To what temperature must a neon gas sample be heated to double the pressure, if the initial volume of the gas at 

$75°\mathrm{C}$  is decreased by $15\%?$

Let          $P_1=P,V_1=V$ and 

             $t_1=75+273=348\mathrm{K}.$

Thus, $P_2=2P,V_2=\frac{(100-15)V}{100}=\frac{85\mathrm{V}}{100}$ and 

                        $T_2=?$ 

                    $[\because$   initial volume is decreased by 15%] 

Using gas equation, we have :  

$\begin{array}{l}\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}\\ {\mathrm{T}}_2=\frac{{\mathrm{P}}_2{\mathrm{V}}_2{\mathrm{T}}_1}{{\mathrm{P}}_1{\mathrm{V}}_1}=2\mathrm{P}\times \frac{85\mathrm{V}}{100}\times \frac{348\mathrm{K}}{\mathrm{P}\times \mathrm{V}}\end{array}$

or   $\begin{array}{r}{T}_2=591.6\mathrm{K}=591.6-273\\ ={318.6}^{\circ }\mathrm{C}\simeq {319}^{\circ }\mathrm{C}\end{array}$

So, the correct answer is  $\left(\mathrm{a}\right).$