# Examples of Gas Laws

We explain that what are examples of gas laws? The **Laws of Gases** are postulates that describe the **relationship between the physical properties of gases** , and have resulted from many experiments carried out over several centuries. It is **Boyle’s Law** , which relates **Pressure and Volume** ; the **Gay-Lussac Law** , which relates the **pressure and temperature** ; and **Charles’s Law** , which relates **Volume and Temperature** .

Together, these three generalizations have had great relevance in the development of many ideas in the field of Chemistry.

In addition, these three Laws allow us to **calculate one of the properties** when the gas is in a change: from the initial state to the final state. For this, all the properties must be in the same System of Units. If **Temperature** is involved in the calculation, as in Gay-Lussac’s Law and Charles’s Law, it **must be on the absolute scale: Kelvin** .

## Boyle’s Law

In the 17th century, **Robert Boyle** studied the behavior of gases. In a series of experiments, Boyle analyzed the relationship between pressure and volume in a sample of a gas.

As the **pressure (P) increased** at constant temperature (T = cte), **the Volume (V) of a quantity of gas decreased** . On the contrary, if the applied pressure decreased, the Volume occupied by the gas increased.

As a conclusion, it was defined that at constant temperature the volume is inversely proportional to the applied pressure. This relationship is known today as **Boyle’s Law** , and is stated:

**“The Pressure of a fixed quantity of a gas at constant temperature is inversely proportional to the volume of the gas”**

You can write a mathematical expression that shows the inverse relationship between Pressure and Volume:

**P α (1 / V)**

Where the symbol α means “proportional to:”. You can change α to the equal sign and write

**P = k * (1 / V)**

Where k is a constant called ” **Constant of proportionality** “. The equation is a mathematical expression **of Boyle’s Law** . You can rearrange the equation to get:

**PV = k**

This form of Boyle’s Law states that the product of Pressure and Volume of a gas at constant temperature and quantity of the gas is a constant.

So it can also be set to compare an initial state with a final state:

**P _{1} V _{1} = P _{2} V _{2}**

## Gay Lussac’s Law

Gay Lussac, in conjunction with Charles, established the Law that bears his name, and which proposes that there is **a proportional relationship between the Temperature of the Gas and the Pressure** it takes inside the container that contains it.

When **the temperature rises** , the gas molecules acquire more kinetic energy, so they collide more with the internal walls of the container. This results in **increased pressure** .

As **the temperature was reduced** , the gas molecules relaxed, hitting the inner walls of the container to a lesser extent. The result is **less pressure** .

The mathematical expression for Gay-Lussac’s Law is developed:

**P α T**

**P = k * T**

**(P / T) = k**

And to compare between two states, initial and final, it is written:

**(P _{1} / T _{1} ) = (P _{2} / T _{2} )**

## Charles Law

In his experiments, Charles discovered that there is a **proportional relationship between the temperature of the gas and the volume it occupies** .

The **higher the temperature** , the gas **occupied more volume** , as its particles were much more agitated, expanding the system that contained it.

If there was a **decrease in temperature** , the gas **occupied less volume** , since its molecules were less restless.

Charles formulated the law that describes the dependence of the volume of a gas with the temperature, given by:

**V α T**

**V = k * T**

**(V / T) = k**

Where k is the constant of proportionality. This law is stated:

**“The volume of a fixed quantity of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas.”**

As was done for the Pressure-Volume relationship of Boyle’s Law, it is possible to compare the two initial and final states to express this law:

**(V _{1} / T _{1} ) = (V _{2} / T _{2} )**

## Examples of Gas Laws

1.- A gas that is at a Pressure of 3 atm covers a Volume of 0.03m ^{3} . Expands to a Volume of 0.10m ^{3} . What pressure does he finally have?

**P _{1} V _{1} = P _{2} V _{2}**

P _{1} = 3 atm

V _{1} = 0.03 m ^{3}

P _{2} =?

V _{2} = 0.10 m ^{3}

(3 atm) * (0.03m ^{3} ) = P _{2} * (0.10m ^{3} )

**P _{2} = 0.9 atm**

2.- A gas that is at a Pressure of 10 atm covers a Volume of 2.40m ^{3} . It is compressed to a Volume of 1m ^{3} . What pressure does he finally have?

