Answer
$2 \%$
Work Step by Step
According to Ideal Gas Law:
\[
\frac{K T}{V}=P
\]
The total differential is given by:
\[
\begin{aligned}
Thus, &-\frac{K T d V}{V^{2}}+\frac{K d T}{V}=d P \\
&\frac{K d T}{P V}-\frac{K T d V}{P V^{2}} \quad \text { divided by } P= \frac{d P}{P}\\
\frac{d P}{P} &=\frac{d T}{P V / K}-\frac{K d V}{V(P V / T)} \quad \because K=P V / T \\
\frac{d P}{P} &=-\frac{d V}{V}+\frac{d T}{T}
\end{aligned}
\]
We are given the percentage increase in $T$ and $V$:
\[
\begin{array}{l}
0.05=\frac{d V}{V} \\
0.03=\frac{d T}{T}
\end{array}
\]
So, we get:
\[
\frac{\Delta P}{P} \approx \frac{d P}{P}=\frac{\Delta T}{T}-\frac{\Delta V}{V}
\]
\[
\frac{\Delta P}{P}=-0.05+0.03=-0.02
\]
$2 \%$
