Proof that $\frac{1}{T}$ is the integrating factor for $dq$

Question

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$

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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.

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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...

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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?

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Reference for integrating factors

Answer 1

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:

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