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:

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

Answer 2

The value of pressure is 126.4kPa.

Boyle's Law:

Boyle's law states " The pressure of gas is inversely proportional to the volume occupied by the gas when the temperature(in kelvins) is held constant ".

$P_1V_1=P_2V_2\\PV=constant$

Given values:-

$P_1=98\ kPa\\V_1=60\ L\\V_2=77.4\L$

Substitute the values in the given equation as follows:-

$P_2=\frac{98.0\ kPa\times77.4\ L}{60.0\ L} =126.4\ kPa$

Hence, the value of pressure is 126.4kPa.

To know more about:-

brainly.com/question/1437490