# Why can we change $dt$ with $(dt/dp)_s dp$?

In my homework assignment there's the following question:

A general thermodynamic system is being compressed isentropically from pressure $$P_i$$ to $$P_f$$ while keeping the number of particles constant. write down the temperature change of the system using an integral. reduce the partial derivative in the integrand to the measurable quantities: $$T,V,\alpha$$, $$C_p$$.

The solution to the question is the following:

My question is: What was used in the transition from $$\Delta T$$ to the first integral over $$dp$$? how would one approach this question and get the correct result as seen above?

The first step is just the following model:

$$f(b)-f(a)=\int_a^b f'(x)\,dx=\int_a^b\frac{df}{dx}(x)\,dx$$

After that it's the triple product rule:

$$\left(\frac{\partial x}{\partial y}\right)\left(\frac{\partial y}{\partial z}\right)\left(\frac{\partial z}{\partial x}\right)=-1$$

That's a big trap compared to the 2D case.

Edit: the starting point is about which variables $$H$$ depends on.

Without going into too much details, let's start with internal energy $$U$$. Its differential form, for a system without chemical reaction or phase transition, is:

$$dU=T\,dS-P\,dV$$

Spotting the $$dS$$ and $$dV$$ terms, it means that $$U$$ is naturally a function of $$S$$ and $$V$$.

Then you define $$H=U+PV$$. Its differential form is:

$$dH=dU+P\,dV+V\,dP=T\,dS+V\,dP$$

which means that $$H$$ is naturally a function of $$S$$ and $$P$$.

So, to answer the question you added in your comment: mathematically speaking, you have to decide from the start which set of variables you're working with. Since you're studying $$H$$, those are $$S$$ and $$P$$, and $$N$$ (I didn't mention it before because $$N$$ is constant as long as there's no chemical reaction or phase transition).

• this clears up some of the confusion but how do we know to that the temperature function is T(p,s,n) and not say T(v,s,n) or any other combination? Commented Jun 10, 2022 at 18:50
• I edited my answer to, hopefully, clarify this point. Commented Jun 10, 2022 at 20:19

This is not how I would have solved this problem. I would first have written: $$ds=\left(\frac{\partial s}{\partial T}\right)_PdT+\left(\frac{\partial s}{\partial P}\right)_TdP=0$$In addition, $$\left(\frac{\partial s}{\partial T}\right)_P=\frac{C_p}{T}$$and, from a Maxwell relationship, $$\left(\frac{\partial s}{\partial P}\right)_T=-\left(\frac{\partial v}{\partial T}\right)_P$$So, combining these equations gives: $$\left(\frac{\partial T}{\partial P}\right)_s=\frac{T}{C_p}\left(\frac{\partial v}{\partial T}\right)_P=\frac{\alpha vT}{C_P}$$