At a given temperature what will be the percentage increase in pressure for a 5% decrease in the volume of the gas ?

According to Boyle’s law

$P_1{V}_1=P_2{V}_2=\text{Constant}$

Let the initial volume of gas be V.

When volume is decreased by 5%, then

$V_2=V-\left(V\times \frac{5}{100}\right)=V-\frac{V}{20}=\frac{19V}{20}$

$\therefore$  From Boyle’s equation,

$P_1\times V=P_2\times \frac{19V}{20}$

$\Rightarrow P_2=\frac{P_1\times V}{\frac{19V}{20}}=\frac{20P_1}{19}$

Increase in pressure $=P_2-P_1=\frac{20P_1}{19}-P_1$ $=\frac{20P_1-19P_1}{19}=\frac{P_1}{19}$

Percentage increase in pressure $=\frac{P_1}{19\times P_1}\times 100$

$=\frac{100}{19}=5.26\%$ .