A bicycle pump contains 250cm3 of air at a pressure of 90kPa, the volume is reduced to 200cm3. What is the pressure of the air inside the pump?
a. 75 kPa
b. 220 kPa
c. 112.5 kPa
d. 240.4 kPa
Assuming the temperature does not change,
90*250 = 200P
Please explain what happen to the volume that was reduced to 200cm3?
PV = kT
So, PV is constant. I just plugged in your numbers
To find the pressure of the air inside the pump after the volume reduction, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature is held constant.
The formula for Boyle's Law is:
P₁V₁ = P₂V₂
Where P₁ is the initial pressure, V₁ is the initial volume, P₂ is the final pressure, and V₂ is the final volume.
Given:
Initial pressure (P₁) = 90 kPa
Initial volume (V₁) = 250 cm³
Final volume (V₂) = 200 cm³
Substituting these values into the formula, we can solve for the final pressure (P₂):
90 kPa * 250 cm³ = P₂ * 200 cm³
Simplifying the equation, we get:
P₂ = (90 kPa * 250 cm³) / 200 cm³
P₂ = 22500 kPa cm³ / 200 cm³
P₂ = 112.5 kPa
Therefore, the pressure of the air inside the pump after the volume reduction is 112.5 kPa. So, the correct answer is option c.