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?

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## 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

decreasethe volume available to it, the pressure should goup. 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

increasethe temperature (usually) the pressure shouldincreaseat constant volume, or alternatively the volume shouldincreaseat 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.

## @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.