# Simplifying $\frac{\partial V}{\partial T} \cdot \frac{\partial T}{\partial P} \cdot \frac{\partial P}{\partial V}$

The problem given is this. If the variables $P, V$, and $T$ are related by the equation $PV = nRT$, where $n$ and $R$ are constants, simplify the expression

$$\frac{\partial V}{\partial T} \cdot \frac{\partial T}{\partial P} \cdot \frac{\partial P}{\partial V}$$

After doing the calculations, we see that we get $-1$ as an answer. My question: why can't we just apply the chain rule to "cancel out" the numerators/denominators and get 1 as an answer?

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Terminological note: this is called the cyclic chain rule or Euler's chain relation. – J. M. Sep 28 '11 at 17:31

## 2 Answers

Because in reality, you have to write

$$\left(\frac{\partial V}{\partial T}\right)_P \cdot \left(\frac{\partial T}{\partial P}\right)_V \cdot \left(\frac{\partial P}{\partial V}\right)_T = -1$$ This is simply due to the fact that the variables are not independent; therefore, we fix one of them and analyze the change of the remaining ones. So these partial derivatives represent the isobaric, isochoric, and isothermal conditions respectively.

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"isovolumetric (?!)" - Isochoric, actually. (Chemist speaking.) ;) – J. M. Sep 28 '11 at 17:29
@J.M. I was tempted to say isometric but kept my mouth shut. (mech. eng. here) ;) – user13838 Sep 29 '11 at 2:11

There are three variable quantities $P$, $V$, $T$ which at any given instant are related by an equation $f(P,V,T)=0$ (the exact form of this equation is not relevant). Assume that $(P_0,V_0,T_0)$ is an admissible triple and that $$a:=f_P(P_0,V_0,T_0)\ne 0, \quad b:=f_V(P_0,V_0,T_0)\ne 0,\quad c:=f_T(P_0,V_0,T_0)\ne 0\ .$$ Under these assumptions the equation $f(P,V,T)=0$ defines each of the variables $P$, $V$, $T$ in the neighborhood of $(P_0,V_0,T_0)$ implicitly as a function of the other two: $$P=\phi(V,T),\quad V=\psi(T,P), \quad T=\chi(P,V)$$ with $\phi(V_0,T_0)=P_0$, $\psi(T_0,P_0)=V_0$, $\chi(P_0,V_0)=T_0$.

Now your ${\partial V\over\partial T}$ is actually the partial derivative ${\partial\psi\over\partial T}$. In order to compute this derivative we note that $$f\bigl(P,\psi(T,P),T\bigr)=0\qquad\forall T,\forall P\ .$$ We now take the partial derivative with respect to $T$ at $(P_0,V_0,T_0)$. Using the chain rule we get $b \>\psi_T(T_0,P_0) + c=0$ and therefore $$\psi_T(T_0,P_0)=-{c\over b}\ .$$ Repeating this calculation with permuted variables one finds that the product of the three partial derivatives in question is indeed $-1$.

There is a "totally linear" version of this phenomenon: Assume that three real quantities $x$, $y$, $z$ are related by an equation $$ax+by+cz =d$$ with $abc\ne0$. Then we can solve for each of the three variables in terms of the two others, e.g., $$z={1\over c}(d-ax-by)\ .$$ Therefore $${\partial z\over\partial x}(x,y)=-{a\over c}\ ,$$ and cyclic permutation of the variables implies $${\partial z\over\partial x}(x,y){\partial x\over\partial y}(y,z){\partial y\over\partial z}(z,x)=(-1)^3{a\over c}{b\over a}{c\over b}=-1\ .$$

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