The adiabatic transition between two states in gases is not among the isoprocesses; nevertheless, it plays an important role not only in various technological processes, but also in nature. In this article, we consider what this process is, and we also give the equations of the adiabat of an ideal gas.
Ideal gas in brief
An ideal gas is one in which there are no interactions between its particles, and their size is zero. In nature, of course, there are no hundred percent perfect gases, because they all consist of measuring molecules and atoms that always interact with each other using at least van der Waals forces. Nevertheless, the described model is often performed with sufficient accuracy for solving practical problems for many real gases.
The main equation for an ideal gas is the Clapeyron-Mendeleev law. It is written in the following form:
P * V = n * R * T.
This equation establishes direct proportionality between the product of pressure P by volume V and the amount of substance n by the absolute temperature T. The quantity R is the gas constant, which plays the role of the proportionality coefficient.
What is an adiabatic process?
An adiabatic process is a transition between the states of a gas system in which energy is not exchanged with the environment. In this case, all three thermodynamic characteristics of the system (P, V, T) are changed, and the amount of substance n remains constant.
There are adiabatic expansion and contraction. Both processes occur only due to the internal energy of the system. So, as a result of expansion, the pressure and especially the temperature of the system fall sharply. Conversely, adiabatic compression leads to a positive jump in temperature and pressure.
To prevent heat exchange between the environment and the system, the latter must have heat-insulated walls. In addition, reducing the duration of the process significantly reduces the heat flux from and to the system.
Poisson equations for the adiabatic process
The first law of thermodynamics is written as follows:
Q = ΔU + A.
In other words, the heat Q communicated to the system is used by the system to perform work A and to increase its internal energy ΔU. To write the adiabatic equation, we should put Q = 0, which corresponds to the definition of the process under study. We get:
ΔU = -A.
In an isochoric process in an ideal gas, all the heat goes to increase internal energy. This fact allows us to write the equality:
ΔU = C V * ΔT.
Where C V - isochoric heat capacity. Work A, in turn, is calculated as follows:
A = P * dV.
Where dV is a small change in volume.
In addition to the Clapeyron-Mendeleev equation, the following equality holds for an ideal gas:
C P - C V = R.
Where C P is the isobaric heat capacity, which is always more isochoric, since it takes into account the gas loss due to expansion.
Analyzing the equalities written above and carrying out integration over temperature and volume, we arrive at the following adiabatic equation:
T * V γ-1 = const.
Here γ is the adiabatic exponent. It is equal to the ratio of isobaric heat capacity to isochoric one. This equality is called the Poisson equation for the adiabatic process. Applying the Clapeyron-Mendeleev law, we can write two more similar expressions, only through the parameters PT and PV:
T * P γ / (γ-1) = const;
P * V γ = const.
The adiabatic graph can be given in various axes. Below it is shown in the axes PV.
The colored lines on the graph correspond to isotherms, the black curve is the adiabat. As can be seen, the adiabat behaves more sharply than any of the isotherms. This fact is easy to explain: for an isotherm, the pressure changes inversely with the volume, for an isobath, the pressure changes faster, since the exponent γ> 1 for any gas system.
Task example
In nature in the mountains, when the air mass moves up the slope, its pressure drops, it increases in volume and cools. This adiabatic process leads to a decrease in the dew point and to the formation of liquid and solid precipitation.
It is proposed to solve the following problem: in the process of lifting the air mass along the slope of the mountain, the pressure dropped by 30% compared with the pressure at the foot. What did its temperature become equal if at the foot it was 25 o C?
To solve the problem, use the following adiabatic equation:
T * P γ / (γ-1) = const.
It is better to write it like this:
T 2 / T 1 = (P 2 / P 1 ) (γ-1) / γ .
If P 1 is taken as 1 atmosphere, then P 2 will be equal to 0.7 atmosphere. For air, the adiabatic exponent is 1.4, since it can be considered an ideal diatomic gas. The temperature T 1 is 298.15 K. Substituting all these numbers in the expression above, we obtain T 2 = 269.26 K, which corresponds to -3.9 o C.