In a isobaric process, the pressure P of a system remains constant. The prefix "iso" comes from the Greek and is used to denote that something remains constant, while "baros", also from the Greek, means weight.
Isobaric processes are very typical both in closed containers and in open spaces, being easy to locate them in nature. By this we mean that it is possible physical and chemical changes on the earth's surface or chemical reactions in containers open to the atmosphere..
Some examples are obtained by heating a balloon filled with air in the sun, cooking, boiling or freezing water, steam generated in boilers, or the process of raising a hot air balloon. Later we will give an explanation of these cases.
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Let us derive an equation for the isobaric process assuming that the system under study is an ideal gas, a fairly suitable model for almost any gas at less than 3 atmospheres of pressure. The ideal gas particles move randomly, occupying the entire volume of the space that contains them without interacting with each other..
If the ideal gas enclosed in a cylinder fitted with a movable piston is allowed to expand slowly, it can be assumed that its particles are in equilibrium at all times. Then the gas exerts on the piston of area TO a force F of magnitude:
F = p.A
Where p is the pressure of the gas. This force exerts work producing an infinitesimal displacement dx on the piston given by:
dW = Fdx = pA.dx
As the product Adx is a volume differential dV, then dW = pdV. It remains to integrate both sides from the initial volume VTO until final volume VB to get the total work done by the gas:
The situation described is experimentally verified by confining a gas inside a cylinder provided with a movable piston, as shown in Figures 2 and 3. A weight of mass M is placed on the piston, the weight of which is directed downwards, while the gas exerts an upward force thanks to the pressure P produced on the piston.
Since the piston is able to move freely, the volume that the gas occupies can change without problem, but the pressure remains constant. Adding the atmospheric pressure Patm, which also exerts a downward force, we have:
Mg - P.A + Patm . A = constant
Therefore: P = (Mg / A) + Patm it does not vary, unless M and thus the weight are modified. By adding heat to the cylinder, the gas will expand by increasing its volume or it will contract when extracting heat..
The ideal gas equation of state relates the variables of importance: pressure P, volume V and temperature T:
P.V = n .R.T
Here n represents the number of moles and R is the ideal gas constant (valid for all gases), which is calculated by multiplying Boltzmann's constant by Avogadro's number, resulting in:
R = 8.31 J / mol K
When the pressure is constant, the equation of state can be written as:
V / T = nR / P
But nR / P is constant, since n, R, and P are. So when the system goes from state 1 to state 2, the following proportion arises, also known as Charles's law:
V1/ T1 = Vtwo/ Ttwo
Substituting in W = PΔV, the work done to go from state 1 to state 2 is obtained, in terms of the constants and the temperature variation, easy to measure with a thermometer:
W1 → 2 = nR (Ttwo - T1) = nR.ΔT
On the other hand, the first law of thermodynamics states that:
∆U = Q - W
This means that adding a certain amount of heat Q to the gas increases the internal energy ∆U and increases the vibrations of its molecules. In this way, the gas expands and does work by moving the piston, as we have said before..
In a monatomic ideal gas and the variation of the internal energy ∆U, which includes both the kinetic energy and the potential energy of its molecules, is:
∆U = (3/2)nR ΔT
Finally, we combine the expressions that we have been obtaining into one:
Q = ∆U + W = (3/2)nR ΔT + nR∆T = (5/2) nR ΔT
Alternatively Q can be rewritten in terms of the mass m, the temperature difference, and a new constant called specific heat of gas at constant pressure, abbreviated cp, whose units are J / mol K:
Q = m cp ∆T
Not all isobaric processes are carried out in closed containers. In fact, innumerable thermodynamic processes of all kinds occur at atmospheric pressure, so isobaric processes are very frequent in nature. This includes physical and chemical changes to the earth's surface, chemical reactions in vessels open to the atmosphere, and much more..
For isobaric processes to occur in closed systems, their boundaries must be flexible enough to allow changes in volume without varying pressure.
