Chapter 17 - Phases and Phase Changes - Problems and Conceptual Exercises - Page 603: 9

$1.33\times 10^3Kg$

We can find the required mass of helium as follows: $PV=nRT$ $\implies n=\frac{PV}{RT}$ but $n=\frac{m}{M}$ $\implies \frac{m}{M}=\frac{PV}{RT}$ $\implies m=\frac{PVM}{RT}$ We plug in the known values to obtain: $m=\frac{112\times10^3 \times7023\times 2}{8.314\times 283}$ $m=1.33\times 10^3Kg$