A quantity of hydrogen gas occupies a volume of $30.0 mL$ at a certain temperature and pressure. What volume would half this mass of hydrogen occupy at triple the above temperature if the pressure were one-ninth that of the original gas?

$\mathrm{PV}=\mathrm{nRT}$

If the pressure were one-ninth the initial gas's pressure and half the amount of hydrogen occupied the space at thrice the absolute temperature,

Because the molar mass is the same both times, the initial number of moles is half that of the final number.

${\mathrm{V}}_1=\frac{{\mathrm{n}}_{1}R{\mathrm{T}}_1}{{\mathrm{P}}_1}=30\mathrm{ml}\cdots 1$

Again,

${\mathrm{P}}_2{\mathrm{V}}_2={\mathrm{n}}_2{\mathrm{RT}}_2$

$\text{ So }\frac{p_1}{9}\times V=\frac{n_1}{2}\times R\times 3T_1\cdots 2$

From 1 and 2 equation 

$\mathrm{V}=\frac{9\times 30\times 3}{2}=405\mathrm{mL}$