = 100 cm^{3}

**Unknown:**

T_{2} = ?

**Formula:**

V_{1} / T_{1} = V_{2} / T_{2}

**Solution**

Applying our data into the formula, we will have:

200 / 300 = 100 / T_{2}

By making T_{2} subject of the formula, we will have

T_{2} = (300 x 100) / 200 = 150 K

Hence, T_{2} = 273 K + t

t = 150 – 273 = -123^{0}C

**Therefore, the temperature is -123 ^{0}C**

### Problem 6: Boyle’s Law

If the volume of a fixed mass of hydrogen in a container is 30 cm^{3} at a pressure of 50 mmHg. Find the volume of the gas if the pressure is 60 mmHg at constant pressure.

#### Answer

The final answer to the question is 25 cm^{3}

**Explanation:**

We need to follow the following steps to solve the problem

**First step (Data):** The available information from the question

Initial volume, V_{1} = 30 cm^{3}

Initial pressure, P_{1} = 50 mmHg

Final Pressure, P_{2} = 60 mmHg

**Unknown: **What we need to find

Final volume, V_{2} = ?

**Formula:** The equation that will help us to solve the problem

We will apply Boyle’s law, P_{1}V_{1} = P_{2}V_{2}

Since we are to find the final volume, we will now make V_{2} subject of the formula

V_{2} = (P_{1}V_{1}) / P_{2}

Therefore, we will use the above equation to solve the problem

**Solution:**

We will start by substituting our formula with available data from the question

**V _{2} = (P_{1}V_{1}) / P_{2} = (50 x 30) / 60 = 25 cm^{3} **

### Problem 7: Boyle’s Law

If the final volume of a gas is one-fifth of the initial volume. Find the ratio of the final pressure to the initial pressure at a constant temperature.

#### Answer

**Data:** Available information from the question

Final volume, V_{2} = (1/5)V_{1}

**Formula:**

We will use Boyle’s law, P_{1}V_{1} = P_{2}V_{2}

**Solution**

Changing our formula with the values in our data, we will obtain:

P_{1}V_{1} = P_{2} x [(1/5)V_{1}]

We can clearly see that V_{1} will cancel each other, leaving us with

5P_{1} = P_{2}

Therefore, to find the ratio of the final to the initial pressure. We divide both sides by P_{1} to obtain:

P_{2} / P_{1} = 5

Thus, we can equally write the above question as

(P_{2} / P_{1}) = (5 / 1)

Therefore, the ratio is

** P _{2} : P_{1} = 5 : 1**

**Thus, the ratio of the final pressure to the initial is 5 to 1**

### Problem 8: Charles’ Law

The volume of a fixed mass of a gas is 200 cm^{3} at 27^{0}C. Calculate its temperature when the volume is reduced 1/2 its original at the same pressure.

#### Answer

The final answer to the question is -123^{0}C

**Explanation:**

**Data:** The information available from the question

Initial volume, V_{1} = 200 cm^{3}

Initial temperature, T_{1} = 27^{0}C = 27 + 273 = 300 K

Final volume, V_{2} = (1/2) V_{1} = 0.5 x 200 cm^{3} = 100 cm^{3}

**Unknown:** What we need to find

Temperature = ?

**Formula:** The equation to solve the problem

V_{1} / T_{1} = V_{2} / T_{2}

**Solution:**

Making T_{2} subject of the formula, and then substituting our data into the formula, we will have

T_{2} = (V_{2}T_{1}) / V_{1} = (100 x 300) / 200 = 150 K

Thus, since T_{2} = 273 + t

We will have:

150 = 273 + t

t = 150 – 273 = -123^{0}C

### Problem 9: Gay-Lussac’s Law

A fixed mass of air at standard temperature pressure is heated such that its volume remains unchanged. Calculate its tmeperature when its pressure is increased to 114 cm of mercury (Hg).

#### Answer

The final answer to the question is 136.5^{0}C

**Explanation:**

We will commence by following these important steps below:

**Data:** The information available from the question

Initial temperature, T_{1} = 0^{0}C = 273 K

The initial pressure, P_{1} = 76 cm Hg

Final Pressure, P_{2} = 114 cm Hg

**Unknown:** What we need to find

Final temperature, T_{2} = ?

