Thermodynamics, finding the fundamental equation of ideal gases

In summary, an ideal monoatomic gas is characterized by the equations PV=NRT and U=\frac{3NRT}{2}, and the fundamental equation for this type of gas can be found by using the equation S=\left ( \frac{1}{T} \right ) U+\left ( \frac{P}{T} \right ) V- \left ( \frac{\mu }{T} \right ) N. By applying the Gibbs-Duhem relation and integrating it, the fundamental equation is found to be S= \frac {5NR}{2}-N \left ( \frac {\mu}{T} \right ) _0 +NR \ln \left [ \left ( \frac {U
  • #1
fluidistic
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Homework Statement


An ideal monoatomic gas is characterized by the two equations [itex]PV=NRT[/itex] and [itex]U=\frac{3NRT}{2}[/itex] in which R is a constant.
Find the fundamental equation corresponding to a monoatomic ideal gas.

Homework Equations


[itex]S=\left ( \frac{1}{T} \right ) U+\left ( \frac{P}{T} \right ) V- \left ( \frac{\mu }{T} \right ) N[/itex].


The Attempt at a Solution


I'm given 2 equations of state, one is missing in order to get the fundamental equation. In the entropy representation the variables of the fundamental equation are [itex]\frac{1}{T}[/itex], [itex]\frac{P}{T}[/itex] and [itex]\frac{\mu }{T}[/itex].
The missing one is [itex]\frac{\mu }{T} (U,V,N)=\frac{\mu}{T} (u,v)[/itex] where u and v in lower script are the molar energy and volume respectively.
So I have that [itex]\frac{1}{T}=\frac{3R}{2u}[/itex] and [itex]\frac{P}{T}=\frac{R}{v}[/itex].
I use Gibbs-Duhem's relation [itex]d \left ( \frac{\mu }{T} \right ) =ud \left ( \frac{1 }{T} \right ) +v d \left ( \frac{P }{T} \right )[/itex]. Replacing the variables in parenthesis and integrating both sides, I reach (like the book): [itex]\frac{\mu}{T}=-\frac{3}{2}R \ln \left ( \frac{u}{u_0} \right ) -R \ln \left ( \frac{v}{v_0} \right ) +\left ( \frac{\mu }{\mu _0} \right ) _0[/itex].
When I replace this equation into the expression for S given above and using the fact that u=N/U, [itex]u_0=\frac{N}{U_0}[/itex], I reach that [itex]S =\frac{3U^2 R}{2N}+\frac{RV^2}{N}+ N \left [ \frac{3R}{2} \ln \left ( \frac{U_0}{U} \right ) +R\ln \left ( \frac{V_0}{V} \right ) - \left ( \frac{\mu }{T} \right ) _0 \right ][/itex]. Using some log property this simplifies to [itex]S=\frac{3U^2R}{2N} +\frac{RV^2}{N} +NR \ln \left [ \left ( \frac{U_0}{U} \right ) ^{3/2} \left ( \frac{V_0}{V} \right ) \right ] -N \left ( \frac{\mu }{\mu _0} \right ) _0[/itex].
However this differs from the answer given in the book: [itex]S= \frac {5NR}{2}-N \left ( \frac {\mu }{T} \right ) _0 +NR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac {N}{N_0} \right ) ^{-5/2} \right ][/itex].
I do not see, for the life of me, what I've done wrong.
 
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  • #2
Hi fluidistic.

Let's start with your integration constant.
It should be ##(\frac{\mu}{T})_0## instead of ##(\frac{\mu}{\mu_0})_0##.

Then you write u=N/U, but that is not right.
Since you defined u to be the molar volume, u=U/N.

Perhaps you can redo it?
 
  • #3
Thanks a lot for helping me!
You are absolutely right, I've redone everything with more care. I reach however [itex]S=\frac{5RN}{2}-N \left ( \frac{\mu}{T} \right ) _0+NR \ln \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \right ][/itex].
Therefore it seems like I'm missing a term in the logarithm.
It seems it comes from the integration of the Gibbs-Duhem relation: [itex]d \left ( \frac{\mu }{T} \right ) =ud \left ( \frac{1 }{T} \right ) +v d \left ( \frac{P }{T} \right )[/itex]. It seems that when I integrate I should get an extra term like "[itex]-R \ln \left ( \frac{N}{N_0} \right )[/itex]. I don't really know how to get that.
 
  • #4
Maybe it's simpler than you think.
What did you fill in for ##u_0##?
 
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  • #5
I like Serena said:
Maybe it's simpler than you think.
What did you fill in for ##u_0##?
Hmm I still don't see it. I took [itex]u_0=\frac{U_0}{N}[/itex].
 
