By Srijan Ghosh


2019-03-11 21:18:03 8 Comments

I can't understand why $\left(\frac{\partial P}{\partial V} \right)_T=-\left(\frac{\partial P}{\partial T} \right)_V\left(\frac{\partial T}{\partial V}\right)_P$. Why does the negative sign arise? I can easily write $\left(\frac{\partial P}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial T}{\partial V}\right)_P$ from the rule of partial derivative, but what's the negative sign for?

4 comments

@knzhou 2019-03-12 21:30:57

As others have said, this is simply the triple product rule.

It is relatively simple to understand how the minus sign comes about. Suppose you have a fixed amount of money, and you allocate $x$, $y$, and $z$ dollars to three things. Then $(\partial x / \partial y)|_z$ is clearly $-1$, because if you keep $z$ constant, then you need to remove one dollar from $y$ to add one dollar to $x$. Similarly, $(\partial y / \partial z)|_x = (\partial z / \partial x)|_y = -1$, and the product of all three is thus $(-1)^3 = -1$. Note how the total money constraint on the three variables is crucial. In thermodynamics, the constraint between $P$, $V$, and $T$ is provided by an equation of state, such as the ideal gas law.

@Ant 2019-03-12 14:13:57

Other answers have given the mathematical reasons, but as these are physical functions it’s always nice to relate back to physical intuition.

Thinking of an ordinary substance – a lump of “stuff”, doesn’t matter whether it’s a solid or a gas or whatever – if you decrease the volume available to it, the pressure should go up. In fact that is exactly how we physically put a sample under pressure. So the left-hand side should be negative.

On the other hand, if you increase the temperature (usually) the pressure should increase at constant volume, or alternatively the volume should increase at constant pressure.1 So both of the factors on the right-hand side should be positive.2

So we had better include a minus sign, to at least get the signs to match on both sides of the equation.


  1. If you’re thinking about a material with negative thermal expansion, then both of these factors should instead be negative, but again the right-hand side of the equation ends up positive.
  2. To be precise, I have used another identity here: that $(\partial T/\partial V)_P = (\partial V/\partial T)_P^{-1}$.

@Chet Miller 2019-03-11 22:04:06

This all starts from the basic relationship $$dP=\left(\frac{\partial P}{\partial T}\right)_VdT+\left(\frac{\partial P}{\partial V}\right)_TdV$$Since, at constant pressure, dP=0, if we solve for dT/dV at constant pressure, we obtain: $$\left(\frac{\partial T}{\partial V}\right)_P=-\frac{\left(\frac{\partial P}{\partial V}\right)_T}{\left(\frac{\partial P}{\partial T}\right)_V}$$

@ACuriousMind 2019-03-11 21:35:12

This is not a simple application of partial derivatives, since the variables that are being held constant vary here, but an instance of the triple product rule, which says that for any three quantities $x,y,z$ depending on each other, the relation $$ \left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z} \right)_x\left(\frac{\partial z}{\partial x}\right)_y = -1$$ holds.

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