Isobaric process for a Van der Waals gas

In summary, the problem at hand involves calculating changes in internal energy, entropy, heat transferred, and work done for various processes, as well as determining the latent heat for a given isotherm. The VdW gas parameters a and b are known and can be used to calculate pressure for the gas. However, for one of the processes, the volume at one point is unknown and would require solving a third degree equation. The latent heat can be calculated using the formula L = TΔS, where T is the temperature at the beginning of the isotherm and ΔS is the change in entropy over the isotherm.
  • #1
Luis Obis
6
0

Homework Statement


[/B]
You are asked to calculate changes in internal energy, entropy, heat transferred and work done for each of the following process. Also you are asked to calculate "the latent heat for the isotherm in the figure".

We know the a and b parameters which characterize the VdW gas since you can get them from the shape of the isotherms assuming you know the volume V0 and the critical temperature Tcri.

Assume ##N = 1 mol##.

VdW.png

Homework Equations



Pressure for the VdW gas (N = 1):

##P = \frac{RT}{V-b}-\frac{a}{V^2}##

The Attempt at a Solution



I have done as asked for every process but for D->A.

The problem is that we don't know the volume at D. We could calculate it using the VdW equation and using the fact that the pressures at A and D are equal, however this will give us a third degree equation which we could solve analitcally but I don't think this would be the best answer. Any ideas how to proceed?

For the latent heat I have done as follows:

##L = T \Delta S##

Where in this case ##T## would be ##T_0## and ##\Delta S## would be the change in entropy from 1.5V0 to 4V0 in the isotherm ##T = T_0##. Is this correct?

Thanks you very much for any help!
 
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  • #2
Why are you scared of solving a third degree equation? People solve quadratics without existential hang ups.

The author of this article certainly has no hang-ups about grappling with solving a cubic equation:

https://en.wikipedia.org/wiki/Van_der_Waals_equation
 

Related to Isobaric process for a Van der Waals gas

1. What is an isobaric process for a Van der Waals gas?

An isobaric process for a Van der Waals gas is a thermodynamic process in which the pressure remains constant while the volume and temperature of the gas change. This type of process is commonly used to study the behavior of real gases, such as Van der Waals gases, which deviate from ideal gas behavior.

2. How does an isobaric process for a Van der Waals gas differ from an ideal gas?

Unlike ideal gases, Van der Waals gases do not behave according to the ideal gas law. They have non-zero molecular volumes and experience intermolecular forces, which affect their behavior. In an isobaric process, these deviations from ideal gas behavior are taken into account, making it a more accurate representation of real gas behavior.

3. What is the equation for an isobaric process for a Van der Waals gas?

The equation for an isobaric process for a Van der Waals gas is: (P + a/V^2)(V - b) = RT, where P is the pressure, V is the volume, a and b are the Van der Waals constants, R is the gas constant, and T is the temperature. This equation takes into account the non-ideal behavior of Van der Waals gases.

4. What are the applications of isobaric processes for Van der Waals gases?

Isobaric processes for Van der Waals gases have various applications in fields such as chemical engineering, physics, and thermodynamics. They are used to study the behavior of real gases and to understand the effects of intermolecular forces on gas properties. Isobaric processes are also used in the design of industrial processes, such as refrigeration and liquefaction of gases.

5. How is an isobaric process for a Van der Waals gas represented on a P-V diagram?

An isobaric process for a Van der Waals gas is represented by a horizontal line on a P-V diagram, as the pressure remains constant during the process. The volume, however, changes as the gas is compressed or expanded. The shape of the line may deviate from the ideal gas behavior, depending on the Van der Waals constants for the gas.

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