This was what happened in the experiment of the piston that moved easily as the gas expanded. It is the same by enclosing a gas in a party balloon or a hot air balloon.
Here we have several examples of isobaric processes:
Boiling water for tea or cooking sauces in open containers are good examples of isobaric processes, since they all take place at atmospheric pressure..
As the water is heated, the temperature and volume increase and if heat continues to be added, the boiling point is finally reached, at which the phase change of the water from liquid to water vapor occurs. While this occurs, the temperature also remains constant at 100ºC..
On the other hand, freezing water is also an isobaric process, whether it takes place in a lake during winter or the home refrigerator..
Another example of an isobaric process is the change in the volume of a balloon inflated with air when it is left exposed to the Sun. First thing in the morning, when it is not very hot yet, the balloon has a certain volume..
As time passes and the temperature increases, the balloon also heats up, increasing its volume and all this occurs at constant pressure. The material of the balloon is a good example of a boundary that is flexible enough so that the air inside it, when heated, expands without modifying the pressure..
The experience can also be carried out by adjusting the uninflated balloon in the spout of a glass bottle filled with a third of water, which is heated in a bain-marie. As soon as the water is heated, the balloon inflates immediately, but care must be taken not to heat too much so that it does not explode..
It is a floating ship without propulsion, which makes use of air currents to transport people and objects. The balloon is usually filled with hot air, which, being colder than the surrounding air, rises and expands causing the balloon to rise..
Although the air currents direct the balloon, it has burners that are activated to heat the gas when it is desired to ascend or maintain altitude, and deactivate when descending or landing. All this happens at atmospheric pressure, assumed constant at a certain height not far from the surface..
Steam is generated in boilers by heating water and keeping the pressure constant. Afterwards, this steam performs a usable job, for example generating electricity in thermoelectric plants or activating other mechanisms such as locomotives and water pumps..
There are 40 liters of gas at a temperature of 27 ºC. Find the increase in volume when heat is added isobarically until reaching 100 ºC.
Charles's law is used to determine the final volume, but attention: the temperatures must be expressed in kelvin, just adding 273 K to each one:
27 ºC = 27 + 273 K = 300 K
100 ºC = 100 + 273 K = 373 K
From:
V1/ T1 = Vtwo/ Ttwo ⇒ Vtwo = Ttwo(V1/ T1) = 373 ºC (40 L / 300 K) = 49.7 L
Finally the volume increase is Vtwo - V1 = 49.7 L - 40 L = 9.7 L.
Supplied 5.00 x 103 J of energy to an ideal gas so that it realizes 2.00 x 103 J working on his environment in an isobaric process. It is asked to find:
a) The change in the internal energy of the gas.
b) The change in volume, if now the internal energy decreases by 4.50 x 103 J and 7.50 x 10 are ejected3 J of the system, considering constant pressure of 1.01 x 105 Pa.
Used ∆U = Q - W and the values given in the statement are substituted: Q = 5.00 x 103 J and W = 2.00 x 103 J:
=5.00 x 103 J - 2.00 x 103 J = 3.00 x 103 J
Therefore the internal energy of the gas increases by 3.00 x 103 J.
The volume change is found in the work done: W = P∆V:
∆U = Q - W = Q - P∆V
The statement states that the internal energy decreases, therefore: ∆U= -4.50 x 103 J. It also tells us that a certain amount of heat is expelled: Q = -7.50 x 103 J. In both cases, the negative sign represents decrease and loss, so:
-4.50 x 103 J = -7.50 x 103 J - P∆V
Where P = 1.01 x 105 Pa. As all units are in the International System, we proceed to solve for the change in volume:
∆V = (-4.50 x 103 J +7.50 x 103 J) / (- 1.01 x 105 Pa) = -2.97 x 10-two m3
As the volume change is negative, it means that the volume decreased, that is, the system contracted.
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