Integrating factor in Caratheodory's formulation

Question

Why is $\dfrac{1}{T}$ chosen as integrating factor for heat differential $\delta Q$ in Caratheodory's formulation of the 2nd law of thermodynamics? It's clear to me why it is that way in Clausius approach but I am not exactly sure how mathematics imposes $\dfrac{1}{T}$ instead of just $T$, for example.

Answer 1

If the expression : $$pdx+qdy\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$$ is not a total differential, i.e.$$pdx+qdy=0\;\;\;\;\;\;\;\;(2)$$ equation (2) always has, according to the existing theorem, a general integral which is written in the form:$$f(x,y)=C$$ the function $f(x,y)$ must verify the relationship $$\frac{\partial f(x,y)}{\partial x}+\frac{\partial f(x,y)}{ \partial y}\frac{dy}{dx}=0$$ where, according to (2), we must replace$\frac{dy}{dx}$ by $-\frac{p}{q}$, that is to say we must have identically $$\frac{\frac{\partial f}{\partial x}}{p}=\frac{\frac{\partial f}{\partial y}}{q}$$

let $\mu$ be the common value of these two ratios, we have $$\frac{\partial f}{\partial x}=\mu p\;,\;\frac{\partial f}{ \partial y}=\mu q$$ that is to say that $\mu$ is the integrating factor of expression (1).

This reasoning shows that every expression $pdx+qdy$ has an integrating factor.

For the function $\mu$ to be an integrating factor of (1), it is necessary and sufficient that we have the relation (Schwarz's theorem): $$\frac{\partial (\mu p)}{\partial y}=\frac{\partial (\mu q)}{ \partial x}\;\;\;\;\;\;\;\;\;\;\;\;\;(3)$$

For one mole of ideal gas, we have the following equations $$PV=RT\;\;,\;\; \delta Q=C_{v}dT+PdV$$ according to these relations, there exists a state function $ S=C $ such that $$ dS=\mu \delta Q $$ $$dS=\mu \left(C_{v}dT+PdV \right)=\mu \left(C_{v}dT+\frac{RT}{V}dV \right)$$

if we assume $\mu=\frac{1}{T}$ the previous equation becomes $$ dS=\frac{C_{v}}{T}dT+\frac{R}{V}dV$$ and condition (3) is verified, i.e. $$\frac{\partial( \frac{C_v}{T})}{\partial V} \Bigg\vert_{T} =\frac{\partial( \frac{R}{V})}{\partial T} \Bigg\vert_{V} $$ $$\frac{1}{T}\frac{\partial C_v}{\partial V} \Bigg\vert_{T} =\frac{1}{V}\frac{\partial R}{\partial T} \Bigg\vert_{V}=0$$ so $$ dS= \frac{ \delta Q}{T} $$

ps: I've never heard of this formulation, but I tried :-)

-The mathematical part see: higher mathematics volume II by V.Smirnov.

(he introduced the relation (3) in the form of Green's formula: $\int \int_{\sigma}\left(\frac{\partial Q }{ \partial x }-\frac{\partial P}{dy}\right)d{\sigma}= \int_{l}Pdx+Qdy=0 =>\frac{\partial P}{dy}=\frac{\partial Q}{ \partial x }\;$ in our case ,$P=\mu p\; , Q=\mu q$ )

-The thermodynamic part (formulas) see (in French) Thermodynamics solved problems Higher and special mathematics 1st university cycle - Hubert Lumbroso.

Answer 2

You can denote the factor as $\beta$, and express everything just as well. However, by comparing the mathematical results with thermodynamics, we can find that integrating factor $\beta$ has to be inversely proportional to ideal gas temperature, or absolute temperature $T$. So sometimes it is denoted $1/T$ right from the start, anticipating the result and avoiding introducing a quantity (inverse temperature) which does not appear in thermodynamics of Clausius and Kelvin (they used temperature $T$).