A bicycle pump contains 250 cm3 of air at a pressure of 90 KPa. If the air is compressed, the volume is reduced to 200 cm3. What is the pressure of the air inside the pump?
use p1v1 = p2v2
Post your work if yo have trouble.
V1 = 90 and V2 =
P1= 250 cm and p9= 200
P1V1 = p2V2
V2=p1V1/p2
V2=250cm3x90kpa/200cm3 =
V2=
It was hard that I thought
To find the pressure of the air inside the pump, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume, provided the temperature remains constant.
Boyle's Law can be expressed as:
P1 * V1 = P2 * V2
Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
Let's substitute the given values into the equation:
P1 = 90 KPa (initial pressure)
V1 = 250 cm^3 (initial volume)
V2 = 200 cm^3 (final volume)
Now we can solve for P2, the final pressure:
P1 * V1 = P2 * V2
90 KPa * 250 cm^3 = P2 * 200 cm^3
22500 KPa·cm^3 = P2 * 200 cm^3
To isolate P2, divide both sides of the equation by 200 cm^3:
P2 = 22500 KPa·cm^3 / 200 cm^3
Simplifying,
P2 = 112.5 KPa
Therefore, the pressure of the air inside the pump is 112.5 KPa.