A given mass of helium(gas)occupies a volume of 3,6m^3 at a pressure of 105 kPa.if the temperature remains constant,calculate volume of the gas if the pressure is decreased to 70 kPa.
P1V1 = P2V2
To solve this question, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional, assuming the temperature remains constant. Boyle's Law can be written mathematically as:
P1 * V1 = P2 * V2
Where:
P1 = Initial pressure (105 kPa)
V1 = Initial volume (3.6 m^3)
P2 = Final pressure (70 kPa)
V2 = Final volume (to be calculated)
We can rearrange the formula to solve for V2:
V2 = (P1 * V1) / P2
Substituting the given values into the equation:
V2 = (105 kPa * 3.6 m^3) / 70 kPa
Now, we can cancel out the units of kPa:
V2 = (105 * 3.6) / 70 m^3
V2 = 5.4 m^3 / 70 m^3
V2 = 0.07714 m^3
Therefore, the volume of the gas at a pressure of 70 kPa is approximately 0.07714 m^3.