Question

what pressure will be needed to reduce the volume of 77.4 l of helium at 98.0 kpa to a volume of 60.0 l? 76 kpa 126 kpa

Answer 1

The system follow the Boyle's Law:

$P_1\cdot V_1 = P_2\cdot V_2\ \to\ P_2 = \frac{P_1\cdot V_1}{V_2} = \frac{98\ kPa\cdot 77.4\ L}{60.0\ L} = \bf 126.4\ kPa$