The problem asks us to show that for the gas law $PV = kT$, where $P$ is pressure, $V$ is volume, $T$ is temperature, and $k$ is a constant, the following equations hold: $V \frac{\partial P}{\partial V} + T \frac{\partial P}{\partial T} = 0$ and $\frac{\partial P}{\partial V} \frac{\partial V}{\partial T} \frac{\partial T}{\partial P} = -1$

1. Problem Description

The problem asks us to show that for the gas law $PV = kT$, where $P$ is pressure, $V$ is volume, $T$ is temperature, and $k$ is a constant, the following equations hold:

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

and

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

2. Solution Steps

First, let's derive the expressions for the partial derivatives of $P$ with respect to $V$ and $T$.

From $PV = kT$, we can express $P$ as:

$P = \frac{kT}{V}$

Now, we can find the partial derivatives:

$\frac{\partial P}{\partial V} = \frac{\partial}{\partial V} (\frac{kT}{V}) = kT \frac{\partial}{\partial V} (\frac{1}{V}) = kT (-\frac{1}{V^2}) = -\frac{kT}{V^2}$

$\frac{\partial P}{\partial T} = \frac{\partial}{\partial T} (\frac{kT}{V}) = \frac{k}{V} \frac{\partial}{\partial T} (T) = \frac{k}{V}$

Now, let's substitute these into the first equation:

$V \frac{\partial P}{\partial V} + T \frac{\partial P}{\partial T} = V (-\frac{kT}{V^2}) + T (\frac{k}{V}) = -\frac{kT}{V} + \frac{kT}{V} = 0$

So, the first equation is satisfied.

Now, let's find $\frac{\partial V}{\partial T}$ and $\frac{\partial T}{\partial P}$.

From $PV = kT$, we can express $V$ as:

$V = \frac{kT}{P}$

Therefore,

$\frac{\partial V}{\partial T} = \frac{\partial}{\partial T} (\frac{kT}{P}) = \frac{k}{P}$

From $PV = kT$, we can express $T$ as:

$T = \frac{PV}{k}$

Therefore,

$\frac{\partial T}{\partial P} = \frac{\partial}{\partial P} (\frac{PV}{k}) = \frac{V}{k}$

Now we compute the product of the partial derivatives:

$\frac{\partial P}{\partial V} \frac{\partial V}{\partial T} \frac{\partial T}{\partial P} = (-\frac{kT}{V^2}) (\frac{k}{P}) (\frac{V}{k}) = -\frac{kT}{V^2} \frac{1}{P} k \frac{V}{k} = -\frac{kT}{V^2} \frac{V}{P} = -\frac{kT}{VP}$

Since $PV = kT$, we can substitute $PV$ with $kT$:

$-\frac{kT}{VP} = -\frac{PV}{VP} = -1$

So, the second equation is satisfied.

3. Final Answer

For the gas law $PV = kT$,

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

and

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