# Find that $d\left(\frac{\mu}{T}\right)=ud\left(\frac{1}{T}\right)+vd\left(\frac{P}{T}\right)$

Find that $$d\left(\frac{\mu}{T}\right)=ud\left(\frac{1}{T}\right)+vd\left(\frac{P}{T}\right)$$

$$U=TS-PV+\mu N\tag{1}\label{1}$$ $$dU=TdS-PdV+\mu d N \tag{2}\label{2}$$ From equation \eqref{1} $$dU=TdS+SdT-PdV-VdP+\mu dN+Nd\mu\tag{3}\label{3}$$

substracting : \eqref{3}-\eqref{2} $$0=SdT-VdP+Nd\mu$$ $$d\mu=\frac{V}{N}dP-\frac{S}{N}dT$$ not good.

I had read that $$\frac{\mu}{T}=\text{function of }u,v$$ so $$\frac{\mu}{T}=\frac{\mu}{T}(u,v)$$

Directly differentiating isn't helpful at all. Even there's no direct derivation of that equation in Callen's Thermodynamics book. He just wrote "by integrating of the Gibbs-Duhem relation [then equation goes]".

(Note : there's no equation with $$G$$ variable (Gibbs-Duhem function?) until chapter 4-6 (not sure) (in Callen's Thermostatics and Thermodynamics book) So I would request not to use that equation)

• My given equations are for only monoatomic.
– Man
Commented Feb 9, 2022 at 15:53
• Have you tried playing around with $d\left(\frac{x}{T}\right)=-\frac{x}{T^2}dT+\left(\frac{1}{T}\right)dx$ (via the product rule and then the chain rule) for various $x$? Commented Feb 9, 2022 at 17:19
• @Chemomechanics Yep! I had. Even I tried it once again. $$d(\frac{\mu}{T})=\mu d\frac{1}{T}+\frac{v}{T}dP-\frac{s}{T}dT$$ since $$d\mu=-sdT+vdP$$ I didn't add it in the question cause, it's too much irrelevant to the given equation.
– Man
Commented Feb 10, 2022 at 8:45

Solving equation $$(2)$$ for $$dS$$ one gets: \begin{align} \frac{\partial{S}}{\partial{U}}&=\frac{1}{T}\\ \frac{\partial{S}}{\partial{V}}&=\frac{P}{T}\\ \frac{\partial{S}}{\partial{N}}&=-\frac{\mu}{T}. \end{align} Then the proof goes exactly as in the case of the usual Gibb-Duhem equation: one has to use Euler's theorem to express the extensive state function $$S$$ in terms of its natural variables $$(U,V,N)$$, and the previous partial derivatives. The final step is to use the differential $$dS$$ coming from this last expression and the one obtained from equation $$(2)$$.
• I have end up with this : $ds=\color{red}{ud(\dfrac{1}{T})}+\dfrac{1}{T}du+\dfrac{P}{T}dv+\color{red}{vd(\dfrac{P}{T})}+\dfrac{\mu}{T}dN+\color{red}{Nd(\dfrac{\mu}{T})}$ I have found expected variables. I just took total differential of $s=\frac{u}{T}+\frac{Pv}{T}+\frac{\mu N}{T}$
• @Man Ok, now subtract from this expression the differential $ds$ coming from your Equation $(2)$. Commented Feb 10, 2022 at 10:56