# Isobaric Process: Formulas, Equations, Experiments, Exercises

In an isobaric process , the pressure P of a system is kept 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 physical and chemical changes on the earth’s surface or chemical reactions in vessels open to the atmosphere are possible .

Some examples are obtained by cheating an air-filled balloon in the sun , cooking, boiling or freezing water, the steam that is generated in boilers or the process of raising a hot air balloon. We will give an explanation of these cases later.

## Formula and equations

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 A a force F of magnitude:

F = pA

Where p is the pressure of the gas. This force does work producing an infinitesimal displacement dx in the piston given by:

dW = Fdx = pA.dx

Since the product Adx is a volume differential dV , then dW = pdV. It remains to integrate both sides from the initial volume V A to the final volume V B to obtain the total work done by the gas:

$W=\int_{V_{A}}^{V_{B}}pdV$

$W=P_{{o}}\int_{V_{A}}^{V_{B}}dV=P_{{o}}\Delta&space;V$If ΔV is positive, the gas expands and the opposite happens when ΔV is negative. The graph of pressure versus the volume (PV diagram) of the isobaric process is a horizontal line joining states A and B, and the work done simply equivalent to the rectangular area under the curve.

## Experiments

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 it exerts an upward force thanks to the pressure P it produces 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 P atm , which also exerts a downward force, we have:

Mg-PA + P atm . A = constant

Therefore: P = (Mg / A) + P atm  does not vary, unless M and thus the weight is modified. By adding heat to the cylinder, the gas will expand by increasing its volume or it will contract as heat is removed.

### Isobaric processes in the ideal gas

The ideal gas equation of state relates the variables of importance: 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, 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 :

V 1 / T 1 = V 2 / T 2

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:

W 1 → 2 = nR (T 2 – T 1 ) = 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 the specific heat of the gas at constant pressure, abbreviated c p , whose units are J / mol K:

Q = m c ∆T

## Examples

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 easily moved 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:

### 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 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 happens, the temperature also remains 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 winter or the home 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 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 one third of water, which is heated in a water bath. 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.

### 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 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.

### Boilers

Steam is generated in boilers by heating water and maintaining constant pressure. Afterwards, this steam performs a useful work, for example generating electricity in thermoelectric plants or operating other mechanisms such as locomotives and water pumps.

## Solved exercises

### Exercise 1

You have 40 liters of gas at a temperature of 27 ºC. Find the volume increase when heat is added isobarically until reaching 100 ºC.

#### Solution

Charles’s law is used to determine the final volume, but be careful : 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:

V 1 / T 1 = V 2 / T 2 ⇒ V 2 = T 2 (V 1 / T 1 ) = 373 ºC (40 L / 300 K) = 49.7 L

Finally the volume increase is V 2 – V 1 = 49.7 L – 40 L = 9.7 L.

### Exercise 2

An ideal gas is supplied with 5.00 x 10 3 J of energy to do 2.00 x 10 3 J of work on its surroundings in an isobaric process. It asks 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 10 3 J and 7.50 x 10 3 J are expelled from the system, considering a constant pressure of 1.01 x 10 5 Pa.

#### Solution to

We use ∆U = Q – W and substitute the values ​​given in the statement: Q = 5.00 x 10 3 J and W = 2.00 x 10 3 J:

= 5.00 x 10 3 J – 2.00 x 10 3 J = 3.00 x 10 3 J

Therefore the internal energy of the gas increases by 3.00 x 10 3 J.

#### Solution b

The change in volume 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 10 3 J. It also tells us that a certain amount of heat is expelled: Q = -7.50 x 10 3 J. In both cases, the sign negative represents decrease and loss, then:

4.50 x 10 3 J = -7.50 x 10 3 J – P∆V

Where P = 1.01 x 10 5 Pa. As all units are in the International System, we proceed to solve for the change in volume:

∆V = (- 4.50 x 10 3 J +7.50 x 10 3 J) / (- 1.01 x 10 5 Pa ) = -2.97 x 10 -2 m 3

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

## References

1. Byjou’s. Isobaric Process. Recovered from: byjus.com.
2. Cengel, Y. 2012. Thermodynamics. 7maEdition. McGraw Hill.