Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 31

The equation $PV = nRT$ represents the Gas Law for an ideal gas at absolute temperature $T$(in Kelvins) pressure $P$(in atmosphere) and volume $V$(in liters).


$
\begin{equation}
\begin{aligned}
\text{where } & n \text{ is the number of moles of the gas.}\\
& R = 0.0821 \text{ is the gas constant.}
\end{aligned}
\end{equation}
$

Suppose that, at a certain instant, $P = 8.0$ atm and is increasing at a rate of $\displaystyle 0.10 \frac{\text{atm}}{\text{min}}$ and $V = 10L$ and is decreasing at a rate of $\displaystyle 0.15 \frac{L}{\text{min}}$. Determine the rate of change of $T$ with respect to time at that instant if $n = 10$ mol.

From the ideal gas equation, $PV = nRT$

$
\text{where, }\\
\begin{equation}
\begin{aligned}
P & = 8 \text{ atm}, &&& \frac{dP}{dt} &= 0.10 \frac{\text{atm}}{\text{min}}\\
\\
V &= 10 L, &&& \frac{dV}{dt} &= - 0.15 \frac{L}{\text{min}}\\
\\
n &= 10 \text{ mol}\\
\\
R &= 0.0821
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\text{so, } \frac{dT}{dt} &= \frac{1}{nR} PV\\
\\
\frac{dT}{dt} &= \frac{1}{nR} \left[ P \cdot \frac{dV}{dt} + \frac{dP}{dt} \cdot V\right] && \text{(Using Product Rule)}\\
\\
\frac{dT}{dt} &= \frac{1}{(10)(0.0821)} \left[ 8(-0.15)+(0.10)(10) \right]\\
\\
\frac{dT}{dt} &= -0.2436 \frac{K}{\text{min}}
\end{aligned}
\end{equation}
$

It means that the rate of change of $T$ with respect to time is decreasing at that instant.