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Consider the first law of thermodynamics,

$$ dU = dq +dw$$

simplfying,

$$ dU + P_{\text{ext}} dV = dq$$

Now we can say that $ q $ is a function of $ U$ and $V$

$ dq(U,V) = dU + P_{\text{ext}} dV$


For a differential $dF(x,y) = A\, dx + B\, dy $ to be exact,

$$ \frac{\partial A}{\partial y} = \frac{\partial B}{\partial x}$$ is a necessary condition.


Clearly the function $q(U,V)$ doesn't obey this definition, and hence, let us multiply both sides by an integrating factor $ \phi(U,V)$ such that the condition of exact differential is satisfied.

$$ \phi(U,V) \, dq = \phi(U,V) \, dU + \phi(U,V) P_{\text{ext}} \, dV$$

For this to be exact,

$$ \frac{ \partial \phi(U,V) P_{\text{ext}} }{\partial U} = \frac{\partial \phi(U,V) }{\partial V}$$

Which leads to:

$$ \left( \frac{\partial P_{\text{ext}} }{ \partial U} \right)_V \phi + P_{\text{ext}}(\frac{\partial \phi}{\partial U})_V =(\frac{\partial \phi}{\partial V})_U $$

Now, I'm not sure how to get a general solution for the above partial differential equation...


My real goal is to derive the expression for entropy at end and prove that $ \frac{1}{T}$ is the integrating factor but I'm a bit stuck. I've seen the proof that $ \frac{1}{T}$ is the integrating factor by people pointing to carnots theorem that the circulation of the differentials over a loop is zero but I wanted to derive it using differential equations.

Now, I'm thinking of how I can include the assumption that the process is reversible because the entropy definition is written used $ dq_{\text{rev}}$; also maybe derive the entropy generation term.

Any hints?


Reference for integrating factors

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The choise of the extensive quantities $(U,V)$ as state variables is appropriate, but of course $Q$ is not a state function. It is better to start from the equation of a reversible adiabatic transformation:

$$ dU + p(U,V) dV = 0 $$

If $p(U, V)$ is a known state function of the thermodynamic system the differential equation is integrable and the integration can be carried out with the method of the integrating factor $F(U,V)$:

$$ {{dU + p(U,V)dV} \over {F(U,V)}} = dS(U,V) \qquad {where:} \quad {\partial{}\over \partial{V}} \left({1}\over{F}\right) = {\partial{}\over\partial{U}} \left({p}\over{F}\right) $$

The differential calculus asserts that integrating factors can always be found and therefore the equation of a reversible adiabatic transformation can be written in the form:

$$ S(U,V) = const $$

where S is a state function such that:

$$ {\partial{S}\over\partial{U}} = {{1}\over{F}} \qquad {\partial{S}\over\partial{V}}={{p}\over{F}} $$

Now stop with math! Physical arguments allow to prove that exists an universal integrating factor (independent of the particular system considered!) called absolute thermodynamic temperature $T$ and that this factor is directly proportional to the absolute temperature defined by the gas thermometer.

For this discussion I must refer to the section 6 of an italian link (unfortunately I haven't had time to translate the pdf into English so far):

http://pangloss.ilbello.com/Fisica/Termodinamica/lavoro_calore.pdf

In this way the goal of defining entropy by differential means is achieved:

$$ dS(U,V) = {{dU + p(U,V)dV}\over{T}} = {dQ \over T} $$

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  • $\begingroup$ Wow +50, but please explain " Physical arguments allow to prove that exists a universal integrating factor".. what physical arguments? $\endgroup$ Oct 5, 2020 at 16:41
  • $\begingroup$ If this $S$ is function of state for adiabatic change, how it is proved that it is function of state for any other process? $\endgroup$ Oct 5, 2020 at 16:42
  • $\begingroup$ @Buraian In my article you see (Fig.9) an adiabatic system consisting of two arbitrary subsystems of variable volume but in thermal equilibrium between them. It's shown that the integrating factor must be the same for all systems having the same temperature: $F (V_A, T) = F (V_B, T) \to F = F (T)$. Subsequently the thermodynamic temperature $F(T)$ is identified with the absolute temperature $T$ of a gas thermometer. In Def.(13) $S(U,V)$ is a state function: $\Delta S$ doesn't depend on the transformation. $\endgroup$
    – Pangloss
    Oct 6, 2020 at 16:28

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