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Often, for thermodynamic processes only a p-V diagram is shown. Even without hard figures, the shape of the curve can be helpful to evaluate the process. However, it is hard to figure out for real processes. Are there good rules of the thumb to arrive at a T-S diagram from a p-V diagram?

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mart, is this account also yours by any chance? –  David Z Apr 23 '12 at 20:05
    
yes. Possible to merge? –  mart Apr 24 '12 at 6:59
    
sure, I'd be happy to do that. Just change the email address on the other account to match the one on this account to confirm the identity, and as soon as you do that I'll merge them together. –  David Z Apr 24 '12 at 15:00
    
can't change the address on the other to match this cause this one is already registered. –  mart Apr 26 '12 at 11:51
    
Oh, OK. Well if you can just edit something into the profile field of the other account to indicate that it's you ("please merge me" or something would work), that would be fine as well. –  David Z Apr 26 '12 at 15:56
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For a general system (unknown equation of state/energy) this will not be possible.

For example a $(p,V)$-curve with points $(p,\frac{c}{p})$ could be an isotherm because maybe $V(p)=\frac{nRT}{p}$. An isotherm corresponds to a horizontal line in a $(T,S)$-diagram. But if you don't know you're dealing with an ideal gas situation then it might look different.


As a general orientation, the stability criteria for reasonable thermodynamic systems and related relations might be helpful to guess the signs of the rates of change.

For example in a typical situation, you'd expect the compressibility

$$\beta=-\frac{1}{V}\frac{\partial V}{\partial p}$$

to be positive. The adiabatic and isothermal compressibilities are defined for curves of constant $S$ and $T$ respectively, so the corresponding adiabatic or isothermal line in the $(T,S)$-diagram will be associated with a curve $(p,V(p))$ in the $(p,V)$-diagram with

$$\frac{\partial V(p)}{\partial p}<0.$$

You see this in the ideal gas example (see picture), where the isotherm is $V\propto \frac{1}{p}$ and the adiabate is also some negative power of pressure $V(p)\propto \frac{1}{p^{\frac{1}{\gamma}}}$, with $\frac{1}{\gamma}>0$.

enter image description here

Positivity of the heat capacities $C_V$ or $C_p$ at constant volume or pressure is a dual example, which relates to properties of $S(T)$ in similar way.

Furthermore, the Maxwell relations cross correlate the slopes.

Lastly, there are some principles, following from the laws of thermodynamics, which forbid some curve configurations. E.g. two different adiabates can never cross twice, because that would mean a possibility for work without heat flow.


But as I said, besides that rough reasoning regarding the shapes of the cyclic processes, if you don't know $T(p,V)$ etc. and you can't compute heat or work, how would you know which line in the $(p,V)$ diagram corresponds to the isotherm or the adiabate? In the other case, i.e. if know know the equation of state or some other relation, then the dependencies can of course be explicitly computed.

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The equations of state of a Gibbsian substance are usually given in the form $$T=f(p,V), \quad S=g(p,V)$$ for suitable functions $f$ and $g$. I understand the query as asking when we can use them to express $p$ and $V$ in terms of $S$ and $T$. Mathematically speaking, this means that you are asking for a representation of the inverse of the corresponding mapping $$(p,V)\mapsto (S,T)$$ of the plane or a part thereof. For simple models, this can be done by solving the resulting equation for $p$ and $V$. Thus for the van der Waals gas in the form $$T=\left (p + \frac 1 {V^2}\right )(V-b), \quad S = \frac 1 {\gamma-1} (\ln p + \gamma \ln V)$$ we can easily compute that $$V=b+e^S T^{\frac 1 {1-\gamma}}\quad p=e^{-S}T^{\frac \gamma {\gamma-1}}-\frac a{(b+e^ST^{\frac 1 {\gamma-1}})^2}.$$ (The ideal gas is the special case $a=b=0$).

For more complicated models, it may not be possible to compute this inverse explicitly (for example, for a model due to Feynman which allows for the fact that the adiabatic exponent depends on temperature) but there are numerical methods available (for example, in Mathematica one has the option NSolve).

On the more theoretical side, the inverse function theorem and the fact that the Jacobian determinant of the above transformation is $1$ (Maxwell relation), ensures that this inverse exists (at least locally).

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