**P _{1} V _{1} = P _{2} V _{2}**

P _{1} = 10 atm

V _{1} = 2.40 m ^{3}

P _{2} =?

V _{2} = 1 m ^{3}

(10 atm) * (2.40 m ^{3} ) = P _{2} * (1 m ^{3} )

**P _{2} = 24 atm**

3.- A gas that is at a Pressure of 0.5 atm covers a Volume of 6m ^{3} . It is compressed to a Volume of 0.01m ^{3} . What pressure does he finally have?

**P _{1} V _{1} = P _{2} V _{2}**

P _{1} = 0.5 atm

V _{1} = 6 m ^{3}

P _{2} =?

V _{2} = 0.01 m ^{3}

(0.5 atm) * (6 m ^{3} ) = P _{2} * (0.01 m ^{3} )

**P _{2} = 300 atm**

4.- A gas that is at a Pressure of 1.32 atm covers a Volume of 0.001m ^{3} . Expands to a Volume of 0.015m ^{3} . What pressure does he finally have?

**P _{1} V _{1} = P _{2} V _{2}**

P _{1} = 1.32 atm

V _{1} = 0.001 m ^{3}

P _{2} =?

V _{2} = 0.015 m ^{3}

(1.32 atm) * (0.001 m ^{3} ) = P _{2} * (0.015 m ^{3} )

**P _{2} = 0.088 atm**

5.- A gas that is at a Temperature of 303 K has a Pressure of 900mmHg. It is heated to a Temperature of 353 K. What is its new Pressure?

**(P _{1} / T _{1} ) = (P _{2} / T _{2} )**

P _{1} = 900 mmHg

T _{1} = 303 K

P _{2} =?

T _{2} = 353 K

(900 mmHg) / (303 K) = P _{2} / (353 K)

**P _{2} = 1048.51 mmHg**

6.- A gas that is at a Temperature of 603 K has a Pressure of 1500mmHg. It heats up to a Temperature of 1203 K. What is its new Pressure?

**(P _{1} / T _{1} ) = (P _{2} / T _{2} )**

P _{1} = 1500 mmHg

T _{1} = 603 K

P _{2} =?

T _{2} = 1203 K

(1500 mmHg) / (603 K) = P _{2} / (1203 K)

**P _{2} = 2992.53 mmHg**

7.- A gas that is at a Temperature of 213 K has a Pressure of 750mmHg. It is heated to a Temperature of 393 K. What is its new Pressure?

**(P _{1} / T _{1} ) = (P _{2} / T _{2} )**

P _{1} = 750 mmHg

T _{1} = 213 K

P _{2} =?

T _{2} = 393 K

(750 mmHg) / (213 K) = P _{2} / (393 K)

**P _{2} = 1383.80 mmHg**

8.- A gas that is at a Temperature of 253 K occupies a Volume of 0.0015 m ^{3} . It is heated to a temperature of 313K. What volume does it occupy at the end?

**(V _{1} / T _{1} ) = (V _{2} / T _{2} )**

V _{1} = 0.0015 m ^{3}

T _{1} = 253 K

V _{2} =?

T _{2} = 313 K

(0.0015 m ^{3} ) / (253 K) = V _{2} / (313 K)

**V _{2} = 0.0018 m ^{3}**

9.- A gas that is at a Temperature of 473 K occupies a Volume of 0.5 m ^{3} . It is heated to a temperature of 603 K. What volume does it occupy at the end?

**(V _{1} / T _{1} ) = (V _{2} / T _{2} )**

V _{1} = 0.5 m ^{3}

T _{1} = 473 K

V _{2} =?

T _{2} = 603 K

(0.5 m ^{3} ) / (473 K) = V _{2} / (603 K)

**V _{2} = 0.6374 m ^{3}**

10.- A gas that is at a Temperature of 803 K occupies a Volume of 0.90 m ^{3} . It cools down to a temperature of 653 K. What volume does it occupy at the end?

**(V _{1} / T _{1} ) = (V _{2} / T _{2} )**

V _{1} = 0.90 m ^{3}

T _{1} = 803 K

V _{2} =?

T _{2} = 653 K

(0.90 m ^{3} ) / (803 K) = V _{2} / (653 K)

**V _{2} = 0.7318 m ^{3}**