At constant temperature, increasing the pressure of a gas by \(5 \%\) its volume will decrease by

According to ideal gas law at constant temperature, \(\mathrm{P} \propto \frac{1}{\mathrm{~V}}\)
\(\begin{array}{ll}
\therefore \quad & \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\
& \mathrm{P}_1=\mathrm{P} \\
& \mathrm{P}_2=\mathrm{P}+\frac{5}{100} \mathrm{P}=1.05 \mathrm{P}
\end{array}\)
Substituting the values
\(\begin{array}{ll}
& \mathrm{PV}_1=1.05 \mathrm{PV}_2 \\
\therefore \quad & \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{1}{1.05} \\
\therefore \quad & \frac{\mathrm{V}_2-\mathrm{V}_1}{\mathrm{~V}_1} \times 100=\frac{1-1.05}{1.05} \times 100 \\
\therefore \quad & \frac{\Delta \mathrm{V}}{\mathrm{V}_1}=-4.76 \% \approx-4.7 \%
\end{array}\)
The negative sign indicates that the volume is decreasing.