Finding the state functions from the fundamental equation

Avogadro's number: \mu=\frac{N_A}{n}\mu' ; \mu is defined as a function of N and the extensive variables S,V instead of a function of n and the intensive variables T,P.
  • #1
fluidistic
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Homework Statement


I know that theoretically when one has the fundamental equation of a system, one can find the state equations and totally solve the system (if I understood well, I could make the analogy in classical mechanics of having the Lagrangian gives you the equations of motion). However I'm not sure how to do so. There are many problems asking you to do this and I'm stuck on this.
From Callen's book (1st edition, page 34): 1)Find the three equations of state for a system with the fundamental equation [itex]u=\left ( \frac {\theta }{R} \right )s^2+ \left ( \frac {R \theta }{v_0 ^2} \right )v^2[/itex].
2)Express [itex]\mu[/itex] as a function of T and P.

Homework Equations


[itex]du=Tds-Pdv[/itex] where [itex]u=U/N[/itex] (N is the number of moles), [itex]s=S/N[/itex] and [itex]v=V/N[/itex].


The Attempt at a Solution


I'm not exactly sure what are the 3 equations of state here. Apparently [itex]T(s,v)[/itex], [itex]P(s,v)[/itex] and [itex]\mu (s,v)[/itex] where mu is the chemical potential.
Also I don't know what are theta and R. Constants? Maybe R is the gas constant?
Can someone could get me started? I'm guessing the problem is very easy, a simple matter of differentiation/integration only and rearanging terms, etc.
Thanks a lot for any help.
 
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  • #2
In my opinion, the solution goes as follows:
[itex]u=\frac{U}{N}[/itex]
[itex]s=\frac{S}{N}[/itex]
[itex]v=\frac{V}{N}[/itex]

If we express the number of moles with the number of particles [itex]n[/itex]

[itex]n=N \:N_{A}[/itex]


where [itex]N_{A}[/itex] is the Avogadro's number, then you have

[itex]U=\frac{N_{A}}{n}\:(\frac{\theta}{R}S^{2}+\frac{R \theta}{v_{0}^{2}}V^{2})[/itex]

Then you can use
[itex]dU=TdS-pdV+\mu dn[/itex]

I think that you should treat [itex]R[/itex] and [itex]\theta[/itex] as constants.

francesco
 
Last edited:
  • #3
francesco85 said:
In my opinion, the solution goes as follows:
[itex]u=\frac{U}{N}[/itex]
[itex]s=\frac{S}{N}[/itex]
[itex]v=\frac{V}{N}[/itex]

If we express the number of moles with the number of particles [itex]n[/itex]

[itex]n=N \:N_{A}[/itex]


where [itex]N_{A}[/itex] is the Avogadro's number, then you have

[itex]U=\frac{N_{A}}{n}\:(\frac{\theta}{R}S^{2}+\frac{R \theta}{v_{0}^{2}}V^{2})[/itex]

Then you can use
[itex]dU=TdS-pdV+\mu dn[/itex]

I think that you should treat [itex]R[/itex] and [itex]\theta[/itex] as constants.

francesco
Thanks a lot. I reached [itex]T=\frac{2N_A \theta S}{nR}[/itex], [itex]P=-\frac{2R\theta V N_A}{nv_0^2}[/itex] and [itex]\mu =0[/itex].
I doubt this is right.
 
  • #4
fluidistic said:
Thanks a lot. I reached [itex]T=\frac{2N_A \theta S}{nR}[/itex], [itex]P=-\frac{2R\theta V N_A}{nv_0^2}[/itex] and [itex]\mu =0[/itex].
I doubt this is right.
Why do you obtain [itex]\mu=0[/itex] ?
In my opinion
[itex]\mu=\frac{\partial U}{\partial n}=-\frac{N_A}{n}U(S,V,n)[/itex] ; then you substitute V(T,P,n) and S(T,P,n); maybe I don't see a macroscopic error in my computation.

