What is an isothermal process? (Examples, exercises)

What is an isothermal process? (Examples, exercises)

The isothermal or isothermal process is a reversible thermodynamic process in which the temperature remains constant. In a gas, there are situations in which a change in the system does not produce variations in temperature, but in physical characteristics.

These changes are phase changes, when the substance changes from solid to liquid, from liquid to gas, or vice versa. In these cases, the substance’s molecules readjust their position, adding or removing thermal energy.

The thermal energy required for a phase change to occur in a substance is called latent heat or heat of transformation.

One way to make a process isothermal is to put the substance that will be the system under study in contact with an external thermal deposit, which is another system with high caloric capacity. In this way, such a slow heat exchange takes place that the temperature remains constant.

This type of process often occurs in nature. For example, in humans when body temperature rises or falls, we feel sick because in our bodies countless life-sustaining chemical reactions take place at a constant temperature. This is true for warm-blooded animals in general.

Other examples include ice that melts in the spring heat and ice cubes that cool the beverage.

 Examples of isothermal processes

-The metabolism of warm-blooded animals is carried out at a constant temperature.

-When water boils, a phase change occurs, from liquid to gas, and the temperature remains constant at approximately 100°C, as other factors can influence the value.

Melting ice is another common isothermal process, as is putting water in the freezer to make ice cubes.

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– Car engines, refrigerators and many other types of machines work correctly within a certain temperature range. Devices called thermostats are used to maintain the proper temperature . Several operating principles are used in its design.

The Carnot Cycle

A Carnot engine is an ideal machine from which work is obtained thanks to fully reversible processes. It is an ideal machine because it does not consider processes that dissipate energy, such as the viscosity of the substance that does the work, nor friction.

The Carnot cycle consists of four stages, two of which are precisely isothermal and the other two adiabatic. Isothermal stages are the compression and expansion of a gas responsible for producing useful work.

An automobile engine operates on similar principles. The movement of a piston inside the cylinder is transmitted to other parts of the car and produces movement. It does not have the behavior of an ideal system such as the Carnot engine, but thermodynamic principles are common.

Calculation of work performed in an isothermal process

To calculate the work done by a system when the temperature is constant, you need to use the first law of thermodynamics, which states:

ΔU = Q – W

This is another way of expressing energy conservation in the system, presented through ΔU or energy change, Q as the supplied heat, and finally W , which is the work done by that system.

Suppose the system in question is an ideal gas contained in the cylinder of a moving piston in area A , which functions when its volume V changes from 1 to 2 .

The ideal gas equation of state is PV = nRT , which relates the volume with the pressure P and temperature T . The values ​​of n and R are constant: n is the number of moles of gas and R is the constant of gases. In the case of an isothermal process, the photovoltaic product is constant.

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Well, the work done is calculated by integrating a small differential work, in which a force F produces a small displacement dx:

dW = Fdx = PAdx

As Adx is exactly the variation of the dV volume , then:

dW = PoE

To get the total work in an isothermal process, the dW expression is integrated:

Pressure P and volume V are plotted on a photovoltaic diagram as shown in the figure and the work done is equivalent to the area under the curve:

Since ΔU = 0, since the temperature remains constant, in an isothermal process, we have to:

Q = W

– Exercise 1

A cylinder with a moving piston contains an ideal gas at 127 °C. If the piston moves to 10 times the initial volume, keeping the temperature constant, find the number of moles of gas contained in the cylinder, if the work done on the gas is 38,180 J.

Data : R = 8.3 J/mol. K

Solution

The statement states that the temperature remains constant, so we are in the presence of an isothermal process. For the work done in the gas, we have the equation deduced above:

127 º C = 127 + 273 K = 400 K

Solve for n, the number of moles:

n = W / RT ln (V2 / V1) = -38 180 J / 8.3 J / mol K x 400 K x ln (V 2 / 10V 2 ) = 5 moles

A minus sign was placed before the job. The attentive reader may have noticed in the previous section that W was defined as “the work done by the system” and has a + sign. Therefore, the “work performed on the system” carries a negative sign.

– Exercise 2

There is air in a cylinder with a piston. Initially, there are 0.4 m 3 of gas under pressure of 100 kPa and 80 º C temperature. The air is compressed to 0.1 m 3, ensuring that the temperature inside the cylinder remains constant during the process.

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Determine how much work is done during this process.

Solution

We use the equation for the work deduced above, but the number of moles is unknown, which can be calculated with the ideal gas equation:

80 º C = 80 + 273 K = 353 K.

1 V 1 = nRT → n = P 1 V 1 / RT = 100000 Pa x 0.4 m 3 / 8.3 J / mol. K x 353 K = 13.65 mol

W = nRT ln (V 2 / V 1 ) = 13.65 mol x 8.3 J / mol. K x 353 K x ln (0.1 / 0.4) = -55,442.26 J

Again, the minus sign indicates that work has been done on the system, which always happens when the gas is compressed.

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