**Formula: **The equation that will help us solve the problem

P_{1 }/ T_{1} = P_{2} / T_{2}

**Solution: **

We will now make T_{2} subject of the formula and then insert our data into the formula

T_{2} = (P_{2}T_{1}) / P_{1} = (114 x 273) / 76 = 409.5

However, T = 273 + t

Therefore,

409.5 = 273 + t

Which implies that

t = 409.5 – 273 = 136.5^{0}C

### Problem 10: Gay-Lussac’s Law

A mass of gas at a pressure of 50 mmHg is heated from 27^{0}C to 97^{0}C. If the volume is maintained at a constant. Calculate the pressure exterted by the gas.

#### Answer

The final answer to the above question is 61.67 mmHg

**Explanation**:

**Data:**

Initial pressure, P_{1} = 50 mmHg

The initial temperature, T_{1} = 27^{0}C = 27 + 273 = 300 K

Final temperature, T_{2} = 97^{0}C = 97 + 273 = 370 K

**Unknown: **

Final pressure, P_{2} = ?

**Formula:**

P_{1 }/ T_{1} = P_{2} / T_{2}

**Solution**:

We will make P_{2} subject of the formula and then insert our data into the formula

P_{2} = P_{1}T_{2} / T_{1} = (50 x 370) / 300 = 61.67 mmHg

**Therefore, the pressure exterted by the gas is 61.67 mmHg**

## 15. Measurement of Gas Pressure

U-tube manometer which contains water is used in measuring gas pressure. One end of the tube is connected to the gas supply, while the other end is opened to the atmospheric pressure. Therefore, when gas is supplied, pressure from the gas causes the water level to rise to a height h, on the other side of the tube. Thus, pressure at A and h are equal since are at the same horizontal level.

Pressure at A = Atmospheric pressure = pressure due to the volume of water at height h.

Also, the Pressure of gas = Atmospheric pressure + ρgh

Where

ρ = density

g = Acceleration due to gravity

h = height

**Note:** In 1643, an Italian scientist named Torricelli invented an instrument called a barometer for measuring the pressure of the air.

## 16. Summary

Summarily, we have seen that general gas laws form the backbone of our understanding of gas behaviour. Boyle’s Law, Charles’s Law, Gay-Lussac’s Law, Combined Gas Law, Avogadro’s Law, the Ideal Gas Law, Dalton’s Law, and Graham’s Law provide valuable insights into the properties and relationships of gases.

By applying these laws, scientists and engineers can make accurate predictions and solve real-world problems across various disciplines.

So, the next time you inflate a balloon or marvel at the wonders of the atmosphere, remember that these fundamental laws are at work!

## 17. Frequently Asked Questions

Q1: What are the units of pressure in gas laws?

A1: Pressure in gas laws is typically measured in units such as Pascals (Pa), atmospheres (atm), or millimetres of mercury (mmHg).

Q2: How do gas laws relate to weather phenomena?

A2: Gas laws, such as changes in pressure and temperature, play a crucial role in understanding and predicting weather phenomena like air masses, fronts, and atmospheric stability.

Q3: Can you provide an example of a real gas?

A3:Yes, gases such as carbon dioxide (CO2) and water vapor (H2O) exhibit real gas behavior under certain conditions, deviating from the ideal gas model.

Q4: What happens to gas molecules at absolute zero?

A4: At absolute zero (-273.15°C or 0 Kelvin), gas molecules theoretically cease all motion and exhibit no pressure or volume.

Q5: How are gas laws used in scuba diving?

A5: Gas laws are essential in scuba diving for understanding the behavior of gases under pressure and determining safe diving depths and decompression schedules.

*You may also like to read:*

How to Calculate Cubic Expansivity with Examples

Linear Expansivity: Definition and Calculations

Density of Water at 20 Degrees Celsius

Also How to Calculate Specific Heat Capacity

And How to Calculate Relative Humidity Using a Formula

How to Calculate Cubic Expansivity with Examples

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## Reference

**Tags**

**temperature and pressure, amount of gas, gas is directly proportional, gay lussacs, ideal gas law, remain constant, ideal gases, law of partial pressures, straight line, volume of the gas, kinetic theory, gas constant**