A gas occupies a volume of 2.5L at  $9\times {10}^{5}N{m}^{-2}.$ The additional pressure required to decrease the volume of the gas to 1.5L, keeping temperature constant is  $Y\times {10}^5$. What is the value Y?  

Here, 
$\begin{array}{l}{P}_1=9\times {10}^{2}N{m}^{-2},P_2=?\\ V_1=2.5L,V_2=1.5L\end{array}$
From Boyle’s law equation  $P_1{V}_1=P_2{V}_2,$ we get
$P_2=\frac{P_1{V}_1}{V_2}=\frac{9.0\times {10}^{5}N{m}^{-2}\times 2.5L}{1.5L}=15\times {10}^{5}N{m}^{-2}$
The additional pressure required is
 $15\times {10}^{5}N{m}^{-2}-9\times {10}^{5}N{m}^{-2}=6\times {10}^{5}N{m}^{-2}$