Gases, from the point of view of thermodynamics, are described by a set of macroscopic characteristics, the main of which are temperature, pressure and volume. The constancy of one of the above parameters and a change in the other two suggests that this or that isoprocess proceeds in the gas. This article will devote a detailed answer to the questions that this is an isochoric process, how it differs from an isothermal and isobaric change in the state of a gas system.
The ideal gas in physics
Before answering the question that this is an isochoric process, one should familiarize oneself with the concept of ideal gas. In physics, it is understood as any gas in which the average kinetic energy of its constituent particles far exceeds the potential energy of their interaction, and the distances between these particles are several orders of magnitude greater than their linear dimensions. Under the indicated conditions, it is possible to ignore the interaction energy between the particles (it is equal to zero) during the calculations, and it can also be assumed that the particles are material points having a certain mass m.
The only process that takes place in an ideal gas is the collision of particles with the walls of the vessel in which the substance is located. In practice, these collisions manifest themselves in the form of the existence of a certain pressure P.
As a rule, any gaseous substance, which consists of relatively chemically inert molecules and which has a slight pressure and high temperatures, can be considered an ideal gas with sufficient accuracy for practical calculations.
The ideal gas equation
Of course, we are talking about the universal law of Clapeyron-Mendeleev, which should be well understood in order to understand that this is an isochoric process. So, the universal equation of state has the following form:
P * V = n * R * T.
That is, the product of pressure P by the volume of gas V is equal to the product of the absolute temperature T by the amount of substance in moles n, where R is the proportionality coefficient. The equation itself was first written by Emil Clapeyron in 1834, and in the 70s of the XIX century D. Mendeleev replaced in it a set of constants of one single universal gas constant R (8.314 J / (mol * K)).
According to the Clapeyron-Mendeleev equation, in a closed system the number of gas particles remains constant, therefore there are only three macroscopic parameters that can vary (T, P and V). The latter fact underlies the understanding of various isoprocesses, which will be discussed below.
What is an isochoric process?
By this process we mean absolutely any change in the state of the system in which its volume is preserved.
If we turn to the universal equation of state, we can say that in an isochoric process in a gas only pressure and absolute temperature change. To understand exactly how the change in thermodynamic parameters occurs, we write the corresponding mathematical expression:
P / T = const.
Sometimes this equality is given in a slightly different form:
P 1 / T 1 = P 2 / T 2 .
Both equalities are called Charles’s law by the name of a French scientist who, at the end of the 18th century, received the noted dependence experimentally.
If we plot the function P (T), then we get a straight-line dependence, which is called an isochore. Any isochore (for all values of n and V) is a straight line.
Energy process description
As noted, an isochoric process is a change in the state of a system that occurs in a closed but not isolated system. We are talking about the possibility of heat exchange between gas and the environment. In the general case, any supply of heat Q to the system leads to two results:
- the internal energy U changes;
- the gas does work A, expanding or contracting.
The last conclusion is mathematically written as follows:
Q = U + A.
The isochoric process of an ideal gas, by its definition, does not imply that the gas performs work, since its volume remains unchanged. This means that all the heat supplied to the system goes to increase its internal energy:
Q = U.
If we substitute an explicit formula for internal energy in this expression, then the heat of the isochoric process can be represented as:
Q = z / 2 * n * R * T.
Here z is the number of degrees of freedom, which is determined by the polyatomicity of the molecules making up the gas. For a monatomic gas, z = 3, for a diatomic gas - 5, and for a triatomic and more - 6. Here, the degrees of freedom mean translational and rotational degrees.
If we compare the efficiency of heating the gas system during isochoric and isobaric processes, then in the first case we will obtain maximum efficiency, since during the isobaric change in the state of the system, the gas expands, and part of the heat input is spent on the work.
Isobaric process
Above, we have described in detail that this is an isochoric process. Now let's say a few words about other isoprocesses. Let's start with isobaric. Based on the name, by it is meant the transition of the system between states at constant pressure. This process is described by Gay-Lussac's law as follows:
V / T = const.
As with the isochore, the isobar V (T) also represents a straight line on the graph.
For any isobaric process, it is convenient to calculate the work performed by the gas, since it is equal to the product of constant pressure and the volume change.
Isothermal process
This is a process in which the system temperature remains constant. It is described by the Boyle-Mariotte law for ideal gas. It is curious to note that this is the first experimentally discovered gas law (second half of the 17th century). His mathematical notation looks like this:
P * V = const.
Isochoric and isothermal processes differ in terms of their graphical representation, since the function P (V) is a hyperbolic, and not a linear dependence.
Problem solving example
We fix the theoretical information provided in the article by their application to solve a practical problem. It is known that pure gaseous nitrogen was in the cylinder at a pressure of 1 atmosphere and a temperature of 25 ° C. After the gas cylinder was heated and the pressure in it was measured, it turned out to be 1.5 atmospheres. What is the temperature of the gas in the cylinder after heating? How much did the internal energy of the gas change if there were 4 mol of nitrogen in the cylinder.
To answer the first question, we use the following expression:
P 1 / T 1 = P 2 / T 2 .
Where do we get:
T 2 = P 2 / P 1 * T 1 .
In this expression, the pressure can be substituted in arbitrary units of measure, since they are reduced, and the temperature is only in kelvins. With that said, we get:
T 2 = 1.5 / 1 * 298.15 = 447.224 K.
The calculated temperature in degrees Celsius is 174 ° C.
Since the nitrogen molecule is diatomic, the change in its internal energy during heating can be determined as follows:
ΔU = 5/2 * n * R * ΔT.
Substituting the known values into this expression, we get the answer to the second question of the problem: ΔU = +12.4 kJ.