Partial derivatives confusion in the equation $PV=nrT$

Question

The question is to exactly: "If the variables $P,V$ and $T$ are related by the equation $PV=nRT$ where $n$ and $R$ are constant, simplify the expression" $$ \frac{\partial V}{\partial T}\frac{\partial T}{\partial P}\frac{\partial P}{\partial V} $$ The first step of the solution confuses me. The solution solves explicitly for $V$ and solves $$ V=\frac{nrT}{P}\Rightarrow \frac{\partial V}{\partial T}=\frac{nr}{P} $$ Treating $P$ as constant with respect to $T$, which it isn't, since $$ P=\frac{nrT}{V} $$ Where is my reasoning flawed?

Answer 1

The partial derivative of

$$V(n,r,T,P)=\frac{nrT}{P}$$

with respect to $T$ is just

$$\frac{\partial V(n,r,T,P)}{\partial T}= V_T(n,r,T,P)=\frac{nr}{P}.$$

Just treat $n$, $r$ and $P$ as constants, i.e. think of it is a function, $v$ of one variable $v(T)=V(n,r,T,P)$.

Answer 2

In the present context the 'understanding' is that when taking derivatives of $V$ the variables are the two others, i.e. for $\partial V/\partial T$ the variables are $P$ and $T$ (and $P$ is held constant). Similarly for the other partial derivatives. From $V=nrT/P$, $T=PV/nr$ and $P=nrT/V$ we therefore get: $$ \frac{\partial V}{\partial T} \frac{\partial T}{\partial P} \frac{\partial P}{\partial V} = \frac{nr}{P} \times \frac{V}{nr} \times \frac{-nrT}{V^2} = \frac{-nrT}{PV} = -1$$ which I presume is the answer given in the text ? Often the notation makes more explicit which variables are fixed. The fact that you get -1 is by the way a general result from differential geometry and independent of the specific form of the law for the ideal gas.