EDIT: sorry, I had to edit because i confused n with N
 
  • #5
Here is how I reached my result:
Since [itex]U=\frac{N_{A}}{n}\:(\frac{\theta}{R}S^{2}+\frac{R \theta}{v_{0}^{2}}V^{2})[/itex], [itex]dU= \left ( \frac{N_A }{n} \right ) \left [ \left ( \frac{\theta }{R} \right ) 2SdS + \left ( \frac{R \theta }{v_0 ^2} \right ) 2VdV \right ][/itex]. But as you said, [itex]dU=TdS-pdV+\mu dn[/itex]. So I look for matching terms. Since in my expression I haven't any "dn", I conclude that [itex]\mu =0[/itex].
 
  • #6
fluidistic said:
Here is how I reached my result:
Since [itex]U=\frac{N_{A}}{n}\:(\frac{\theta}{R}S^{2}+\frac{R \theta}{v_{0}^{2}}V^{2})[/itex], [itex]dU= \left ( \frac{N_A }{n} \right ) \left [ \left ( \frac{\theta }{R} \right ) 2SdS + \left ( \frac{R \theta }{v_0 ^2} \right ) 2VdV \right ][/itex]. But as you said, [itex]dU=TdS-pdV+\mu dn[/itex]. So I look for matching terms. Since in my expression I haven't any "dn", I conclude that [itex]\mu =0[/itex].

In my opinion one should treat n as an independent variable, like S and V; the reason why it is not usually done is that n is considered constant; the chemical potential takes into account the fact that it can change.
So
[itex]dU=\frac{\partial U}{\partial S}dS+\frac{\partial U}{\partial V}dV+\frac{\partial U}{\partial n} dn[/itex] so that
[itex]\frac{\partial U}{\partial n}=\mu[/itex]
 
  • #7
Thank you immensely. :smile:
I get that [itex]\mu =-\frac{N_A}{n^2} \left [ \left ( \frac{\theta }{R} \right ) S^2 + \left ( \frac{R \theta }{v_0 ^2} \right ) V^2 \right ][/itex].
 
  • #8
fluidistic said:
Thank you immensely. :smile:
I get that [itex]\mu =-\frac{N_A}{n^2} \left [ \left ( \frac{\theta }{R} \right ) S^2 + \left ( \frac{R \theta }{v_0 ^2} \right ) V^2 \right ][/itex].

You are welcome :)
 
  • #9
2)[itex]\mu (T,P)=-\frac{T^2}{4N_A \theta R^2}-\frac{P^2v_0^2}{4N_AR\theta }[/itex].
 
  • #10
Hmm I am confused now. My book says [itex]dU=TdS-pdV+\mu dN[/itex], not [itex]dU=TdS-pdV+\mu dn[/itex]. Where N is the number of moles and n is the number of particles.
 
  • #11
fluidistic said:
Hmm I am confused now. My book says [itex]dU=TdS-pdV+\mu dN[/itex], not [itex]dU=TdS-pdV+\mu dn[/itex]. Where N is the number of moles and n is the number of particles.

It is a matter of definitions, in my opinion; the two [itex]\mu[/itex]'s are related by multiplication of [itex]N_A[/itex]
 

Related to Finding the state functions from the fundamental equation

1. What are state functions?

State functions are thermodynamic properties that depend only on the current state of a system, such as temperature, pressure, and volume. They do not depend on the path taken to reach that state, only the initial and final conditions.

2. How do you determine state functions from the fundamental equation?

To determine state functions from the fundamental equation, you need to identify which variables in the equation are state functions. These are typically quantities that are not affected by changes in the system's path. The remaining variables are path functions, which depend on the specific process or path taken to reach a certain state.

3. Can state functions be measured directly?

No, state functions cannot be measured directly. They can only be determined by measuring other properties, such as temperature or pressure, and using mathematical equations to calculate the state function.

4. How do state functions relate to thermodynamic processes?

State functions are important in thermodynamics because they allow us to analyze and understand thermodynamic processes without needing to know every detail of the process. State functions only depend on the initial and final states, making them useful for determining the overall change in a system during a process.

5. What is the significance of state functions in thermodynamics?

State functions are important in thermodynamics because they allow us to predict and analyze the behavior of a system without needing to know the exact path or process. They also help us to understand the relationship between different thermodynamic variables and how they change during a process.

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