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
The system follow the Boyle's Law:
\(P_1\cdot V_1 = P_2\cdot V_2\ \to\ P_2 = \frac{P_1\cdot V_1}{V_2} = \frac{98\ kPa\cdot 77.4\ L}{60.0\ L} = \bf 126.4\ kPa\)
The value of pressure is 126.4kPa.
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 ".
\(P_1V_1=P_2V_2\\PV=constant\)
\(P_1=98\ kPa\\V_1=60\ L\\V_2=77.4\L\)
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.
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