Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

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

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.

share|cite|improve this answer
"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 $${\partial z\over\partial x}(x,y){\partial x\over\partial y}(y,z){\partial y\over\partial z}(z,x)=(-1)^3{a\over c}{b\over a}{c\over b}=-1\ .$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.