  • #6
fluidistic said:
Hmm I still don't see it. I took [itex]u_0=\frac{U_0}{N}[/itex].

Note that N is also a variable that changes.
It has an initial value too, just like ##u_0## and ##U_0##.
 
  • #7
I like Serena said:
Note that N is also a variable that changes.
It has an initial value too, just like ##u_0## and ##U_0##.

Yes I know this but when should I make use of this?
Here are my steps:
1)[itex]\frac{1}{T}=\frac{3NR}{2U}=\frac{3R}{2u}[/itex]
2)[itex]\frac{P}{T}=\frac{NR}{V}=\frac{R}{v}[/itex]
I use Gibbs-Duhem's relation [itex]d \left ( \frac{\mu }{T} \right ) =ud \left ( \frac{1 }{T} \right ) +v d \left ( \frac{P }{T} \right )[/itex].
3)[itex]ud \left ( \frac{1 }{T} \right ) =-\frac{3Rdu}{2u}[/itex]
4)[itex]v d \left ( \frac{P }{T} \right ) =-R\frac{dv}{v}[/itex].
Integrating the GD relations gives me [itex]\frac{\mu}{T}=-\frac{3}{2}R \ln \left ( \frac{u}{u_0} \right ) -R \ln \left ( \frac{v}{v_0} \right ) +\left ( \frac{\mu }{T} \right ) _0[/itex]. The book (Callen's page 52) reaches exactly the same result so far.
He then says that inserting this into the Euler equation [itex]S=\left ( \frac{1}{T} \right ) U+\left ( \frac{P}{T} \right ) V- \left ( \frac{\mu }{T} \right ) N[/itex] one reaches [itex]S= \frac {5NR}{2}-N \left ( \frac {\mu }{T} \right ) _0 +NR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac {N}{N_0} \right ) ^{-5/2} \right ][/itex].
So my error is when I insert what I obtained into the Euler equation or it's in integrating GD relation (in which case the book would also be wrong) or it's in either of the 4 steps I denoted by 1), etc.?
I really don't see it. :/
 
  • #8
I got to your book result substituting ##u_0=\frac{U_0}{N_0}## and ##v_0=\frac{V_0}{N_0}## in the integrated GD result.
 
  • #9
I like Serena said:
I got to your book result substituting ##u_0=\frac{U_0}{N_0}## and ##v_0=\frac{V_0}{N_0}## in the integrated GD result.

Ah! That explains everything :biggrin: In my case those were variables, I didn't even notice. Now I reach the desired result... thank you very much for all.
 
  • #10
Just out of curiosity, were my previous hints too cryptic?
It is not my intention to make a thread a cryptic puzzle...
Perhaps you have an idea what kind of hint I might have given that would have helped you better?
 
  • #11
I like Serena said:
Just out of curiosity, were my previous hints too cryptic?
It is not my intention to make a thread a cryptic puzzle...
Perhaps you have an idea what kind of hint I might have given that would have helped you better?

You mean post 6? No, not cryptic. In fact now I understand your "hint" that I should have used N_0 instead of N. I did not realize that with my definition of u_0, it wasn't even a constant.
So in my case pointing out in post 6 that my definition of u_0 was wrong and that it's not even a constant would have made me understand where the problem lies. I did not realize it, because I already knew that N was a variable and I thought that was what you wanted me to know.
But I'm maybe a "weird learner" :) Anyway, thank you very much for having went through all of this with me.
 
  • #12
Okay. Thanks.
 

What is thermodynamics?

Thermodynamics is the branch of physics that deals with the relationships between heat, energy, and work. It studies the behavior of systems that involve temperature, heat transfer, and energy conversion.

What is the fundamental equation of ideal gases?

The fundamental equation of ideal gases is known as the Ideal Gas Law, which is PV = nRT. This equation relates the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas.

What is an ideal gas?

An ideal gas is a theoretical gas that follows the Ideal Gas Law and has the following properties: the particles have no volume, there are no intermolecular forces between the particles, and the collisions between particles are perfectly elastic.

How is the fundamental equation of ideal gases derived?

The fundamental equation of ideal gases can be derived from the Kinetic Molecular Theory, which describes the behavior of gases at a molecular level. By assuming the properties of an ideal gas, the Ideal Gas Law can be derived from the Kinetic Molecular Theory.

What is the significance of the fundamental equation of ideal gases?

The fundamental equation of ideal gases is significant because it allows us to predict the behavior of ideal gases in various situations. It is used in many applications, such as in the design of gas storage tanks and in the study of atmospheric gases. It also serves as the basis for understanding the behavior of real gases.

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