Problem 31 states that for an ideal gas, the pressure $P$, temperature $T$, and volume $V$ are related by $PV = kT$, where $k$ is a constant. We need to find the rate of change of pressure with respect to temperature, $\frac{dP}{dT}$, when the temperature $T = 300$ K and the volume $V = 100$ cubic inches.

1. Problem Description

Problem 31 states that for an ideal gas, the pressure $P$, temperature $T$, and volume $V$ are related by $PV = kT$, where $k$ is a constant. We need to find the rate of change of pressure with respect to temperature, $\frac{dP}{dT}$, when the temperature $T = 300$ K and the volume $V = 100$ cubic inches.

2. Solution Steps

We are given the ideal gas law:

$PV = kT$

Since we are asked to find the rate of change of pressure with respect to temperature $T$, we differentiate the equation with respect to $T$, while keeping $V$ constant.

$\frac{d}{dT}(PV) = \frac{d}{dT}(kT)$

Since $V$ is constant,

$V \frac{dP}{dT} = k$

Therefore,

$\frac{dP}{dT} = \frac{k}{V}$

From the ideal gas law, we can write $k = \frac{PV}{T}$.

Substitute this into the expression for $\frac{dP}{dT}$:

$\frac{dP}{dT} = \frac{PV}{TV} = \frac{P}{T}$

Now we need to find the pressure $P$ when $T = 300$ K and $V = 100$ cubic inches.

From the equation $PV = kT$, we have $P = \frac{kT}{V}$. However, we don't know the value of $k$.

We can write $P = \frac{kT}{V}$. Also $\frac{dP}{dT} = \frac{k}{V}$.

When $T=300$, we have $PV=k(300)$. Therefore, $k=\frac{PV}{300}$.

So $\frac{dP}{dT}=\frac{k}{V}=\frac{PV}{300V}=\frac{P}{300}$

The question is missing a key element: an initial pressure value.

I'll assume that we want to find an expression for $\frac{dP}{dT}$ in terms of variables instead of finding a numerical value.

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

$\frac{dP}{dT} = \frac{k}{V}$.

Given $V = 100$ cubic inches,

$\frac{dP}{dT} = \frac{k}{100}$.

Since $PV = kT$, at $T = 300 K$, we have $P(100) = k(300)$, thus $k = \frac{P(100)}{300}$.

Then $\frac{dP}{dT} = \frac{P(100)/300}{100} = \frac{P}{300}$.

3. Final Answer

$\frac{dP}{dT} = \frac{P}{300}$ (pounds per square inch per Kelvin). The problem does not have enough information to compute a numeric value. The rate of change of pressure with respect to temperature is $\frac{P}{300}$.