The problem asks us to show that for the gas law of problem 31, the following two equations hold: $V \frac{\partial P}{\partial V} + T \frac{\partial P}{\partial T} = 0$ and $\frac{\partial P}{\partial V} \frac{\partial V}{\partial T} \frac{\partial T}{\partial P} = -1$ Given that from Problem 31 we have the gas law $P = \frac{RT}{V-b} - \frac{a}{V^2}$.

1. Problem Description

The problem asks us to show that for the gas law of problem 31, the following two equations hold:

$V \frac{\partial P}{\partial V} + T \frac{\partial P}{\partial T} = 0$

and

$\frac{\partial P}{\partial V} \frac{\partial V}{\partial T} \frac{\partial T}{\partial P} = -1$

Given that from Problem 31 we have the gas law $P = \frac{RT}{V-b} - \frac{a}{V^2}$.

2. Solution Steps

First, let's derive the first equation $V \frac{\partial P}{\partial V} + T \frac{\partial P}{\partial T} = 0$.

We need to find the partial derivatives of $P$ with respect to $V$ and $T$.

We are given $P = \frac{RT}{V-b} - \frac{a}{V^2}$.

$\frac{\partial P}{\partial V} = \frac{\partial}{\partial V} (\frac{RT}{V-b} - \frac{a}{V^2}) = RT \frac{\partial}{\partial V} (V-b)^{-1} - a \frac{\partial}{\partial V} V^{-2} = RT(-1)(V-b)^{-2} - a(-2)V^{-3} = \frac{-RT}{(V-b)^2} + \frac{2a}{V^3}$.

$\frac{\partial P}{\partial T} = \frac{\partial}{\partial T} (\frac{RT}{V-b} - \frac{a}{V^2}) = \frac{R}{V-b}$.

Now substitute these into the first equation:

$V \frac{\partial P}{\partial V} + T \frac{\partial P}{\partial T} = V(\frac{-RT}{(V-b)^2} + \frac{2a}{V^3}) + T(\frac{R}{V-b}) = \frac{-VRT}{(V-b)^2} + \frac{2a}{V^2} + \frac{RT}{V-b}$.

$= \frac{-VRT + RT(V-b)}{(V-b)^2} + \frac{2a}{V^2} = \frac{-VRT + VRT - bRT}{(V-b)^2} + \frac{2a}{V^2} = \frac{-bRT}{(V-b)^2} + \frac{2a}{V^2}$.

This doesn't appear to easily simplify to

0. Let's reexamine the gas law: $P = \frac{RT}{V-b} - \frac{a}{V^2}$, then $\frac{RT}{V-b} = P + \frac{a}{V^2}$. Plugging this into our expression:

$\frac{-bRT}{(V-b)^2} + \frac{2a}{V^2} = \frac{-b (V-b)(P + \frac{a}{V^2})}{(V-b)^2} + \frac{2a}{V^2} = \frac{-b(P + \frac{a}{V^2})}{(V-b)} + \frac{2a}{V^2}$

There may be an error in the problem statement. Let's move on to the second equation.

For the second equation, we can use the cyclic rule for partial derivatives:

$\frac{\partial P}{\partial V} \frac{\partial V}{\partial T} \frac{\partial T}{\partial P} = -1$

The cyclic rule for partial derivatives is $\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y = -1$.

Here we have $x=P$, $y=V$, and $z=T$. So,

$\frac{\partial P}{\partial V} \frac{\partial V}{\partial T} \frac{\partial T}{\partial P} = -1$.

This equation is an identity due to the cyclic permutation rule. Thus, we have shown that the second equation holds.

3. Final Answer

$V \frac{\partial P}{\partial V} + T \frac{\partial P}{\partial T} = 0$ is not generally true based on the given equation of state.

$\frac{\partial P}{\partial V} \frac{\partial V}{\partial T} \frac{\partial T}{\partial P} = -1$ is true based on the cyclic rule for partial derivatives.