Thermodynamics - How to find an "adiabat"?

[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...

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

and used the quadratic formula

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

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?

 

[DrClaude]

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?

 

[Javier0289]

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

Hi, I put the other results below

 

[Jerkov]

The best way to get the adiabat is to find an equation U as a function of V and P. Start with your EoS for P/T. Substitute in your EoS for 1/T on the LHS. Then you get an equation for P with a mess of U,V,N on the RHS. Keep working on getting a common denominator on the RHS and you are left with P=U/V. Use this to get dU=PdV+VdP and also recognize dU=del Q + del W = 0 - PdV. Then separate P and V variables on both sides to get PV^2=constant.

 

[TSny]

Callen points out in part (d) of the problem that "An adiabat is a locus of constant entropy". From the given fundamental equation $\dfrac{s}{R} = uv- \dfrac{1}{uv}$, show that the only way that $s$ can remain constant for this system is for $uv$ to be constant. From this and $P = -(\dfrac{\partial u}{\partial v})_s\,$, you can derive the form of the adiabats in the $P$-$v$ plane.

 

[Jerkov]

Callen points out in part (d) of the problem that "An adiabat is a locus of constant entropy". From the given fundamental equation $\dfrac{s}{R} = uv- \dfrac{1}{uv}$, show that the only way that $s$ can remain constant for this system is for $uv$ to be constant. From this and $P = -(\dfrac{\partial u}{\partial v})_s\,$, you can derive the form of the adiabats in the $P$-$v$ plane.

impressive

 

[ohwilleke]

Just in the interest of clarifying the term for future readers in slightly more plain language:

An adiabat is a line on a thermodynamic chart relating the pressure and temperature of a substance (such as air) that is undergoing a transformation in which no heat is exchanged with its environment.

(Source)