# Thermodynamics - How to find an "adiabat"?

• Javier0289
In summary: It seems like the equations of state are homogeneous of zero order and the temperature is always positive. I also found the mechanical equation of state for P(T,v), but I'm not sure if it guarantees constant entropy. In summary, the conversation discussed the fundamental relation and how to obtain a simplified expression using substitutions. The quadratic formula was used and the result was then derived. The conversation also mentioned finding the mechanical equation of state for P(T,v) and the concern about constant entropy.
Javier0289
Homework Statement
Hi, I've been struggling to find the form of the adiabat of this problem:
Given the fundamental relation, find the form of the adiabats in the P-V plane
(problem 2.3-5. Thermodynamics and an Introduction to Thermostatistics, Callen, pp 42)
Relevant Equations
$${S \over R} = {UV \over N} - {N^3 \over UV}$$
I tried this... but I'm not sure if I'm doing it right or maybe there's a simpler way. Thanks for your time or help :)

The fundamental relation is:

$${S \over R} = {UV \over N} - {N^3 \over UV}$$

but I used $$s=S/N, u=U/N, v=V/N$$ to obtain

$${s \over R} = uv - {1 \over uv}$$

then I did some algebra...

$$u^2 - ({s \over Rv})u - {1 \over v^2} = 0$$

$$u = ({1 \over 2Rv})({s \pm \sqrt{s^2+4R^2}})$$

next I derived the previous expression

$$(\frac{\partial u}{\partial v})_s = -P$$

$$P = ({1 \over 2Rv^2})({s \pm \sqrt{s^2+4R^2}})$$

And my doubt is... how do I get rid of s?

I also tried

$$(\frac{\partial s}{\partial v})_u = {P \over T}$$

and got

$${P \over T} = R(u+{1 \over uv^2})$$

but then how do I get rid of u?

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You need the answers from the preceding sub-questions. What did you get?

Javier0289
Hi! from a) all the equations of state are homogeneous of zero order
$${1 \over T} = R({V \over N} + {N^3 \over U^2V})$$
$${P \over T} = R({U \over N} + {N^3 \over UV^2})$$
$${μ \over T} = R({UV \over N^2} + {3N^2 \over UV})$$
b) the temperature is always positive
c) the mechanical equation of state P(T,v)
$$P(T,v) = {RT \over v}[({1 \over vRT}-1)^{-1 \over 2}+({1 \over vRT}-1)^{1 \over 2}]$$
But then I realized that in the last equation is not guaranteed that the entropy will be kept constant, or is it?

DrClaude said:
You need the answers from the preceding sub-questions. What did you get?
Hi, I put the other results below

## 1. What is an adiabat in thermodynamics?

An adiabat in thermodynamics is a thermodynamic process in which there is no heat transfer between the system and its surroundings. This means that the system's internal energy remains constant.

## 2. How do you find an adiabat on a thermodynamic diagram?

To find an adiabat on a thermodynamic diagram, you need to plot the points where the pressure and volume of the system remain constant. This will create a curve on the diagram, which represents the adiabatic process.

## 3. What is the difference between an adiabat and an isotherm?

An adiabat is a thermodynamic process in which there is no heat transfer, while an isotherm is a process in which the temperature remains constant. This means that adiabats are represented by curved lines on a thermodynamic diagram, while isotherms are represented by straight lines.

## 4. How does an adiabatic process affect the internal energy of a system?

In an adiabatic process, the internal energy of a system remains constant. This is because there is no heat transfer, so the energy of the system is not changing. However, the temperature and pressure of the system may change due to work being done on or by the system.

## 5. What are some real-life examples of adiabatic processes?

Some real-life examples of adiabatic processes include the compression or expansion of a gas in a piston, the movement of air in a weather system, and the compression of air in a diesel engine. In all of these examples, there is no heat transfer, so the process can be considered adiabatic.

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