The general gas law It is the result of combining the Boyle-Mariotte law, the Charles law and the Gay-Lussac 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 in industry 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 it can even be used to explain the processes of cloud formation..
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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:
P1 ∙ V1 / T1 = Ptwo ∙ Vtwo / Ttwo
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..
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 = Ktwo ∙ 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 = K1
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 = K3 ∙ T
In said expression K1, Ktwo and K3 represent different constants.
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 specific particles. 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 is most similar to that of ideal gases are monatomic gases when they are 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).
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. Since 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; this reduces its density and the balloon can rise.
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.
Substituting in the following expression:
P1 ∙ V1 / T1 = Ptwo ∙ Vtwo / Ttwo
you have to:
3 ∙ 2 / 208.25 = 6 ∙ 1 / Ttwo
Clearing, you get to Ttwo = 208.25 ºK
Given a gas subjected to a pressure of 600 mm of Hg, occupying a volume of 670 ml and at a temperature of 100 ºC, determine what its pressure will be at 473 ºK if at that temperature it occupies a volume of 1500 ml.
In the first place it is advisable (and in general, necessary) to transform all the data into units of the international system. Thus, you have to:
P1 = 600/760 = 0.789473684 atm approximately 0.79 atm
V1 = 0.67 l
T1 = 373 ºK
Ptwo = ?
Vtwo = 1.5 l
Ttwo = 473 ºK
Substituting in the following expression:
P1 ∙ V1 / T1 = Ptwo ∙ Vtwo / Ttwo
you have to:
0.79 ∙ 0.67 / 373 = Ptwo ∙ 1.5 / 473
Solving for Ptwo you get to:
Ptwo = 0.484210526 approximately 0.48 atm
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