State equations for a thermodynamic substance/system

[cianfa72]

TL;DR

About the state equations for a thermodynamic substance/system.

Hi, as follow up to this thread I believe for any substance/thermodynamic system there exists actually a set of 3 state equations between the 5 variables $(U,T,S,p,V)$.

For example in the case of ideal gas which are the 3 equations ? Thanks.

 

[Chestermiller]

What are your thoughts on this? Please also articulate your understanding of the definition of a "state equation."

 

[cianfa72]

Please also articulate your understanding of the definition of a "state equation."

For an ideal gas, I'm aware of there are two state equations, namely $$pV=nRT$$ and $$U=\frac 3 2 nRT$$ From that thread it should be another equation in which enters the entropy $S$. What is this third equation ?

 

[Chestermiller]

For an ideal gas, I'm aware of there are two state equations, namely $$pV=nRT$$ and $$U=\frac 3 2 nRT$$ From that thread it should be another equation in which enters the entropy $S$. What is this third equation ?

How about $$dS=\frac{3}{2}nR\frac{dT}{T}+nR\frac{dV}{V}$$

 

[cianfa72]

Yes, integrating it we get $$S=\frac 3 2 nR \,lnT + nR\, lnV$$

 

[vanhees71]

That cannot be true, because you have dimensionful quantities in the logarithm. The correct Sackur-Tetrode formula for the entropy of an ideal gas is
$$S=\frac{5}{2} k_{\text{B}} N +k_{\text{B}} N \ln \left [ \frac{V}{N} \left (\frac{m U}{3 \pi \hbar^2 N} \right)^{3/2}\right].$$

 

[cianfa72]

That cannot be true, because you have dimensionful quantities in the logarithm

Sorry, we cannot simply integrate the differential form in post #4 ?

 

[vanhees71]

[EDIT: Correted typos in formulae in view of #9]

From this you can only get the entropy differences, i.e.,
$$S-S_0=\frac{3}{2} n R \ln(T/T_0)+n R \ln(V/V_0)=n R \ln \left [\frac{V}{V_0} \left (\frac{T}{T_0} \right)^{3/2}\right] .$$
Now $U=3 N k_{\text{B}} T/2$ and $n R=N k_{\text{B}}$. So you can write the above result as
$$S-S_0=N k_{\text{B}} \ln \left [\frac{V}{V_0} \left (\frac{U}{U_0} \right)^{3/2} \right].$$
Thus this is, of course, consistent with the Sackur-Tetrode formula for the absolute entropy, but the latter can only be derived by semi-classical quantum considerations, not from phenomenological classical thermodynamics.

You need in addition to the "classical fundamental Laws 0-2 of thermodynamics" also Nernst's theorem (3rd Law) as well as the indistinguishability of particles and the "natural measure" for phase-space volumes, which is determined by QT in terms of Planck's action constant, $h=2 \pi \hbar$.

For details of a semi-classical argument for the entropy, see Sect. 1.5 in

https://itp.uni-frankfurt.de/~hees/publ/kolkata.pdf

 

[cianfa72]

So you can write the above result as
$$S-S_0=N k_{\text{B}} \ln \left [\frac{V}{V_0} \left (\frac{U}{U_0} \right) \right].$$

From what you said, it should be actually: $$S-S_0=N k_{\text{B}} \ln \left [\frac{V}{V_0} \left (\frac{U}{U_0} \right)^{\frac 3 2} \right]$$

 

[vanhees71]

Of course. I correct it in the original posting.