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
