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 in nature. By this we mean that physical and chemical changes on the earth’s surface or chemical reactions in containers open to the atmosphere are possible.

Some examples are obtained by cblowing a balloon full of air to the sun, cooking, boiling or freezing water, the 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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**formula and equations**

Let us derive an equation for the isobaric process assuming that the system under study is an ideal gas, a fairly adequate model for almost any gas at less than 3 atmospheres of pressure. Ideal gas particles move randomly, they occupy the entire volume of the space that contains them without interacting with each other.

If an ideal gas enclosed in a cylinder fitted with a movable piston is allowed to expand slowly, its particles can be assumed to be in equilibrium at all times. Then the gas exerts on the piston of area *TO* a force *F* of magnitude:

*F = pA*

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*

like the product *adx* is a volume differential *dV*so *dW = pdV. *Subtract to integrate both sides from the initial volume *GOES* until the final volume *VB* To get the total work done by the gas:

When the system pressure remains constant at a value Po, it goes outside the integral, in which case:

If ΔV is positive, the gas expands and the opposite occurs when ΔV is negative. The graph of pressure versus volume (PV diagram) of the isobaric process is a horizontal line joining states A and B, and the work done simply equals the rectangular area under the curve.

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**experiments**

The situation described is verified experimentally 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, whose weight is directed downwards, while the gas exerts an upward force due to the pressure P it produces on the piston.

Since the piston is free to move, the volume occupied by the gas can change without problem, but the pressure remains constant. Adding the atmospheric pressure Patm, which also exerts a downward force, we have:

*Mg–PA + Patm. A = constant*

Therefore: P = (Mg/A) + Patm does not change, unless M is modified and with it the weight. By adding heat to the cylinder, the gas will either expand by increasing its volume or contract by removing heat.

**Isobaric processes in the ideal gas**

The ideal gas equation of state relates the important variables: pressure P, volume V and temperature T:

* **PV= n .RT*

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, giving as a result:

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 constant. So when the system goes from state 1 to state 2, the following proportion arises, also known as Charles’s law:

* **V1/T1 = V2/T2*

substituting in *W = PΔV*the work done to go from state 1 to 2 is obtained, in terms of the constants and the temperature variation, easy to measure with a thermometer:

*W1→2 = nR (T2 – 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 displacing 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 a single 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 = mcp ∆T*

**examples**

Not all isobaric processes take place in closed vessels. In fact, countless thermodynamic processes of all kinds occur at atmospheric pressure, so isobaric processes are very common in nature. This includes physical and chemical changes on 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 changing pressure.

This was what happened in the experiment with the piston moving easily when the gas expanded. The same is true of enclosing a gas in a party balloon or a hot air balloon.

Here we have several examples of isobaric processes:

**boil water and cook**

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 water is heated, the temperature and volume increase, and if you continue to add heat, you eventually reach the boiling point, at which the phase change of water from liquid to water vapor occurs. While this is happening, the temperature is also kept constant at 100ºC.

**freeze the water**

On the other hand, freezing water is also an isobaric process, whether it takes place in a lake during the winter or the domestic refrigerator.

**Heating a balloon filled with air in the sun**

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 yet very hot, 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 balloon material is a good example of a boundary that is flexible enough that the air inside it, when heated, expands without changing the pressure.

The experiment can also be carried out by fitting the uninflated balloon to the mouth of a glass bottle filled with one 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 get too hot or it will burst.

**The aerostatic balloon**

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 cooler 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 you want to ascend or maintain altitude, and are deactivated when descending or landing. All this happens at atmospheric pressure, assumed to be constant at a certain height not far from the surface.

**boilers**

In boilers, steam is generated by heating water and maintaining constant pressure. This steam then performs usable work, for example generating electricity in thermoelectric plants or driving other mechanisms such as locomotives and water pumps.

**solved exercises**

**Exercise 1**

There are 40 liters of gas at a temperature of 27ºC. Find the increase in volume when heat is added isobarically up to 100°C.

**Solution**

Charles’ law is used to determine the final volume, but be careful: the temperatures must be expressed in kelvin, it is enough to add 273 K to each one:

27ºC = 27 + 273K = 300K

100ºC = 100 + 273K = 373K

From:

*V1/T1 = V2/T2 ⇒ V2 = T2(V1/T1)= 373 ºC (40 L/300 K) = 49.7 L*

Finally the volume increase is V2 – V1 = 49.7 L – 40 L = 9.7 L.

**Exercise 2**

5.00 x 103 J of energy is supplied to an ideal gas to do 2.00 x 103 J of work on its surroundings in an isobaric process. You are 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 103 J are expelled from the system, considering a constant pressure of 1.01 x 105 Pa.

**Solution to**

Used *∆U = Q – W* and substitute the values given in the statement: *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.

**solution b**

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. Since all the units are in the International System, we proceed to clear the change of volume:

* **∆V = (-**4.50 x 103 J +7.50 x 103 J)/ (- 1.01 x 105 Pa*) = -2.97 *x 10-2 m3*

Since the volume change is negative, it means that the volume decreased, that is, the system contracted.

**References**

Byjou’s. Isobaric Process. Recovered from: byjus.com.

Cengel, Y. 2012. Thermodynamics. 7ma Edition. McGraw Hill.

process xyz. Learn more about the isobaric process. Recovered from: 10process.xyz.

Serway, R., Vulle, C. 2011. Fundamentals of Physics. 9na Ed. Cengage Learning.

Wikipedia. Gas Laws. Recovered from: es.wikipedia.org.