GENERAL GAS LAW: FORMULAS, APPLICATIONS AND EXERCISES - CHEMISTRY - 2025

The general gas law is the result of combining Boyle-Mariotte's law, Charles's law, and Gay-Lussac's law; in fact, these three laws can be considered particular cases of the general gas law. In turn, the general gas law can be considered as a particularization of the ideal gas law.

The general gas law establishes a relationship between the volume, pressure and temperature of a gas. In this way he affirms that, given a gas, the product of its pressure by the volume it occupies divided by the temperature at which it is found always remains constant.

Gases are present in different processes in nature and in a great multitude of applications, both industrial and in everyday life. Therefore, it is not surprising that the general gas law has multiple and diverse applications.

For example, this law makes it possible to explain the operation of different mechanical devices such as air conditioners and refrigerators, the operation of hot air balloons, and can even be used to explain the processes of cloud formation.

Formulas

The mathematical formulation of the law is as follows:

P ∙ V / T = K

In this expression P is the pressure, T represents the temperature (in degrees Kelvin), V is the volume of the gas, and K represents a constant value.

The previous expression can be replaced by the following:

P ₁ ∙ V ₁ / T ₁ = P ₂ ∙ V ₂ / T ₂

This last equation is quite useful to study the changes that gases undergo when one or two of the thermodynamic variables (pressure, temperature and volume) are modified.

Boyle-Mariotte's Law, Charles's Law, and Gay-Lussac's Law

Each of the aforementioned laws relates two of the thermodynamic variables, in the case that the third variable remains constant.

Charles's law states that volume and temperature are directly proportional as long as the pressure remains unchanged. The mathematical expression of this law is the following:

V = K ₂ ∙ T

For its part, Boyle's law establishes that pressure and volume have an inverse relationship with each other when the temperature remains constant. Boyle's law is mathematically summarized as follows:

P ∙ V = K ₁

Finally, Gay-Lussac's law states that temperature and pressure are directly proportional for cases in which the volume of the gas does not vary. Mathematically the law is expressed as follows:

P = K ₃ ∙ T

In this expression K ₁, K ₂ and K ₃ represent different constants.

Ideal gas law

The general gas law can be obtained from the ideal gas law. The ideal gas law is the equation of state of an ideal gas.

An ideal gas is a hypothetical gas made up of particles with a point character. The molecules of these gases do not exert any gravitational force with each other and their collisions are characterized by being totally elastic. In this way, the value of its kinetic energy is directly proportional to its temperature.

The real gases whose behavior most closely resembles that of ideal gases are monatomic gases at low pressures and high temperatures.

The mathematical expression of the ideal gas law is as follows:

P ∙ V = n ∙ R ∙ T

This equation n is the number of moles and R is the universal constant of ideal gases whose value is 0.082 atm ∙ L / (mol ∙ K).

Applications

Both the general gas law and the Boyle-Mariotte, Charles, and Gay-Lussac laws can be found in a multitude of physical phenomena. In the same way, they serve to explain the operation of many and varied mechanical devices of everyday life.

For example, in a pressure cooker you can observe Gay Lussac's Law. In the pot the volume remains constant, so if the temperature of the gases that accumulate in it increases, the internal pressure of the pot also increases.

Another interesting example is that of the hot air balloon. Its operation is based on Charles Law. Given that atmospheric pressure can be considered practically constant, what happens when the gas that fills the balloon is heated is that the volume it occupies increases; thus its density is reduced and the balloon can ascend.

Solved exercises

First exercise

Determine the final temperature of gas whose initial pressure of 3 atmospheres doubles until reaching a pressure of 6 atmospheres, while reducing its volume from a volume of 2 liters to 1 liter, knowing that the initial temperature of the gas was 208, 25 ºK.

Solution

Substituting in the following expression:

P ₁ ∙ V ₁ / T ₁ = P ₂ ∙ V ₂ / T ₂

you have to:

Solving for, we get that T ₂ = 208.25 ºK

Second exercise

Given a gas subjected to a pressure of 600 mm Hg, occupying a volume of 670 ml and at a temperature of 100 ºC, determine what its pressure will be at 473ºK if at that temperature it occupies a volume of 1500 ml.

Solution

First of all, it is advisable (and in general, necessary) to transform all the data into units of the international system. Thus, you have to:

P ₁ = 600/760 = 0.789473684 atm approximately 0.79 atm

V ₁ = 0.67 l

T ₁ = 373 ºK

P ₂ =?

V ₂ = 1.5 l

T ₂ = 473 ºK

Substituting in the following expression:

P ₁ ∙ V ₁ / T ₁ = P ₂ ∙ V ₂ / T ₂

you have to:

0.79 ∙ 0.67 / 373 = P ₂ ∙ 1.5 / 473

Solving for P ₂ we get:

P ₂ = 0.484210526 approximately 0.48 atm

References

1. Schiavello, Mario; Vicente Ribes, Leonardo Palmisano (2003). Fundamentals of Chemistry. Barcelona: Editorial Ariel, SA
2. Laider, Keith, J. (1993). Oxford University Press, ed. The World of Physical Chemistry.
3. General gas law. (nd). On Wikipedia. Retrieved on May 8, 2018, from es.wikipedia.org.
4. Gas laws. (nd). In Wikipedia. Retrieved on May 8, 2018, from en.wikipedia.org.
5. Zumdahl, Steven S (1998). Chemical Principles. Houghton Mifflin Company.